GQB_ME.pdf

GATE QUESTION BANK for

Mechanical Engineering By

GATE QUESTION BANK

Contents

Contents #1.

#2.

#3.

#4.

Subject Name Mathematics

Topic Name

Page No. 1-148

1 2 3 4 5 6 7

Linear Algebra Probability & Distribution Numerical Methods Calculus Differential Equations Complex Variables Laplace Transform

1 – 28 29 – 57 58 – 73 74 – 112 113 – 131 132 – 143 144 – 148

Engineering Mechanics

149 – 162

8 9

149 – 155 156 – 162

Strength of Materials

163 – 187

10 11 12 13 14 15 16 17

163 – 168 169 – 171 172 – 174 175 – 179 180 – 182 183 – 184 185 186 – 187

#6.

Simple Stress and Strain Shear Force and Bending Moment Stresses in Beams Deflection of Beams Torsion Mohr's Circle Strain Energy Method Columns and Struts

Thermodynamics 18 19 20 21 22 23 24 25

#5.

Statics Dynamics

188 – 225 Basic Thermodynamics Irreversibility & Availability Properties of Pure Substances Work, Heat & Entropy Psychrometrics Power Engineering Refrigeration Internal Combustion Engines

188 – 195 196 197 – 198 199 – 204 205 – 209 210 – 217 218 – 222 223 – 225

Theory of Machines

226 – 248

26 27 28 29

226 – 233 234 – 237 238 – 240 241 – 248

Mechanisms Gear Trains Flywheel Vibration

Machine Design 30 31

249 – 267 Design of Static Loading Design for Dynamic Loading th

249 – 251 252 – 254 th

th

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30 Cross, 10 Main, Jayanagar 4 Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page I

GATE QUESTION BANK

32 33 34 35 36

#7.

255 – 258 259 260 -261 262 – 264 265 – 267

268 - 291 Fluid Properties Fluid Statics Fluid Kinematics Fluid Dynamics Boundary Layer Flow through Pipes Hydraulic Machines

268 269 – 270 271 – 275 276 – 279 280 – 282 283 – 286 287 – 291

Heat Transfer 44 45 46 47

#9.

Design of Joints Design of Shaft and Shaft Component Design of Bearing Design of Brakes and Clutches Design of Spur Gears

Fluid Mechanics 37 38 39 40 41 42 43

#8.

Contents

292 – 311 Conduction Convection Radiation Heat Exchanger

292 – 298 299 – 303 304 – 307 308 – 311

Manufacturing Engineering

312 – 350

48 49 50 51 52 53 54

312 – 314 315 – 320 321 – 325 326 – 329 330 – 343 344 – 347 348 – 350

Engineering Materials Casting Forming process Joining Process Machining and Machine Tool Operations Metrology and Inspection Computer Integrated Manufacturing (CIM)

#10. Industrial Engineering 55 56 57

351 – 371

Production Planning and Control Inventory Control Operation Research

th

th

351 – 353 354 – 359 360 – 371

th

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30 Cross, 10 Main, Jayanagar 4 Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page II

GATE QUESTION BANK

Mathematics

Linear Algebra ME – 2005 1. Which one of the following is an Eigenvector of the matrix[

(A) [

]

(B) [ ]

2.

5.

]?

(C) [

]

(D) [

]

A is a 3 4 real matrix and Ax=B is an inconsistent system of equations. The highest possible rank of A is (A) 1 (C) 3 (B) 2 (D) 4

ME – 2006 3. Multiplication of matrices E and F is G. Matrices E and G are os sin E [ sin ] and os G

4.

[

sin os

sin (B) [ os

os sin

]

os (C) [ sin

sin os

]

sin (D) [ os

os sin

7.

Eigenvectors of 0

1 is

(A) 0 (B) 1

(C) 2 (D) Infinite

If a square matrix A is real and symmetric, then the Eigenvalues (A) are always real (B) are always real and positive (C) are always real and non-negative (D) occur in complex conjugate pairs

]

ME – 2008 8.

The Eigenvectors of the matrix 0

1 are

written in the form 0 1 and 0 1. What is a + b? (A) 0 (B) 1/2

]

Eigen values of a matrix 0

ME – 2007 6. The number of linearly independent

]. What is the matrix F?

os (A) [ sin

S

Match the items in columns I and II. Column I Column II P. Singular 1. Determinant is not matrix defined Q. Non-square 2. Determinant is matrix always one R. Real 3. Determinant is symmetric zero matrix S. Orthogonal 4. Eigen values are matrix always real 5. Eigen values are not defined (A) P - 3 Q - 1 R - 4 S - 2 (B) P - 2 Q - 3 R - 4 S - 1 (C) P - 3 Q - 2 R - 5 S - 4 (D) P - 3 Q - 4 R - 2 S - 1

9.

(C) 1 (D) 2

The matrix [

] has one Eigenvalue p equal to 3. The sum of the other two Eigenvalues is (A) p (C) p – 2 (B) p – 1 (D) p – 3

1are 5 and 1. What are the

Eigenvalues of the matrix = SS? (A) 1 and 25 (C) 5 and 1 (B) 6 and 4 (D) 2 and 10 th

th

th

THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30 Cross, 10 Main, Jayanagar 4 Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 1

GATE QUESTION BANK

10.

For what value of a, if any, will the following system of equations in x, y and z have a solution x y x y z x y z (A) Any real number (B) 0 (C) 1 (D) There is no such value

11.

ME – 2012 15.

For a matrix,M-

*

x

√

(B) (√ )

1 is

(A) 2 (B) 2 3

3

(C) 2 3 (D) 2

3

ME – 2011 13. Consider the following system equations: x x x x x x x The system has (A) A unique solution (B) No solution (C) Infinite number of solutions (D) Five solutions 14.

of

Eigen values of a real symmetric matrix are always (A) Positive (C) Negative (B) Real (D) Complex

(D) ( ) √

√

of the matrix is equal to the inverse of the ,M- . The value of x is matrix ,Mgiven by ) (A) ( (C) ⁄ ( ⁄ ) (B) (D) ⁄

0

1 , one of the

(C) (√ )

(A) (√ )

+, the transpose

ME – 2010 12. One of the Eigenvectors of the matrix

For the matrix A=0

normalized Eigenvectors is given as

16.

ME – 2009

Mathematics

x + 2y + z =4 2x + y + 2z =5 x–y+z=1 The system of algebraic equations given above has (A) a unique algebraic equation of x = 1, y = 1 and z = 1 (B) only the two solutions of ( x = 1, y = 1, z = 1) and ( x = 2, y = 1, z = 0) (C) infinite number of solutions. (D) No feasible solution.

ME – 2013 17. The Eigenvalues of a symmetric matrix are all (A) Complex with non –zero positive imaginary part. (B) Complex with non – zero negative imaginary part. (C) Real (D) Pure imaginary. 18.

Choose correct set of functions, which are linearly dependent. (A) sin x sin x n os x (B) os x sin x n t n x (C) os x sin x n os x (D) os x sin x n os x

ME – 2014 19. Given that the determinant of the matrix [

] is

12 , the determinant of

the matrix [ (A) th

] is (B)

th

(C) th

(D)


GATE QUESTION BANK

20.

One of the Eigenvectors of the matrix 0

21.

22.

2.

Consider a non-homogeneous system of linear equations representing mathematically an over-determined system. Such a system will be (A) consistent having a unique solution (B) consistent having many solutions (C) inconsistent having a unique solution (D) inconsistent having no solution

3.

Consider the matrices , - . The order of , (

1 is

(A) {– }

(C) 2

(B) {– }

(D) 2 3

3

Consider a 3×3 real symmetric matrix S such that two of its Eigenvalues are with respective Eigenvectors x y [x ] [y ] If then x y + x y +x y x y equals (A) a (C) ab (B) b (D) 0 Which one of the following equations is a correct identity for arbitrary 3×3 real matrices P, Q and R? (A) ( ) ) (B) ( ( ) (C) et et et ) (D) (

CE – 2005 1. Consider the system of equations ( ) is s l r Let ( ) ( ) where ( ) e n Eigen -pair of an Eigenvalue and its corresponding Eigenvector for real matrix A. Let I be a (n × n) unit matrix. Which one of the following statement is NOT correct? (A) For a homogeneous n × n system of linear equations,(A ) X = 0 having a nontrivial solution the rank of (A ) is less than n. (B) For matrix , m being a positive integer, ( ) will be the Eigen pair for all i. (C) If = then | | = 1 for all i. (D) If = A then is real for all i.

Mathematics

,

-

,

-

and

- will be ) (C) (4 × 3) (D) (3 × 4

(A) (2 × 2) (B) (3 × 3

CE – 2006 4. Solution for the system defined by the set of equations 4y + 3z = 8; 2x – z = 2 and 3x + 2y = 5 is (A) x = 0; y =1; z = ⁄ (B) x = 0; y = ⁄ ; z = 2 (C) x = 1; y = ⁄ ; z = 2 (D) non – existent

5.

For the given matrix A = [

],

one of the Eigen values is 3. The other two Eigen values are (A) (C) (B) (D) CE – 2007 6. The minimum and the maximum Eigenvalue of the matrix [

]are 2

and 6, respectively. What is the other Eigenvalue? (A) (C) (B) (D) 7.

For what values of and the following simultaneous equations have an infinite of solutions? X + Y + Z = 5; X + 3Y + 3Z = 9; X+2Y+ Z (A) 2, 7 (C) 8, 3 (B) 3, 8 (D) 7, 2 th

th

th


GATE QUESTION BANK

8.

The inverse of the (A)

0

(B)

0

1 1

m trix 0

(A) (B)

1

is

(D)

0

1

( )

0

( )

0

( )

0

( )

0

11.

is

15. (C) (D)

i

i

i

i i

i

i

i i

i

i

i i

i i

1

1

1

1

i

i

i i

i

1

CE – 2012

1 are and 8 and 5

The inverse of the matrix 0

0

The Eigenvalue of the matrix [P] = 0

14.

(C)

CE – 2008 9. The product of matrices ( ) (A) (C) (B) (D) PQ 10.

1 is

Mathematics

n n

The following simultaneous equation x+y+z=3 x + 2y + 3z = 4 x + 4y + kz = 6 will NOT have a unique solution for k equal to (A) 0 (C) 6 (B) 5 (D) 7

CE – 2009 12. A square matrix B is skew-symmetric if (C) (A) (D) (B) CE – 2011 13. [A] is square matrix which is neither symmetric nor skew-symmetric and , is its transpose. The sum and difference of these matrices are defined as [S] = [A] + , - and [D] = [A] , - , respectively. Which of the following statements is TRUE? (A) Both [S] and [D] are symmetric (B) Both [S] and [D] are skew-symmetric (C) [S] is skew-symmetric and [D] is symmetric (D) [S] is symmetric and [D] is skew symmetric

The Eigenvalues of matrix 0 (A) (B) (C) (D)

1 are

2.42 and 6.86 3.48 and 13.53 4.70 and 6.86 6.86 and 9.50

CE – 2013 16. There is no value of x that can simultaneously satisfy both the given equations. Therefore, find the ‘le st squares error’ solution to the two equations, i.e., find the value of x that minimizes the sum of squares of the errors in the two equations. 2x = 3 and 4x = 1 17.

What is the minimum number of multiplications involved in computing the matrix product PQR? Matrix P has 4 rows and 2 columns, matrix Q has 2 rows and 4 columns, and matrix R has 4 rows and 1 column. __________

CE – 2014 18.

Given the matrices J = [ K

19.

[

] n

], the product K JK is

The sum of Eigenvalues of the matrix, [M] is, where [M] = [

]

(A) 915 (B) 1355 th

th

(C) 1640 (D) 2180 th


GATE QUESTION BANK

4. 20.

The determinant of matrix [

Let A be a 4x4 matrix with Eigenvalues –5, –2, 1, 4. Which of the following is an I Eigenvalue of 0 1, where I is the 4x4 I identity matrix? (A) (C) (B) (D)

]

is ____________ 21.

The

rank

[

of

the

matrix

] is ________________

CS – 2005 1. Consider the following system of equations in three real variables x x n x x x x x x x x x x This system of equation has (A) no solution (B) a unique solution (C) more than one but a finite number of solutions (D) an infinite number of solutions 2.

What are the Eigenvalues of the following 2 2 matrix? 0 (A) (B)

1 n n

(C) (D)

n n

CS – 2006 3. F is an n x n real matrix. b is an n real vector. Suppose there are two nx1 vectors, u and v such that u  v , and Fu=b, Fv=b. Which one of the following statement is false? (A) Determinant of F is zero (B) There are infinite number of solutions to Fx=b (C) There is an x  0 such that Fx=0 (D) F must have two identical rows

Mathematics

CS – 2007 5. Consider the set of (column) vectors defined by X={xR3 x1+x2+x3=0, where XT =[x1, x2, x3]T }. Which of the following is TRUE? (A) {[1, 1, 0]T, [1, 0, 1]T} is a basis for the subspace X. (B) {[1, 1, 0]T, [1, 0, 1]T} is a linearly independent set, but it does not span X and therefore, is not a basis of X. (C) X is not the subspace for R3 (D) None of the above CS – 2008 6. The following system of x x x x x x x x x Has unique solution. The only possible value (s) for is/ are (A) 0 (B) either 0 or 1 (C) one of 0,1, 1 (D) any real number except 5 7.

How many of the following matrices have an Eigenvalue 1? 0

1 0

1 n 0

1 0

(A) One (B) two

1

(C) three (D) four

CS – 2010 8. Consider the following matrix A=[

] x y If the Eigen values of A are 4 and 8, then (A) x = 4, y = 10 (C) x = 3, y = 9 (B) x = 5, y = 8 (D) x = 4, y = 10 th

th

th


GATE QUESTION BANK

CS – 2011 9. Consider the matrix as given below [

13.

The value of the dot product of the Eigenvectors corresponding to any pair of different Eigenvalues of a 4-by-4 symmetric positive definite matrix is __________.

14.

If the matrix A is such that

]

Which one of the following options provides the CORRECT values of the Eigenvalues of the matrix? (A) 1, 4, 3 (C) 7, 3, 2 (B) 3, 7, 3 (D) 1, 2, 3

[

CS – 2013 11. Which one of x x equal [ y y z z x(x y(y (A) | z(z x (B) | y z x y (C) | y z z x y (D) | y z z

15.

-

The product of the non – zero Eigenvalues of the matrix

is __________. [ 16.

the following does NOT ] ) x ) y | ) z x | y z x y y z | z x y y z | z

],

Then the determinant of A is equal to __________.

CS – 2012 10. Let A be the 2

2 matrix with elements and . Then the Eigenvalues of the matrix are (A) 1024 and (B) 1024√ and √ (C) √ n √ (D) √ n √

Mathematics

]

Which one of the following statements is TRUE about every n n matrix with only real eigenvalues? (A) If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative. (B) If the trace of the matrix is positive, all its eigenvalues are positive. (C) If the determinant of the matrix is positive, all its eigenvalues are positive. (D) If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive.

ECE – 2005 1. Given an orthogonal matrix

CS – 2014 12. Consider the following system of equations: x y x z x y z x y z The number of solutions for this system is __________.

A= [

]. ,

-

is

⁄ (A) [

⁄

]

⁄ ⁄

th

th

th


GATE QUESTION BANK

Mathematics

⁄ ⁄

(B) [

6.

]

⁄

The rank of the matrix [

⁄ (C) [

(A) 0 (B) 1

] ⁄ ⁄

(D) [

]

⁄ ⁄

2.

Let,

A=0

1 and

Then (a + b)= (A) ⁄ (B) ⁄ 3.

= 0

1.

⁄ ⁄

(C) (D)

Given the matrix 0

⁄

Eigenvector is (C) 0

1

(B) 0 1

(D) 0

1

ECE – 2006 4.

For the matrix 0 corresponding 0

the

ECE – 2007 7. It is given that X1 , X2 …… M are M nonzero, orthogonal vectors. The dimension of the vector space spanned by the 2M vector X1 , X2 … XM , X1 , X2 … XM is (A) 2M (B) M+1 (C) M (D) dependent on the choice of X1 , X2 … XM.

9.

All the four entries of the 2 x 2 matrix p p P = 0p p 1 are non-zero, and one of its Eigenvalues is zero. Which of the following statements is true? (A) p p p p (B) p p p p (C) p p p p (D) p p p p

Eigenvector

1 is

(A) 2 (B) 4 5.

1 , the Eigenvalue to

(C) 6 (D) 8

The Eigenvalues and the corresponding Eigenvectors of a 2 2 matrix are given by Eigenvalue Eigenvector =8

v =0 1

=4

(C) 2 (D) 3

ECE – 2008 8. The system of linear equations 4x + 2y = 7, 2x + y = 6 has (A) a unique solution (B) no solution (C) an infinite number of solutions (D) exactly two distinct solutions

1 the

(A) 0 1

]

v =0

1

ECE – 2009 10. The Eigen values of the following matrix are [

The matrix is (A) 0

1

(C) 0

1

(B) 0

1

(D) 0

1

]

(A) 3, 3 + 5j, 6 j (B) 6 + 5j, 3 + j, 3 j (C) 3 + j, 3 j, 5 + j (D) 3, 1 + 3j, 1 3j

th

th

th


GATE QUESTION BANK

ECE – 2010 11. The Eigenvalues of a skew-symmetric matrix are (A) Always zero (B) Always pure imaginary (C) Either zero or pure imaginary (D) Always real ECE – 2011 12. The system of equations x y z x y z x y z has NO solution for values of given by (A) (C) (B) (D)

Mathematics

ECE – 2014 16. For matrices of same dimension M, N and scalar c, which one of these properties DOES NOT ALWAYS hold? (A) (M ) M (M) (B) ( M ) (C) (M N) M N (D) MN NM 17.

A real (4 × 4) matrix A satisfies the equation I where 𝐼 is the (4 × 4) identity matrix. The positive Eigenvalue of A is _____.

18.

Consider the matrix

n

J ECE\EE\IN – 2012 13.

Given that A = 0

1 and I = 0

the value of A3 is (A) 15 A + 12 I (B) 19A + 30

(C) 17 A + 15 I (D) 17A +21

ECE – 2013 14. The minimum Eigenvalue of the following matrix is [

19.

The determinant of matrix A is 5 and the determinant of matrix B is 40. The determinant of matrix AB is ________.

20.

The system of linear equations

]

(A) 0 (B) 1 15.

[ ] Which is obtained by reversing the order of the columns of the identity matrix I . Let I J where is a nonnegative real number. The value of for which det(P) = 0 is _____.

1,

(C) 2 (D) 3

(

Let A be a m n matrix and B be a n m matrix. It is given that ) determinant Determinant(I (I ) where I is the k k identity matrix. Using the above property, the determinant of the matrix given below is

(A) 2 (B) 5

(

)h s

(A) a unique solution (B) infinitely many solutions (C) no solution (D) exactly two solutions 21.

[

)4 5

] (C) 8 (D) 16

th

Which one of the following statements is NOT true for a square matrix A? (A) If A is upper triangular, the Eigenvalues of A are the diagonal elements of it (B) If A is real symmetric, the Eigenvalues of A are always real and positive th

th


GATE QUESTION BANK

(C) If A is real, the Eigenvalues of A and are always the same (D) If all the principal minors of A are positive, all the Eigenvalues of A are also positive 22.

The maximum value of the determinant among all 2×2 real symmetric matrices with trace 14 is ___.

EE – 2005 1.

5.

If R = [

] , then top row of

(A) , (B) ,

2.

-

(C) , (D) ,

(B) [

] [

] [

]

(C) [

] [

] [

]

(D) [

] [

] [

]

-

(A) [ ] (B) [

]

(C) [

(B) [

]

(D) [ ]

]

In the matrix equation Px = q, which of the following is necessary condition for the existence of at least one solution for the unknown vector x (A) Augmented matrix [P/Q] must have the same rank as matrix P (B) Vector q must have only non-zero elements (C) Matrix P must be singular (D) Matrix P must be square

] ,R=[

(C) [ ] ]

(D) [

]

EE – 2007 6. X = [x , x . . . . x - is an n-tuple non-zero vector. The n n matrix V = X (A) Has rank zero (C) Is orthogonal (B) Has rank 1 (D) Has rank n 7.

The linear operation L(x) is defined by the cross product L(x) = b x, where b =[0 1 0- and x =[x x x - are three dimensional vectors. The matrix M of this operation satisfies x L(x) = M [ x ] x Then the Eigenvalues of M are (A) 0, +1, 1 (C) i, i, 1 (B) 1, 1, 1 (D) i, i, 0

8.

Let x and y be two vectors in a 3 dimensional space and denote their dot product. Then the determinant xx xy det 0 y x yy 1 (A) is zero when x and y are linearly independent (B) is positive when x and y are linearly independent (C) is non-zero for all non-zero x and y (D) is zero only when either x or y is zero

EE – 2006 Statement for Linked Answer Questions 4 and 5.

4.

]

is

] , one of

(A) [

] ,Q=[

] [

-

For the matrix p = [

P=[

(A) [

The following vector is linearly dependent upon the solution to the previous problem

the Eigenvalues is equal to 2 . Which of the following is an Eigenvector?

3.

Mathematics

] are

three vectors An orthogonal set of vectors having a span that contains P,Q, R is th

th

th


GATE QUESTION BANK

Statement for Linked Questions 9 and 10. Cayley-Hamilton Theorem states that a square matrix satisfies its own characteristic equation. Consider a matrix. A=0

A satisfies the relation (A) A + 3 + 2 =0 2 (B) A + 2A + 2 = 0 (C) (A+ ) (A 2) = 0 (D) exp (A) = 0

10.

equals (A) 511 A + 510  (B) 309 A + 104  (C) 154 A + 155  (D) exp (9A)

EE – 2008 11. If the rank of a ( ) matrix Q is 4, then which one of the following statements is correct? (A) Q will have four linearly independent rows and four linearly independent columns (B) Q will have four linearly independent rows and five linearly independent columns (C) Q will be invertible (D) Q will be invertible 12.

13.

(A) A A+ A = A (B) (AA+ ) = A A+ 14.

The characteristic equation of a ( ) matrix P is defined as () = | P| =    =0 If I denotes identity matrix, then the inverse of matrix P will be (A) ( I) (B) ( I) (C) ( I) (D) ( I)

(C) A+ A =  (D) A A+ A = A+

Let P be a real orthogonal matrix. x⃗ is a real vector [x x - with length ⃗x (x x ) . Then, which one of the following statements is correct? (A) x⃗ x⃗ where at least one vector satisfies x⃗ x⃗ (B) x⃗ x⃗ for all vectors x⃗ (C) x⃗ x⃗ where at least one vector satisfies x⃗ x⃗ (D) No relationship can be established between x⃗ and x⃗

1

9.

Mathematics

EE – 2009 15. The trace and determinant of a matrix are known to be –2 and –35 respe tively It’s Eigenv lues re (A) –30 and –5 (C) –7 and 5 (B) –37 and –1 (D) 17.5 and –2 EE – 2010 16. For the set of equations x x x x =2 x x x x =6 The following statement is true (A) Only the trivial solution x x x x = 0 exists (B) There are no solutions (C) A unique non-trivial solution exists (D) Multiple non-trivial solutions exist

17.

An Eigenvector of

[

(A) , (B) ,

(C) , (D) ,

-

] is -

EE – 2011 18.

The matrix[A] = 0

1 is decomposed

into a product of a lower triangular matrix [L] and an upper triangular matrix [U]. The properly decomposed [L] and [U] matrices respectively are

A is m n full rank matrix with m > n and  is an identity matrix. Let matrix A+ = ( ) , then, which one of the following statements is FALSE?

(A) 0 th

th

1 and 0

1 th


GATE QUESTION BANK

(B) 0 (C) 0 (D) 0

1 and 0 1 and 0 1 and 0

23.

1 1 1

EE – 2013 19.

The equation 0

x 1 0x 1

0 1 has

0 1.

Eigenvector of the matrix A = 0

(C) Non – zero unique solution (D) Multiple solution 20.

(A) [ 1 1]T (B) [3 1]T

A matrix has Eigenvalues – 1 and – 2. The corresponding Eigenvectors are 0 0

1 respectively. The matrix is

(A) 0

1

(C) 0

1

(B) 0

1

(D) 0

1

Which one of the following statements is true for all real symmetric matrices? (A) All the eigenvalues are real. (B) All the eigenvalues are positive. (C) All the eigenvalues are distinct. (D) Sum of all the eigenvalues is zero.

1?

(C) [1 1]T (D) [ 2 1]T

Let A be a 3 3 matrix with rank 2. Then AX = 0 has (A) only the trivial solution X = 0 (B) one independent solution (C) two independent solutions (D) three independent solutions

2.

1 and

EE – 2014 21. Given a system of equations: x y z x y z Which of the following is true regarding its solutions? (A) The system has a unique solution for any given and (B) The system will have infinitely many solutions for any given and (C) Whether or not a solution exists depends on the given and (D) The system would have no solution for any values of and 22.

Two matrices A and B are given below: p q pr qs p q [ ] 0 1 r s pr qs r s If the rank of matrix A is N, then the rank of matrix B is (A) N (C) N (B) N (D) N

IN – 2005 1. Identify which one of the following is an

(A) No solution x (B) Only one solution 0x 1

Mathematics

IN – 2006 Statement for Linked Answer Questions 3 and 4 A system of linear simultaneous equations is given as Ax=B where [

] n

[ ]

3.

The rank of matrix A is (A) 1 (C) 3 (B) 2 (D) 4

4.

Which of the following statements is true? (A) x is a null vector (B) x is unique (C) x does not exist (D) x has infinitely many values

5.

For a given that 0

1

matrix A, it is observed 0

1 n

0

1

0

1

Then matrix A is

th

th

th


GATE QUESTION BANK

 2 1  1 0   1 1     1 1  0 2 1 2

10.

(A) A  

1

1  1 0  2

1

1  1

1

(B) A      1 2 0 2 1 1 (C) A    1 2  0

02 1 2 1 1

0 2

(D) A    1 3 IN – 2007 6. Let A = [ ] i j n with n = i. j. Then the rank of A is (A) (C) n (B) (D) n 7.

n

Let A be an n×n real matrix such that = I and y be an n- dimensional vector. Then the linear system of equations Ax=Y has (A) no solution (B) a unique solution (C) more than one but finitely many independent solutions (D) Infinitely many independent solutions

The matrix P =[

12.

9.

The Eigenvalues of a (2 2) matrix X are 2 and 3. The Eigenvalues of matrix ( I) ( I) are (A) (C) (B) (D)

(

)(

)

(D) n IN – 2011 13.

The matrix M = [

] has

Eigenvalues . An Eigenvector corresponding to the Eigenvalue 5 is , - . One of the Eigenvectors of the matrix M is (A) , (C) , √ (B) , (D) ,

] rotates a vector

(C) (D)

A real n × n matrix A = [ ] is defined as i i j follows: { otherwise The summation of all n Eigenvalues of A is (A) n(n ) (B) n(n ) (C)

about the axis[ ] by an angle of (A) (B)

Let P 0 be a 3 3 real matrix. There exist linearly independent vectors x and y such that Px = 0 and Py = 0. The dimension of the range space of P is (A) 0 (C) 2 (B) 1 (D) 3

IN – 2010 11. X and Y are non-zero square matrices of size n n. If then (A) |X| = 0 and |Y| 0 (B) |X| 0 and |Y| = 0 (C) |X| = 0 and |Y| = 0 (D) |X| 0 and |Y| 0

IN – 2009 8.

Mathematics

IN – 2013 14. The dimension of the null space of the

15.

matrix [

] is

(A) 0 (B) 1

(C) 2 (D) 3

One of Eigenvectors corresponding to the two Eigenvalues of the matrix 0 (A) [

j

] 0

(B) 0 1 0 th

th

j

1 is

(C) [ ] 0 1 j j (D) [ ] 0 1 j

1 1 th


GATE QUESTION BANK

Mathematics

IN – 2014 16. For the matrix A satisfying the equation given below, the eigenvalues are , -[

]

[

(A) ( 𝑗,𝑗) (B) (1,1,0)

] (C) ( ) (D) (1,0,0)

Answer Keys and Explanations ME 1.

and G = [

[Ans. A] [

Now E × F = G

]

h r teristi equ tions is | I| ( )( )( ) ∴ Real eigenvalues are 5, 5 other two are complex Eigenvector corresponding to is ( I) (or) →( ) Verify the options which satisfies relation (1) Option (A) satisfies. [Ans. B] Given

n

in onsistent

4.

5.

[Ans. A]

6.

[Ans. B] 1 Eigenv lues re 2, 2 I)

(

I)

No. of L.I Eigenvectors ( (no of v ri les)

( ⁄ )

7.

matrix be A = 0 sin os

.

/ I)

[Ans. A] ( I) . olving for , Let the symmetric and real

[Ans. C] os Given , E = [ sin

]

matrix, if Eigenvalues are … … … … … then for matrix, the Eigenvalues will be , , ……… For S matrix, if Eigenvalues are 1 and 5 then for matrix, the Eigenvalues are 1 and 25.

No (

3.

sin os

[Ans. A] For S

0

( ) n ( ⁄ ) ( ( ) minimum of m n) For inconsistence ( ) ( ⁄ ) ∴ he highest possi le r nk of is

os [ sin

,E-

∴

2.

]

]

th

1

Now |

|

Which gives ( ⟹

)

th

th


GATE QUESTION BANK

⟹ Hence real Eigen value. 8.

Mathematics

x [

][

]

0

1

x

[Ans. B] Let

0

eigenv lues re

1

12.

Eigen vector corresponding to is ( I) x . / .y/ . / By simplifying K . / . / y t king K

⁄

Equating the elements x

n

[Ans. A] 0

1 → Eigenv lues re

Eigenve tor is x 13.

Eigen vector corresponding to =2 is ( I) x . / .y/ . / K By simplifying ( ) 4 5 by ⁄ K

[Ans. C] [

]

[

→

[

]

→

taking K

x verify the options

( )

[

]

]

infinite m ny solutions

⁄ ⁄ 9.

10.

[Ans. C] Sum of the diagonal elements = Sum of the Eigenvalues ⟹ 1 + 0 + p = 3+S ⟹ S= p 2

[Ans. B] Eigenvalues of a real symmetric matrix are always real

15.

[Ans. B] 0

If

1 eigenv lues v lue

Eigen vector will be .

/

Norm lize ve tor

[Ans. B] ( ⁄ )

11.

14.

[

]

√( )

(

)

[

]

[√( )

(

) ]

→ →

[

→

[

*

]

]

16.

system will h ve solution

[Ans. A] iven M

M

→ MM

I

th

⁄ √ + ⁄ √

[Ans. C] The given system is x y z x y z x y z Use Gauss elimination method as follows Augmented matrix is

th

th


GATE QUESTION BANK

, | -

[

→

→

| ] [

So, |

|

[

|

|

[Ans. C] Suppose the Eigenvalue of matrix A is ( i )(s y) and the Eigenvector is ‘x’ where s the onjug te p ir of Eigenvalue and Eigenvector is ̅ n x̅. So Ax = x … ① and x̅ ̅x̅……② king tr nspose of equ tion ② x̅ x̅ ̅ … ③ [( ) n ̅ is s l r ] ̅ x̅ x x̅ x x̅ x x̅ ̅x … , x̅ x x̅ ̅ x ̅ (x̅ x) ( ̅ re s l r ) (x̅ x) ̅

20.

[Ans. C] We know that os x os x sin x ( ) os x sin x ( ) os x Hence 1, 1 and 1 are coefficients. They are linearly dependent.

1 eigen v lues

Eigenve tor is

verify for oth n

21.

[Ans. D] We know that the Eigenvectors corresponding to distinct Eigenvalues of real symmetric matrix are orthogonal. x y x y [ ][ ] x y x y x y y x

22.

[Ans. D] ( ) In case of matrix PQ

CE 1.

QP (generally)

[Ans. C] If = i.e. A is orthogonal, we can only s y th t if is n Eigenv lue of then

also will be an Eigenvalue of A,

which does not necessarily imply that | | = 1 for all i. 2.

[Ans. A] In an over determined system having more equations than variables, it is necessary to have consistent unique solution, by definition

3.

[Ans. A] With the given order we can say that order of matrices are as follows:  3×4 Y  4×3  3×3

[Ans. A] |

[Ans. D] 0

nnot e zero )

Hence Eigenvalue of a symmetric matrix are real

19.

|

(Taking 2 common from each row) ( )

]

( x x̅ re Eigenve tors they i i i 0

18.

|

]

nk ( ) nk ( | ) So, Rank (A) = Rank (A|B) = 2 < n (no. of variables) So, we have infinite number of solutions 17.

Mathematics

|

th

th

th


GATE QUESTION BANK

( )  3×3 P 2×3  3×2 P( )  (2×3) (3×3) (3×2) 2×2 ( ( ) )  2×2

Using Gauss elimination we reduce this to an upper triangular matrix to find its rank | ]→

[

→ 4.

[Ans. D] The augmented matrix for given system is [

| ]→

[

| ]→

| ]

[

8.

| ]

[

[

|

[

|

]

( ⁄ ) ( ) ( ) ( ⁄ ) ∴ olution is non – existent for above system. 5.

6.

7.

[Ans. B] ∑ = Trace (A) + + = Trace (A) = 2 + ( 1) + 0 = 1 Now = 3 ∴3+ + =1  Only choice (B) satisfies this condition. [Ans. B] ∑ = Trace (A) + + =1+5+1=7 Now = 2, = 6 ∴ 2+6+ =7 =3

0

1

∴0

]

1 is (

1

)

(

) 0

9.

10.

0

1

0

1 1

[Ans. B] ( ) P=( ( )( ) =( ) (I) =

)P

[Ans. B] A=0

1

Characteristic equation of A is |

|=0

 (4 )( 5 ) 2 × 5 =0  + 30 = 0 6, 5 11.

[Ans. A] The augmented matrix for given system is [

]

[Ans. A] Inverse of 0

→

|

Now for infinite solution last row must be completely zero ie –2=0 n –7=0  n

Then by Gauss elimination procedure [

Mathematics

| ]

th

[Ans. D] The augmented matrix for given system is x [ | ] 6y7 [ ] z k Using Gauss elimination we reduce this to an upper triangular matrix to find its rank

th

th


GATE QUESTION BANK

| ]→

[

17.

[Ans. 16] , , M trix , The product of matrix PQR is , - , - , The minimum number of multiplications involves in computing the matrix product PQR is 16

18.

[Ans. 23]

k [

| ]

[

| ]

→

Now if k Rank (A) = rank (A|B) = 3 ∴ Unique solution If k = 7, rank (A) = rank (A|B) = 2 which is less than number of variables ∴ When K = 7, unique solution is not possible and only infinite solution is possible 12.

[Ans. A] A square matrix B is defined as skewsymmetric if and only if = B

13.

[Ans. D] By definition A + is always symmetric is symmetri is lw ys skew symmetri is skew symmetri

Mathematics

[

][

]

[

,

K JK

]

-[ ,

]

,

[

] -

-

19.

[Ans. A] Sum of Eigenvalues = Sum of trace/main diagonal elements = 215 + 150 + 550 = 915

20.

[Ans. 88] The determinant of matrix is [

]

→

14.

[Ans. B] 1 =(

0

i

∴ 0

15.

i

i

i

,( =

0

0

)

i)( i i

[

1

→

1 i -

i) i i

0

i i

i i

1

[

1

[

]

1 Interchanging Column 1& Column 2 and taking transpose

Sum of the Eigenvalues = 17 Product of the Eigenvalues = From options, 3.48 + 13.53 = 17 (3.48)(13.53) = 47 16.

]

→

[Ans. B] 0

]

[

[Ans. 0.5] 0.5

]

|

th

th

|

th


GATE QUESTION BANK * (

)

= ( 21.

(

(

)+

= 1, 6 ∴ The Eigenvalues of A are 1 and 6

[Ans. 2] ]

3.

[Ans. D] Given that Fu =b and Fv =b If F is non singular, then it has a unique inverse. Now, u = b and v= b Since is unique, u = v but it is given th t u v his is contradiction. So F must be singular. This means that (A) Determinant of F is zero is true. Also (B) There are infinite number of solution to Fx= b is true since |F| = 0 (C) here is n su h the is also true, since X has infinite number of solutions., including the X = 0 solution (D) F must have 2 identical rows is false, since a determinant may become zero, even if two identical columns are present. It is not necessary that 2 identical rows must be present for |F| to become zero.

4.

[Ans. C] It is given that Eigenvalues of A is 5, 2, 1, 4 I Let P = 0 1 I Eigenvalues of P : | I| I | | I ( ) I I I Eigenvalue of P is ( 5 +1 ), ( 2+ 1), (1+ 1), (4+1 ), ( 5 1 ), ( 2 1 ),(1 1), (4 1) = 4, 1, 2, 5, 6, 3,0,3

5.

[Ans. B] |x X= {x x x = ,x x x - then,

→ [ ( )

(

)

( ) ]

( )

( )

[

] ( )

no. of non zero rows = 2

[Ans. B] The augmented matrix for the given system is [

| ]

Using elementary transformation on above matrix we get, [

| ]

→

⁄ | ] ⁄ ⁄

[

→

[

|

]

Rank ([A B]) = 3 Rank ([A]) = 3 Since Rank ([A B]) = Rank ([A]) = number of variables, the system has unique solution. 2.

[Ans. B] 0

1

The characteristic equation of this matrix is given by | I| |

)

)

[

CS 1.

)(

Mathematics

+

| th

th

th


GATE QUESTION BANK

{ [1, 1, 0]T , [1,0, 1 ]T } is a linearly independent set because one cannot be obtained from another by scalar multiplication. However (1, 1, 0) and (1,0, 1) do not span X, since all such combinations (x1, x2, x3) such that x1+ x2+ x3 =0 cannot be expressed as linear combination of (1, 1,0) and (1,0, 1) 6.

7.

Only one matrix has an Eigenvalue of 1 which is 0

| ] →

→

[

[

1

Correct choice is (A) 8.

[Ans. D] |

| x y ( )( y) When ( y) x y x When ( y) x y x x y Solving (1) & (2) x y

[Ans. D] The augmented matrix for above system is [

Mathematics

| ] | ]

x

( )

( )

Now as long as – 5 0, rank (A) =rank (A|B) =3 ∴ can be any real value except 5. Closest correct answer is (D).

9.

[Ans. A] The Eigenvalues of a upper triangular matrix are given by its diagonal entries. ∴ Eigenvalues are 1, 4, 3 only

[Ans. A]

10.

[Ans. D]

Eigenvalues of 0 |

1

0

| =0

Eigenvalues of 0 |

Eigenvalues of the matrix (A) are the roots of the characteristic polynomial given below.

=0,1 1

|

| =0 =0

1

|

(√ )

)( ) =0 = –1, 1

n

√

n ( √ ) n

1

n

| =0

( (

) )

√ Eigenvalues of A are √ respectively So Eigenvalues of

) =0 ) = i or 1 = 1 –i or 1 + i

Eigenvalues of 0

)( )(

(

|= 0

( (

|

(

= 0, 0

Eigenvalues of 0 |

1

√

) =0

th

th

n

√

th


GATE QUESTION BANK

11.

12.

[Ans. A] → p q nd Since 2 & 3rd columns have been swapped which introduces a –ve sign Hence (A) is not equal to the problem

[

]

→

[

] →

→

[

] →

16.

[

]

( ) ( ) no of v ri ∴ nique solution exists

14.

[ ] x x Let X = x e eigen ve tor x [x ] By the definition of eigenvector, AX = x x x x x x x x [ ] [x ] [x ] x x x x x x x x x x x x x x x x x x x x x x n x x x x x x (I) If s yx x x x x x x x x x (2) If Eigenv lue ∴ Three distinct eigenvalues are 0, 2, 3 Product of non zero eigenvalues = 2 × 3 = 6

]

→ →

les

[Ans. 0] The Eigenvectors corresponding to distinct Eigenvalues of real symmetric matrix are orthogonal

ECE 1.

2.

[Ans. A] If the trace or determinant of matrix is positive then it is not necessary that all eigenvalues are positive. So, option (B), (C), (D) are not correct

[Ans. C] Since, ,

] (

=I

16

0

[

-

[Ans. A] We know,

[Ans. 0]

| |

[Ans. 6] Let A =

[Ans. 1] x y x z x y z x y z ugmente m trix is [

13.

15.

Mathematics

0

7=0 1

0

1 1

1 b ,  a 60 10 1 1 21 7   a+b =   3 60 60 20

)

Or 2a 0.1b=0,  2a 

th

th

th


GATE QUESTION BANK

3.

[Ans. C]

8.

0

[Ans. B] Approach 1: Given 4x + 2y =7 and 2x + y =6

1

 (A I)=0  ( 4 ) (3 ) 2 4=0  2 + 20=0  = 5, 4

4 2 x  7 2 1y   6      0 0 x  5 2 1y   6      

 x1   x2 

Putting = 5, 0

1   =0

 x + 2x = 0 x = 2x

On comparing LHS and RHS 0= 5, which is irrelevant and so no solution. Approach 2: 4x + 2y =7

x x  1= 2 2 1 Hence, 0 4.

1 is Eigenvector.

[Ans. C]

Then Eigenvector is x Verify the options (C) 5.

or 2x  y=

1 We know th t it is Eigenvalue

0

We know

0

1

|I A|=0

|

|

 2 –I2 +32 =0  = 4, 8 (Eigenvalues) For

= 4, ( I

)=0

1

)=0

1

9.

[Ans. C] Matrix will be singular if any of the Eigenvalues are zero. |  |= 0 For = 0, P = 0 p p  |p p | =0  p p p p

10.

[Ans. D] Approach1: Eigenvalues exists as complex conjugate or real Approach 2: Eigenvalues are given by

v =0 1 For

= 8, ( I

v =0 6.

1

[Ans. C] [

] [

|

]

[Ans. C] There are M non-zero, orthogonal vectors, so there is required M dimension to represent them ’

| =0

( 

( ) 7.

7 2

2x+y=6 Since both the linear equation represent parallel set of straight lines, therefore no solution exists. Approach 3: Rank (A)=1; rank (C)=2, As Rank (A) rank (C) therefore no solution exists.

x

[Ans. A] or m trix

Mathematics

11.

th

)(( ,

)=0

) j

j

[Ans. C] Eigenvalue of skew – symmetric matrix is either zero or pure imaginary. th

th


GATE QUESTION BANK

12.

13.

[Ans. B] Given equations are x y z x y z and x y z If and , then x y z have Infinite solution If and , then x y z ( ) no solution x y z If n x y z will have solution x y z and will also give solution

et of , -

et of [

]

16.

[Ans. D] Matrix multiplication is not commutative in general.

17.

[Ans. *] Range 0.99 to 1.01 Let ‘ ’ e Eigenv lue of ‘ ’ hen ‘ e Eigenv lue of ‘ ’ A. =I= Using Cauchey Hamilton Theorem,

[Ans. B] 0

Mathematics

’ will

1

Characteristic Equations is 18. By Cayley Hamilton theorem I ∴ ( I) I 14.

I | | [

[Ans. A] [

]

→

(

[

[Ans. *] Range 199 to 201 From matrix properties we know that the determinant of the product is equal to the product of the determinants. That is if A and B are two matrix with determinant | | n | | respectively, then | | | | | | ∴| | | | | |

20.

[Ans. B]

) ]

]

19.

| |

| | Product of Eigenvalues = 0 ∴ Minimum Eigenv lue h s to e ‘ ’ 15.

[Ans. *] Range 0.99 to 1.01 I J I J

[Ans. B] ,

Let

-

[ ]

[

I

I

[

Then AB = [4]; BA Here m = 1, n = 4 ) And et(I

]

→

[

→ →

[

[

]

th

les

[Ans. B] onsi er

)

]

]

( ) ( | ) no of v r Infinitely many solutions 21.

et(I

]

th

0

1 th


GATE QUESTION BANK

whi h is re l symmetri m trix h r teristi equ tion is | I| ( ) ∴ (not positive) ( ) is not true (A), (C), (D) are true using properties of Eigenvalues 22.

EE 1.

2.

[Ans. B] ] j( ) | |

=[

]

∴ Top row of

=,

-

[Ans. D] Since matrix is triangular, the Eigenvalues are the diagonal elements themselves namely  = 3, 2 & 1. Corresponding to Eigenvalue  = 2, let us find the Eigenvector [A - ] x̂ = 0 x [ ][x ] [ ] x Putting in above equation we get, x [ ][x ] [ ] x Which gives the equations, 5x x x =0 . . . . . (i) x =0 . . . . . (ii) 3x = 0 . . . . . (iii) Since eqa (ii) and (iii) are same we have 5x x x =0 . . . . . (i) x =0 . . . . . (ii) Putting x = k, we get x = 0, x = k and 5x k =0

[Ans. *] Range 48.9 to 49.1 Real symmetric matrices are diagnosable Let the matrix be x 0 1 s tr e is x So determinant is product of diagonal entries So | | x x ∴ M ximum v lue of etermin nt x x ∴| |

R= [

Mathematics

, of tor( )| |

x = k | |=|

|

∴ Eigenvectorss are of the form x k x [ ] * k + x

= 1(2 + 3) – 0(4 + 2) – 1 (6 – 2) = 1 Since we need only the top row of , we need to find only first column of (R) which after transpose will become first row adj(A). cof. (1, 1) = + |

|=2+3=5

cof. (2, 1) =

|= 3

|

cof. (2, 1) = + |

i.e. x x x = k : k : 0 = :1:0 =2:5:0 x x ∴ [ ]=[ ] is an Eigenvector of matrix p. x

|= +1 3.

∴ cof. (A) = [

[Ans. A] Rank [P|Q] = Rank [P] is necessary for existence of at least one solution to x q.

]

Adj (A) =, of ( )=[

]

Dividing by |R| = 1 gives th

th

th


GATE QUESTION BANK

4.

  (

[Ans. A] We need to find orthogonal vectors, verify the options. Option (A) is orthogonal vectors (

)(

[Ans. B] The vector ( ) is linearly dependent upon the solution obtained in - and , Q. No. 4 namely , We can easily verify the linearly dependence as |

6.

7.

i

[Ans. B] xy xx | yx

xy xx x n xy yx xy x xy y y | |y x y | (x y) x y = Positive when x and y are linearly independent.

Option (B), (C), (D) are not orthogonal 5.

) i

8.

)

Mathematics

9.

[Ans. A] A=0

1

|A – | = 0  |

|

[Ans. B] hen n n m trix xx x x x x x x x x x x x x * + x x x x x x Take x common from 1st row, x common from 2nd row …… x common from nth row. It h s r nk ‘ ’

| =0  A will satisfy this equation according to Cayley Hamilton theorem i.e. I=0 Multiplying by on oth si es we get I=0 I =0 10.

[Ans. A] To calculate Start from derived above

I = 0 which has I

[Ans. D] ⃗ k L(x) = |

(

| x

x

= (x )

I)(

x (

⃗( k

)

(

x )

I) I

x = x

⃗ =[ x k

(

x L(x) = M [x ] x Comparing both , we get,

(

|

)

I

I) (

I) I

| (

I)

I) I

(

Hence Eigenvalue of M : | M



I)( I

]

|

I

] x

M=[

I) I

(

) th

th

th


GATE QUESTION BANK

11.

12.

13.

[Ans. A] If rank of (5 6 ) matrix is 4,then surely it must have exactly 4 linearly independent rows as well as 4 linearly independent columns.

= A is correct =A[( ) -A = A[( ) Put =P Then A [ ] = A. = A Choice (C) =  is also correct since =( ) = I 14.

os

x in )

|| x⃗ || = √x

(x in

x

x

→

[Ans. C] Trace = Sum of Principle diagonal elements.

16.

[Ans. D] On writing the equation in the form of AX =B

+

, *

+

nk ( ) nk( ) Number of variables = 4 Since, Rank (A) = Rank(C) < Number of variables Hence, system of equations are consistent and there is multiple non-trivial solution exists. 17.

[Ans. B] Characteristic equation | |

I|

|

 (1 ) ( )( )  Eigenve tors orrespon ing to ( I) x [ ] [x ] [ ] x 2x x  x x At x x x x  x x At x ,x

is

Eigenvectors = c[ ]{Here c is a constant}

os ) 18.

[Ans. D] , - ,L-, - ⟹ Options D is correct

19.

[Ans. D] x x … (i) } (i) n (ii) re s me x x … (ii) ∴x x So it has multiple solutions.

|| x⃗ || = || x̅|| for any vector x̅ 15.

* +

Argument matrix C =*

[Ans. B] Let orthogonal matrix be os in P=0 1 in os By Property of orthogonal matrix A I x os x in So, x⃗ = [ ] x in x os || x⃗ || = √(x

x x + *x + x

*

[Ans. D] If characteristic equation is    =0 Then by Cayley – Hamilton theorem, I=0 = Multiplying by on both sides, = I = ( I) [Ans. D] Choice (A) Since

Mathematics

th

th

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GATE QUESTION BANK

20.

[Ans. D] Eigen value

|A

Eigenvectors 0

1 n 0

Let matrix 0 x

1

x 10

1

0

1

0

10

1

0

1

I|= |

|

i.e., (1 ) (2 ) 2 Thus the Eigenvalue are 1, 2. If x, y, be the component of Eigenvectors corresponding to the Eigenv lues we have x [A- I- 0 1 0y1=0

1

0

Mathematics

For =1, we get the Eigenvector as 0 Hence, the answer will be ,

21.

22.

23.

IN 1.

1

0

[Ans. B] AX=0 and (A) = 2 n=3 No. of linearly independent solutions = n r = 3 =1

3.

[Ans. C] There are 3 non-zero rows and hence rank (A) = 3

4.

[Ans. C] Rank (A) = 3 (This is Co-efficient matrix) Rank (A:b) =4(This is Augmented matrix) s r nk( ) r nk ( ) olution oes not exist.

5.

[Ans. C] We know Hen e from the given problem, Eigenvalue & Eigenvector is known.

1

[Ans. B] Since there are 2 equations and 3 variables (unknowns), there will be infinitely many solutions. If if then x y z x y z x z y For any x and z, there will be a value of y. ∴ Infinitely many solutions [Ans. A] For all real symmetric matrices, the Eigenvalues are real (property), they may be either ve or ve and also may be same. The sum of Eigenvalues necessarily not be zero. [Ans. C] p q 0 1 r s ( pplying → p q →r s element ry tr nsform tions) p q pr qs [ ] pr qs r s ∴ hey h ve s me r nk N

1 X1    , X2  1

1 2 , 1  1, 2  2  

We also know that

, where

1 1

P  X1 X2     1 2

1 0  1 0     0 2   0 2

[Ans. B] Given:

-

2. Solving 0

1

& D=  0

1 Hence

Characteristic equation is,

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GATE QUESTION BANK

 1 1  1 0   2 1  A       1 2  0 2 1 1 6.

12.

[Ans. B] A= [

]=[

A=[

[Ans. B] Given I Hence rank (A) = n Hence AX= Y will have unique solution

8.

[Ans. C]

9.

[Ans. C] Approach 1:

13.

14.

Assume,

0

(

∴A

(

0

1

0

10

Now | I

0

[Ans. B] Dim of null space [A]= nullity of A.

0

[

]

1

| )(

]

Apply row operations 1

1

- is also vector

For given A = [

1

I) 0

[Ans. B] If AX = → From this result [1, 2, for M

|

| (

I

1

I)

]

n For diagonal matrix Eigenvalues are diagonal elements itself. n(n ) ∴ n

]

Hence, rank (A) =1 7.

[Ans. A] A=[ ] i if i j = 0 otherwise. For n n matrix

]

Using elementary transformation [

Mathematics

)=0

[Ans. D]

11.

[Ans. C] A null matrix can be obtained by multiplying either with one null matrix or two singular matrices.

[

→

[

] ]

∴ ( ) By rank – nullity theorem Rank [A]+ nullity [A]= no. of columns[A] Nullity [A]= 3 ∴ Nullity , -

Approach 2: Eigenvalues of ( I) is = 1, 1/2 Eigenvalues of (X+5I) is = 3, 2 Eigenvalues of ( I) (X+5I) is = , 10.

→

15.

[Ans. A] A=|

|

Characteristics equation | |

I|

| j j

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GATE QUESTION BANK

j

[

j

x ] 0x 1

Mathematics

0 1

x x

j j

[

j j

x ] 0x 1

x

0 1

j

x 16.

[Ans. C]

A[

]=[

→| | |

|

] |

|

→| | (

|

|

|

| two rows ounter lose thus | |

| |) =Product of eigenvalues Verify options Options (C) correct answer

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th

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GATE QUESTION BANK

Mathematics

Probability and Distribution ME - 2005 1. A single die is thrown twice. What is the probability that the sum is neither 8 nor 9? (A) ⁄ (C) ⁄ ⁄ (B) ⁄ (D)

ME - 2008 6. A coin is tossed 4 times. What is the probability of getting heads exactly 3 times? (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄

2.

ME - 2009 7. The standard deviation of a uniformly distributed random variable between 0 and 1 is (A) (C) ⁄√ √ (B) (D) √ √

A lot has 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective is (A) 0.0036 (C) 0.2234 (B) 0.1937 (D) 0.3874

ME - 2006 3. Consider a continuous random variable with probability density function f(t) = 1 + t for 1  t  0 = 1 t for 0  t  1 The standard deviation of the random variable is: (C) ⁄ (A) ⁄√ (D) ⁄ (B) ⁄√ 4.

A box contains 20 defective items and 80 non-defective items. If two items are selected at random without replacement, what will be the probability that both items are defective? ⁄ (A) ⁄ (C) ⁄ ⁄ (B) (D)

ME - 2007 5. Let X and Y be two independent random variables. Which one of the relations between expectation (E), variance (Var) and covariance (Cov) given below is FALSE? (A) E (XY) = E (X) E (Y) (B) Cov (X, Y) = 0 (C) Var (X + Y) = Var (X) + Var (Y) (D)

(X Y )

( (X)) ( (Y))

8.

If three coins are tossed simultaneously, the probability of getting at least one head is (A) 1/8 (C) 1/2 (B) 3/8 (D) 7/8

ME - 2010 9. A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is (A) 2/315 (C) 1/1260 (B) 1/630 (D) 1/2520 ME - 2011 10. An unbiased coin is tossed five times. The outcome of each toss is either a head or a tail. The probability of getting at least one head is________ ⁄ (A) ⁄ (C) ⁄ ⁄ (B) (D) ME - 2012 11. A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set has one red ball and two black balls is (A) 1/20 (C) 3/10 (B) 1/12 (D) 1/2 th

th

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GATE QUESTION BANK

ME - 2013 12. Let X be a normal random variable with mean 1 and variance 4. The probability (X ) is (A) 0.5 (B) Greater than zero and less than 0.5 (C) Greater than 0.5 and less than 1.0 (D) 1.0 13.

The probability that a student knows the correct answer to a multiple choice

the probability of obtaining red colour on top face of the dice at least twice is _______ 17.

A group consists of equal number of men and women. Of this group 20% of the men and 50% of the women are unemployed. If a person is selected at random from this group, the probability of the selected person being employed is _______

18.

A machine produces 0, 1 or 2 defective pieces in a day with associated probability of 1/6, 2/3 and 1/6, respectively. The mean value and the variance of the number of defective pieces produced by the machine in a day, respectively, are (A) 1 and 1/3 (C) 1 and 4/3 (B) 1/3 and 1 (D) 1/3 and 4/3

19.

A nationalized bank has found that the daily balance available in its savings accounts follows a normal distribution with a mean of Rs. 500 and a standard deviation of Rs. 50. The percentage of savings account holders, who maintain an average daily balance more than Rs. 500 is _______

20.

The number of accidents occurring in a plant in a month follows Poisson distribution with mean as 5.2. The probability of occurrence of less than 2 accidents in the plant during a randomly selected month is (A) 0.029 (C) 0.039 (B) 0.034 (D) 0.044

question is . If the student dose not know the answer, then the student guesses the answer. The probability of the guessed answer being correct is . Given that the student has answered the questions correctly, the conditional probability that the student knows the correct answer is (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄ ME - 2014 14. In the following table x is a discrete random variable and P(x) is the probability density. The standard deviation of x is x 1 2 3 P(x) 0.3 0.6 0.1 (A) 0.18 (C) 0.54 (B) 0.3 (D) 0.6 15.

16.

Box contains 25 parts of which 10 are defective. Two parts are being drawn simultaneously in a random manner from the box. The probability of both the parts being good is ( )

( )

( )

( )

Consider an unbiased cubic dice with opposite faces coloured identically and each face coloured red, blue or green such that each colour appears only two times on the dice. If the dice is thrown thrice,

Mathematics

CE - 2005 1. Which one of the following statements is NOT true? (A) The measure of skewness is dependent upon the amount of dispersion

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GATE QUESTION BANK

(B) In a symmetric distribution the value of mean, mode and median are the same (C) In a positively skewed distribution mean > median > mode (D) In a negatively skewed distribution mode > mean > median CE - 2006 2. A class of first years B. Tech students is composed of four batches A, B, C and D each consisting of 30 students. It is found that the sessional marks of students in Engineering Drawing in batch C have a mean of 6.6 and standard deviation of 2.3. The mean and standard deviation of the marks for the entire class are 5.5 and 4.2 respectively. It is decided by the course instruction to normalize the marks of the students of all batches to have the same mean and standard deviation as that of the entire class. Due to this, the marks of a student in batch C are changed from 8.5 to (A) 6.0 (C) 8.0 (B) 7.0 (D) 9.0 3.

There are 25 calculators in a box. Two of them are defective. Suppose 5 calculators are randomly picked for inspection (i.e. each has the same chance of being selected). What is the probability that only one of the defective calculators will be included in the inspection? (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄

CE - 2007 4. If the standard deviation of the spot speed of vehicles in a highway is 8.8 kmph and the mean speed of the vehicles is 33 kmph, the coefficient of variation in speed is (A) 0.1517 (C) 0.2666 (B) 0.1867 (D) 0.3646

Mathematics

CE - 2008 5. If probability density function of a random variable x is x for x nd f(x) { for ny other v lue of x Then, the percentage probability P.

x

/ is

(A) 0.247 (B) 2.47 6.

(C) 24.7 (D) 247

A person on a trip has a choice between private car and public transport. The probability of using a private car is 0.45. While using the public transport, further choices available are bus and metro out of which the probability of commuting by a bus is 0.55. In such a situation, the probability, (rounded upto two decimals) of using a car, bus and metro, respectively would be (A) 0.45, 0.30 and 0.25 (B) 0.45, 0.25 and 0.30 (C) 0.45, 0.55 and 0.00 (D) 0.45, 0.35 and 0.20

CE - 2009 7. The standard normal probability function can be approximated as (x )

|x | ) exp( Where x = standard normal deviate. If mean and standard deviation of annual precipitation are 102 cm and 27 cm respectively, the probability that the annual precipitation will be between 90 cm and 102 cm is (A) 66.7% (C) 33.3% (B) 50.0% (D) 16.7% CE - 2010 8. Two coins are simultaneously tossed. The probability of two heads simultaneously appearing is (A) 1/8 (C) 1/4 (B) 1/6 (D) 1/2

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GATE QUESTION BANK

CE - 2011 9. There are two containers with one containing 4 red and 3 green balls and the other containing 3 blue and 4 green balls. One ball is drawn at random from each container. The probability that one of the balls is red and the other is blue will be (A) 1/7 (C) 12/49 (B) 9/49 (D) 3/7 CE - 2012 10. The annual precipitation data of a city is normally distributed with mean and standard deviation as 1000mm and 200 mm, respectively. The probability that the annual precipitation will be more than 1200 mm is (A) < 50 % (C) 75 % (B) 50 % (D) 100 % 11.

14.

A traffic office imposes on an average 5 number of penalties daily on traffic violators. Assume that the number of penalties on different days is independent and follows a poisson distribution. The probability that there will be less than 4 penalties in a day is ____.

15.

A fair (unbiased) coin was tossed four times in succession and resulted in the following outcomes: (i) Head (iii) Head (ii) Head (iv) Head The prob bility of obt ining ‘T il’ when the coin is tossed again is (A) 0 (C) ⁄ (B) ⁄ (D) ⁄

16.

An observer counts 240 veh/h at a specific highway location. Assume that the vehicle arrival at the location is Poisson distributed, the probability of having one vehicle arriving over a 30-second time interval is ____________

In an experiment, positive and negative values are equally likely to occur. The probability of obtaining at most one negative value in five trials is (A)

(C)

(B)

(D)

CE - 2013 12. Find the value of such that the function f(x) is a valid probability density function ____________________ (x )( f(x) x) for x otherwise CE - 2014 13. The probability density function of evaporation E on any day during a year in a watershed is given by f( )

{

mm d y

Mathematics

CS - 2005 1. Box P has 2 red balls and 3 blue balls and box Q has 3 red balls and 1 blue ball. A ball is selected as follows: (i) select a box (ii) choose a ball from the selected box such that each ball in the box is equally likely to be chosen. The probabilities of selecting boxes P and Q are 1/3 and 2/3 respectively. Given that a ball selected in the above process is red, the probability that it comes from box P is (A) 4/19 (C) 2/9 (B) 5/19 (D) 19/30 2.

Let f(x) be the continuous probability density function of a random variable X. The probability that a  X  b , is (A) f(b a) (C) ∫ f(x)dx

otherwise The probability that E lies in between 2 and 4 mm/day in a day in the watershed is (in decimal) ______

(B) f(b)

th

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f( )

(D) ∫ x f(x)dx

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GATE QUESTION BANK

CS - 2006 3. For each element in a set of size 2n, an unbiased coin is tossed. The 2n coin tosses are independent. An element is chosen if the corresponding coin toss were head. The probability that exactly n elements are chosen is (A) ( n ⁄ ) (C) ( ⁄ n ) (D) ⁄ (B) ( n ⁄ ) CS - 2007 Linked Data for Q4 & Q5 are given below. Solve the problems and choose the correct answers. Suppose that robot is placed on the Cartesian plane. At each step it is easy to move either one unit up or one unit right, i.e if it is at (i,j) then it can move to either (i+1,j) or (i,j+1) 4. How many distinct paths are there for the robot to reach the point (10,10) starting from the initial position (0,0)? (C) 210 (A) 20 (D) None of these (B) 2 5.

Suppose that the robot is not allowed to traverse the line segment from (4,4) to (5,4). With this constraint, how many distinct paths are there for the robot to reach (10,10) starting from (0,0)? (A) 29 (B) 219 (C) . / . (D) .

6.

/

/ . / .

/

Suppose we uniformly and randomly select a permutation from the 20! ermut tions of ………… Wh t is the probability that 2 appears at an earlier position than any other even number in the selected permutation? (A) ⁄ (C) ⁄ (B) ⁄ (D) none of these

Mathematics

CS - 2008 7. Let X be a random variable following normal distribution with mean +1 and variance 4. Let Y be another normal variable with mean of 1 and variance unknown If (X ) (Y≥ ) the standard deviation of Y is (A) 3 (C) √ (B) 2 (D) 1 8.

Aishwarya studies either computer science or mathematics every day. If she studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday? (A) 0.24 (C) 0.4 (B) 0.36 (D) 0.6

CS - 2009 9. An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 90% of the probability that the face value is even. The probability of getting any even numbered face is the same. If the probability that the face is even given that it is greater than 3 is 0.75, which one of the following options is closest to the probability that the face value exceeds 3? (A) 0.453 (C) 0.485 (B) 0.468 (D) 0.492 CS - 2010 10. Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company therefore subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q. What is the probability of a computer being declared faulty? th

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GATE QUESTION BANK

(A) (B) (C) (D) 11.

12.

pq+(1 – p)(1 – q) (1 – q)p (1 – p)q pq

What is the probability that a divisor of is a multiple of ? (A) 1/625 (C) 12/625 (B) 4/625 (D) 16/625 If the difference between the expectation of the square if a random variable ( ,x -) and the square if the exopectation of the random variable ( ,x-) is denoted by R, then (A) R = 0 (C) R≥ (B) R< 0 (D) R > 0

CS - 2011 13. A deck of 5 cards (each carrying a distinct number from 1 to 5) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second card? (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄ 14.

Consider a finite sequence of random values X = [x1, x2 … xn].Let be the me n nd σx be the standard deviation of X. Let another finite sequence Y of equal length be derived from this as yi, a*xi+b, where a and b are positive constants. Let μy be the me n nd σy be the standard deviation of this sequence. Which one of the following statements is INCORRECT? (A) Index position of mode of X in X is the same as the index position of mode of Y in Y. (B) Index position of median of X in X is the same as the index position of median of Y in Y. (C) μy μx + b (D) σy σx + b

15.

Mathematics

If two fair coins flipped and at least one of the outcomes is known to be a head, what is the probability that both outcomes are heads? (A) 1/3 (C) 1/4 (B) 1/2 (D) 2/3

CS - 2012 16. Suppose a fair six – sided die is rolled once. If the value on the die is 1,2, or 3 the die is rolled a second time. What is the probability that the some total of value that turn up is at least 6? (A) 10/21 (C) 2/3 (B) 5/12 (D) 1/6 17.

Consider a random variable X that takes values +1 and 1 with probability 0.5 each. The values of the cumulative distribution function F(x) at x = and +1 are (A) 0 and 0.5 (C) 0.5 and 1 (B) 0 and 1 (D) 0.25 and 0.75

CS - 2013 18. Suppose p is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and p has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval? ⁄ e (A) ⁄ e (C) ⁄ e (B) ⁄ e (D) CS - 2014 19. Suppose you break a stick of unit length at a point chosen uniformly at random. Then the expected length of the shorter stick is ________ . 20.

th

Four fair six – sided dice are rolled. The probability that the sum of the results being 22 is x/1296. The value of x is ____________

th

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GATE QUESTION BANK

21.

The security system at an IT office is composed of 10 computers of which exactly four are working. To check whether the system is functional, the officials inspect four of the computers picked at random (without replacement). The system is deemed functional if at least three of the four computers inspected are working. Let the probability that the system is deemed functional be denoted by p. Then 100p = _____________.

22.

Each of the nine words in the sentence “The quick brown fox jumps over the l zy dog” is written on sep r te piece of paper. These nine pieces of paper are kept in a box. One of the pieces is drawn at random from the box. The expected length of the word drawn is _____________. (The answer should be rounded to one decimal place.)

23.

The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is __________.

24.

Let S be a sample space and two mutually exclusive events A and B be such that ∪ S If ( ) denotes the prob bility of the event, the maximum value of P(A) P(B) is _______

ECE - 2006 3. A probability density function is of the ). form (x) e || x ( The value of K is (A) 0.5 (C) 0.5a (B) 1 (D) A 4.

Three Companies X, Y and Z supply computers to a university. The percentage of computers supplied by them and the probability of those being defective are tabulated below Company % of Probability computers of being supplied defective X 60% 0.01 Y 30% 0.02 Z 10% 0.03 Given that a computer is defective, the probability that it was supplied by Y is (A) 0.1 (C) 0.3 (B) 0.2 (D) 0.4

ECE - 2007 5. If E denotes expectation, the variance of a random variable X is given by (A) E[X2] E2[X] (C) E[X2] (B) E[X2] + E2[X] (D) E2[X] 6.

An examination consists of two papers, Paper1 and Paper2. The probability of failing in Paper1 is 0.3 and that in Paper2 is 0.2.Given that a student has failed in Paper2, the probability of failing in paper1 is 0.6. The probability of a student failing in both the papers is (A) 0.5 (C) 0.12 (B) 0.18 (D) 0.06

ECE - 2005 1. A fair dice is rolled twice. The probability that an odd number will follow an even number is

2.

( )

( )

( )

( )

Mathematics

The value of the integral

I

 x2  1  exp   dx is 2 0  8

(A) 1 (B)

(C) 2 (D) th

th

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GATE QUESTION BANK

ECE - 2008 7. The probability density function (PDF) of a random variable X is as shown below.

(x) exp( |x|) exp( |x|) is the probability density function for the real random variable X, over the entire x axis. M and N are both positive real numbers. The equation relating M and N is

8.

PDF PDF

1

Mathematics

(A) 1

0

(B) 2M

x 11

The -1 corresponding cumulative 0 distribution function (CDF) has the form

(A)

1

(C) M + N = 1 (D) M + N = 3 ECE - 2009 9. Consider two independent random variables X and Y with identical distributions. The variables X and Y take value 0, 1 and 2 with probabilities

CDF

1

N=1

x

and respectively. What is the 1

1

0

(B)

x

conditional probability (x y ) |x y| (A) 0 (C) ⁄ ⁄ (B) (D) 1

CD F C

1

D F

1

10. 0

1 -1

(C)

x

1

2

(B)

11. 1

0

x

1

1 1 1

0 0 1

1  2

(C)   2

0

(D)

10

1  2

(A)  

CDF 1

0

1

A fair coin is tossed 10 times. What is the probability that ONLY the first two tosses will yield heads?

CDF

1 1

x

th

10

10

1 C2    2

(D)

10

1 C2    2

A discrete random variable X takes values from 1 to 5 with probabilities as shown in the table. A student calculates the mean of X as 3.5 and her teacher calculates the variance of X as 1.5. Which of the following statements is true? k P(X=k) 1 0.1 2 0.2 3 0.4 4 0.2 5 0.1

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GATE QUESTION BANK

(A) Both the student and the teacher are right (B) Both the student and the teacher are wrong (C) The student is wrong but the teacher is right (D) The student is right but the teacher is wrong ECE - 2010 12. A fair coin is tossed independently four times. The prob bility of the event “the number of times heads show up is more th n the number of times t ils show up” is (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄ ECE - 2011 13. A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is (A) 2/36 (C) 5/12 (B) 2/6 (D) 1/2 ECE\EE\IN - 2012 14. A fair coin is tossed till a head appears for the first time probability that the number of required tosses is odd , is (A) 1/3 (C) 2/3 (B) 1/2 (D) 3/4 ECE - 2013 15. Let U and V be two independent zero mean Gaussian random variables of variances ⁄ and ⁄ respectively. The probability ( V ≥ U) is (A) 4/9 (C) 2/3 (B) 1/2 (D) 5/9 16.

Consider two identically distributed zeromean random variables U and V . Let the cumulative distribution functions of U and 2V be F(x) and G(x) respectively. Then, for all values of x (x)) (A) ( (x) (B) ( (x)

(C) ( (x) (D) ( (x)

Mathematics

(x)) x (x)) x ≥

ECE - 2014 17. In a housing society, half of the families have a single child per family, while the remaining half have two children per family. The probability that a child picked at random, has a sibling is _____ 18.

Let X X nd X , be independent and identically distributed random variables with the uniform distribution on [0, 1]. The probability P{X is the largest} is _____

19.

Let X be a random variable which is uniformly chosen from the set of positive odd numbers less than 100. The expectation, E[X], is __________.

20.

An unbiased coin is tossed an infinite number of times. The probability that the fourth head appears at the tenth toss is (A) 0.067 (C) 0.082 (B) 0.073 (D) 0.091

21.

A fair coin is tossed repeatedly till both head and tail appear at least once. The average number of tosses required is _______.

22.

Let X X and X be independent and identically distributed random variables with the uniform distribution on [0, 1]. The probability P{X X X } is ______.

23.

Let X be a zero mean unit variance Gaussian random variable. ,|x|- is equal to __________

24.

Parcels from sender S to receiver R pass sequentially through two post-offices. Each post-office has a probability

of

losing an incoming parcel, independently of all other parcels. Given that a parcel is lost, the probability that it was lost by the second post-office is ____________.

(x)) ≥ th

th

th


GATE QUESTION BANK

EE - 2005 1. If P and Q are two random events, then the following is TRUE (A) Independence of P and Q implies that probability (P Q) = 0 (B) Probability (P ∪ Q)≥ Probability (P) +Probability (Q) (C) If P and Q are mutually exclusive, then they must be independent (D) Probability (P Q) Probability (P) 2.

A fair coin is tossed three times in succession. If the first toss produces a head, then the probability of getting exactly two heads in three tosses is (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄

EE - 2006 3. Two f ir dice re rolled nd the sum “ r ” of the numbers turned up is considered (A) Pr (r > 6) = (B) Pr (r/3 is an integer) = (C) Pr (r = 8|r/4 is an integer) = (D) Pr (r = 6|r/5 is an integer) = EE - 2007 4. A loaded dice has following probability distribution of occurrences Dice Value Probability ⁄ 1 2

⁄

3

⁄

4

⁄

5

⁄

⁄ 6 If three identical dice as the above are thrown, the probability of occurrence of values, 1, 5 and 6 on the three dice is (A) same as that of occurrence of 3, 4, 5 (B) same as that of occurrence of 1, 2, 5 (C) 1/128 (D) 5/8

Mathematics

EE - 2008 5. X is a uniformly distributed random variable that takes values between 0 and 1. The value of E{X } will be (A) 0 (C) 1/4 (B) 1/8 (D) 1/2 EE - 2009 6. Assume for simplicity that N people, all born in April (a month of 30 days), are collected in a room. Consider the event of atleast two people in the room being born on the same date of the month, even if in different years, e.g. 1980 and 1985. What is the smallest N so that the probability of this event exceeds 0.5? (A) 20 (C) 15 (B) 7 (D) 16 EE - 2010 7. A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is (A) 1/3 (C) 1/2 (B) 3/7 (D) 4/7 ECE\EE\IN - 2012 8. Two independent random variables X and Y are uniformly distributed in the interval , -. The probability that max , - is less than 1/2 is (A) 3/4 (C) 1/4 (B) 9/16 (D) 2/3 EE - 2013 9. A continuous random variable x has a probability density function + is f(x) e x . Then *x (A) 0.368 (C) 0.632 (B) 0.5 (D) 1.0

th

th

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GATE QUESTION BANK

EE - 2014 10. A fair coin is tossed n times. The probability that the difference between the number of heads and tails is (n – 3) is (C) (A) (B) (D) 11.

12.

13.

14.

IN - 2005 1. The probability that there are 53 Sundays in a randomly chosen leap year is (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄ 2.

A mass of 10 kg is measured with an instrument and the readings are normally distributed with respect to the mean of 10 kg. Given that

Consider a dice with the property that the probability of a face with n dots showing up is proportional to n. The probability of the face with three dots showing up is _______________ Let x be a random variable with probability density function for |x| f(x) { |x| for otherwise The probability P(0.5 < x < 5) is_________ Lifetime of an electric bulb is a random variable with density f(x) kx , where x is measured in years. If the minimum and maximum lifetimes of bulb are 1 and 2 years respectively, then the value of k is__________ The mean thickness and variance of silicon steel laminations are 0.2 mm and 0.02 respectively. The varnish insulation is applied on both the sides of the laminations. The mean thickness of one side insulation and its variance are 0.1 mm and 0.01 respectively. If the transformer core is made using 100 such varnish coated laminations, the mean thickness and variance of the core respectively are (A) 30 mm and 0.22 (B) 30 mm and 2.44 (C) 40 mm and 2.44 (D) 40 mm and 0.24

Mathematics

exp .

∫

√

/ d =0.6

and that 60per cent of the readings are found to be within 0.05 kg from the mean, the standard deviation of the data is (A) 0.02 (C) 0.06 (B) 0.04 (D) 0.08 3.

The measurements of a source voltage are 5.9V, 5.7V and 6.1V. The sample standard deviation of the readings is (A) 0.013 (C) 0.115 (B) 0.04 (D) 0.2

IN - 2006 4. You have gone to a cyber-cafe with a friend. You found that the cyber-café has only three terminals. All terminals are unoccupied. You and your friend have to make a random choice of selecting a terminal. What is the probability that both of you will NOT select the same terminal? (A) ⁄ (C) ⁄ (B) ⁄ (D) 1 5.

Probability density function p(x) of a random variable x is as shown below. The value of  is p(x) α

0

th

α

α b

α c

(A)

c

(C)

(B)

c

(D)

th

th

(

)

(

)


GATE QUESTION BANK

6.

Mathematics

Two dices are rolled simultaneously. The probability that the sum of digits on the top surface of the two dices is even is (A) 0.5 (C) 0.167 (B) 0.25 (D) 0.125

measurements, it can be expected that the number of measurement more than 10.15 mm will be (A) 230 (C) 15 (B) 115 (D) 2

IN - 2007 7. Assume that the duration in minutes of a telephone conversation follows the

IN - 2011 12. The box 1 contains chips numbered 3, 6, 9, 12 and 15. The box 2 contains chips numbered 6, 11, 16, 21 and 26. Two chips, one from each box, are drawn at random. The numbers written on these chips are multiplied. The probability for the product to be an even number is (A) ⁄ (C) ⁄ ⁄ (B) ⁄ (D)

exponential distribution f(x) =

e ,x≥

The probability that the conversation will exceed five minutes is (A) e (C) (B) e (D) e IN - 2008 8. Consider a Gaussian distributed random variable with zero mean and standard deviation . The value of its cummulative distribution function at the origin will be (A) 0 (C) 1 (B) 0.5 (D) σ 9.

A random variable is uniformly distributed over the interval 2 to 10. Its variance will be ⁄ ⁄ (A) (C) (B) 6 (D) 36

IN - 2013 13. A continuous random variable X has probability density f(x) = . Then P(X > 1) is (A) 0.368 (C) 0.632 (B) 0.5 (D) 1.0 IN - 2014 14. Given that x is a random variable in the r nge , - with prob bility density function

the value of the constant k is

___________________ IN - 2009 10. A screening test is carried out to detect a certain disease. It is found that 12% of the positive reports and 15% of the negative reports are incorrect. Assuming that the probability of a person getting a positive report is 0.01, the probability that a person tested gets an incorrect report is (A) 0.0027 (C) 0.1497 (B) 0.0173 (D) 0.2100

15.

IN - 2010 11. The diameters of 10000 ball bearings were measured. The mean diameter and standard deviation were found to be 10 mm and 0.05mm respectively. Assuming Gaussian distribution of

The figure shows the schematic of production process with machines A,B and C. An input job needs to be preprocessed either by A or by B before it is fed to C, from which the final finished product comes out. The probabilities of failure of the machines are given as:

Assuming independence of failures of the machines, the probability that a given job is successfully processed (up to third decimal place)is ______________

th

th

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GATE QUESTION BANK

Mathematics


4. [Ans. D] The number of ways coming 8 and 9 are (2,6),(3,5),(4,4),(5,3),(6,2),(3,6),(4,5), (5,4),(6,3) Total ways =9 So Probability of coming 8 and 9 are

[Ans. D]

5.

[Ans. D] X and Y are independent ∴ ( ) ( ) ( ) re true Only (D) is odd one

6.

[Ans. A] Number of favourable cases are given by HHHT HHTH HTHH THHH Total number of cases = 2C1 2C1 2C1 2C1 =16

So probability of not coming these

2.

[Ans. B] Probability of defective item = Probability of not defective item = 1 0.1 = 0.9 So, Probability that exactly 2 of the chosen items are defective = ( ) ( )

3.

[Ans. B]

∴ Probability = 7.

[Ans. A] A uniform function

Mean (t)̅ = ∫ t f(t) dt ∫ t( t 6

t)dt t

t 6

7

[

]

∫ t(

[

t

t)dt

t)dt

=∫ (t =0

t )dt 1

0

7

density

Density function

 1  f(x)   b  a  0

∫ t (

t)dt

a,x  b a  x,x  b

 Mean E(x)=

t)dt

b

x(F(x))  x a

ab 2

 Variance = F(x)2  f(x)

2

1

2

b   x F(x)  xF(x) x a  x a  b

= Standard deviation = √v ri nce =

and

0,x  a x  a  f(x)   f  x dx   , axb 0 b  a  xb 0,

]

∫ t (

distribution

x

Variance = ∫ t f(t)dt =∫ t (

( oth defective) S mple sp ce

( oth defective)

2

Put the value of F(x), we get √

2

1 1 b  dx  x. dx  Variance  x ba x a  x a b  a  b

th

th

2

th


GATE QUESTION BANK b

 x3   xL      3(b  a) a  2 b  a  

Mathematics 3

2

1 7  (3  3  1)    2 8



b3  a3 (b2  a2 )2  3(b  a) 4 b  a 2



(b  a)(b2  ab  a2 ) (b  a)2(b  a)2  2 3(b  a) 4 b  a



4b2  4ab  4a2  3a2  3b2  6ab 12



b2  a2  2ab 12

9.

[Ans. C] Probability of drawing 2 washers, first followed by 3 nuts, and subsequently the 4 bolts

10.

[Ans. D] Required probability =

(b  a)2  12

. / . /

Standard deviation = √v ri nce

(b  a)2 12 (b  a)  12



11.

[Ans. D] Given 4R and 6B , -

12.

[Ans. C]

Given: b=1, a=0

 Standard deviation =

8.

10 1  12 12

[Ans. D] Let probability of getting atleast one head = P(H) then P (at least one head) = 1 P(no head)  P(H)=1 P(all tails) But in all cases, 23=8

1 7 8 8

X=0

 P (H) = 1  

(X ) is Below X (X ) has to be less than 0.5 but greater than zero

Alternately Probability of getting at least one head ( ) ( )

13.

1 7 1   8 8 Alternately From Binomial theorem Probability of getting at least one head pq ( )

( )

X=1

[Ans. D] A event that he knows the correct answer B event that student answered correctly the question P(B) = ? ( )

(

)

( )

th

th

th


GATE QUESTION BANK

( )

he knows correct nswer

(

) (

( ⁄ )

14.

No. of employed men = 80% of men = 80 No. of employed women = 50% of women = 50 Probability if the selected one person being employed = probability of one employed women +probability of one employed man

⏟

⏟

e does not know correct nswer so he guesses

( ) ( ⁄ ) ) ⁄ ( ) ⁄

[Ans. D] x 1 2 P(x) 0.3 0.6 (x) (x) x

18.

V(x) x (

σ

(x )

[Ans. A]

3 0.1 So from figure Mean value = 1 V ri nce : μ me n x defective pieces (x μ) σ ) n(n ( ) ( ) ( ) ( )

(x) x (x) σ

Mathematics

, (x)-

(x) ( x (x)) ) ( )

√ ( )

15.

[Ans. A] 19.

16.

[Ans. *](Range 49 to 51)

[Ans. *] Range 0.25 to 0.27 p orm l distribution

q

Given that μ σ x μ x z σ ere x μ , s x gre ter th n z ) ence prob bility (z

Using Binomial distribution (x ≥ )

17.

( ) ( )

( ) ( )

-

∫ e dz σ√ ∴ of s ving ccount holder

[Ans. *] Range 0.64 to 0.66 Let number of men = 100 Number of women = 100 th

th

th


GATE QUESTION BANK

20.

[Ans. B] Mean m = np = 5.2 me (x ) e

25 Calculators

m

23 Non-defective

2 Defective

e

5 Calculators

e (x

Mathematics

) 4 Non-defective

1 Defective

CE 1.

2.

[Ans D] A, B, C are true (D) is not true. Since in a negatively skewed distribution mode > median > mean [Ans. D] Let the mean and standard deviation of the students of batch C be μ and σ respectively and the mean and standard deviation of entire class of first year students be μ and σ respectively Now given, μ σ and μ σ In order to normalise batch C to entire class, the normalize score must be equated since Z = Z =

=

Now Z =

p( defective in c lcul tors)

4.

[Ans. C] σ μ

5.

[Ans. B] Given f(x) = x for x = 0 else where (

)

∫ f(x)dx

∫ x dx

=0 1 The probability expressed in percentage P= = 2.469% = 2.47% 6.

[Ans. A] Given P(private car) = 0.45 P(bus 1 public transport) = 0.55 Since a person has a choice between private car and public transport P(public transport) = 1 – P(private car) = 1 – 0.45 =0.55 P(bus) = P(bus public transport) (bus public tr nsport) (public tr nsport) = 0.55 × 0.55 = 0.3025 ≃ 0.30 Now P(metro) = 1 [P(private car) + P(bus)] = 1 (0.45 + 0.30) = 0.25

=

Equation these two and solving, we get =  x = 8.969 ≃ 9.0 3.

x

[Ans. B] Since population is finite, hypergeometric distribution is applicable

th

th

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GATE QUESTION BANK

∴ P(private car) = 0.45 P(bus) = 0.30 and P(metro) = 0.25 7.

12.

[Ans. D] ere μ cm; σ ( x 102) =P.

[Ans. 6] ∫ f(x)dx ( x

f(x)

{

∴∫

( x

cm 6

x

Mathematics

)

x

)dx

x

x

x

(

)

x otherwise

x7

/ [

=P( x ) This area is shown below:

[

(

(

)]

]

[

-0.44

)

]

The shades area in above figure is given by F(0) –F ( 0.44) =

( )

(

(

)(

)

= 0.5 – 0.3345 = 1.1655 ≃ 16.55% Closest answer is 16.7% 8.

)

13.

[Ans. 0.4] (

)

∫ f( )d

[Ans. C] ( )|

P(2 heads) = 9.

[Ans. C] P(one ball is Red & another is blue) = P(first is Red and second is Blue)

14.

= 10.

[Ans. A] Given μ = 1000, σ = 200 We know that Z When X= 1200, Z Req. Prob = P (X (Z ) ( Z Less than 50%

11.

∫ d

[Ans. D] (X ) ( )

(X

)

(

[Ans. *] Range 0.26 to 0.27 Avg= 5 Let x denote penalty (x ) (x ) (x ) (x ) (x ) e ew (x n) x e e e ) p(x e

[

)

e

]

)

)

(X

15.

[Ans. B] S * T+ n( ) ( ) n(S)

16.

[Ans. *] Range 0.25 to 0.28 ( t) e (n t) n

)

( )

th

th

th


GATE QUESTION BANK

no of vehicles (

)

veh km

e

.

m ke ex ctly ‘ ’ moves nd ‘U’ moves in any order. Similarly to reach (10, 10) from (0,0) the robot h s to m ke ‘ ’ moves nd ‘U’ moves in any order. The number of ways this can be done is same as number of permutations of a word consisting of 10 ‘ ’ s nd ’U’s Applying formula of permutation with limited repetitions we get the answer as

/

= 2.e = 0.2707 CS 1.

[Ans. A] P: Event of selecting Box P, Q: Event of selecting Box P P(P)=1/3, P(Q)=2/3 P(R/P)=2/5, P(R/Q)=3/4

P(R/P).P(P) P(R/P).P(P)  P(R/Q)P(Q) 2/51/3   4/19 2/51/3  3/ 4  2/3 P(P/R)=

2.

5.

[Ans. D] The robot can reach (4,4) from (0,0) in 8C ways as argued in previous problem. 4 Now after reaching (4,4) robot is not allowed to go to (5,4) Let us count how many paths are there from (0,0) to (10,10) if robot goes from (4,4) to (5,4) and then we can subtract this from total number of ways to get the answer. Now there are 8C4 ways for robot to reach (4,4) from (0,0) and then robot takes the ‘U’ move from ( ) to ( ) ow from (5,4) to (10,10) the robot has to make 5 ‘U’ moves nd ‘ ’ moves in ny order which can be done in 11! ways = 11C5 ways Therefore, the number of ways robot can move from (0,0) (10,10) via (4,4) – (5,4) move is

[Ans. C] If f (x) is the continuous probability density function of a random variable X then, ( x b) P( x b) b

= f  x  dx

 a

3.

4.

[Ans. A] The probability that exactly n elements are chosen =The probability of getting n heads out of 2n tosses =

(

) . /

= =

(

) (

Mathematics

(Binomial formula) )

8C 4

[Ans. A] Consider the following diagram (3,3)

11C 5

 8   11   4 5 

=   

No. of ways robot can move from (0,0) to (10,10) without using (4,4) to (5,4) move is

 20  8   11        ways  10   4   5 

(0,0) The robot can move only right or up as defined in problem. Let us denote right move by ‘ ’ nd up move by ‘U’ ow to reach (3, 3), from (0,0) , the robot has to

which is choice (D)

th

th

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GATE QUESTION BANK

6.

[Ans. D] umber of permut tions with ‘ ’ in the first position =19! Number of permutations with ‘ ’ in the second position = 10 18! (Fill the first space with any of the 10 odd numbers and the 18 spaces after the 2 with 18 of the remaining numbers in 18! ways) umber of permut tions with ‘ ’ in rd position =10 9 17! (Fill the first 2 place with 2 of the 10 odd numbers and then the remaining 17 places with remaining 17 numbers) nd so on until ‘ ’ is in th place. After that it is not possible to satisfy the given condition, since there are only 10 odd numbers v il ble to fill before the ‘ ’ So the desired number of permutations which satisfies the given condition is

8.

[Ans. C] Let C denote computes science study and M denotes maths study. P(C on Monday and C on Wednesday) = P(C on Monday, M on Tuesday and C on Wednesday) + P(C on Monday, C on Tuesday and C on Wednesday) =1 0.6 0.4+ 1 0.4 0.4 = 0.24 + 0.16 = 0.40

9.

[Ans. B] It is given that P (odd) = 0.9 P (even) Now since 𝜮P(x) = 1 ∴ P (odd) + P (even) = 1 0.9 P (even) + P (even) = 1 P(even) =

…

∴ P(2) = P(4) = P(6) = P(even) )

=

Which are clearly not choices (A), (B) or (C) 7.

/ = P (z ≥

.z

/ = P (z ≥

(z

) = P (z ≥

(

)

)

10. _____(i)

)

(

)

(

)

P(f ce )

) (

) (

(0.5263)

= 0.1754 It is given that P(even | face > 3) = 0.75

[Ans. A] Given μ = 1, σ = 4 σ =2 and μ = 1, σ is unknown Given, P(X ) = P (Y ≥ 2 ) Converting into standard normal variates, .z

= 0.5263

Now, it is given that P(any even face) is same i.e. P(2) = P(4) = P(6) Now since, P(even) = P(2) or P(4) or P(6) = P(2) + P(4) + P(6)

… Now the probability of this happening is given by = (

Mathematics

= 0.75

= 0.75 ( )

)

( )

=1

decl red f ulty

f ulty

p

q p

σ =3

not f ulty

decl red not f ulty decl red f ulty

q

q

decl red not f ulty

From above tree (decl red f ulty) th

=0.468

[Ans. A] The tree diagram of probabilities is shown below q

Now since we know that in standard normal distribution P (z ) = P (z ≥ 1) _____(ii) Comparing (i) and (ii) we can say that

=

th

pq th

(

q)(

p)


GATE QUESTION BANK

11.

[Ans. A] If b c … Then, no. of divisors of (x )(y )(z )… iven ∴ o of ivisors of ( )( ( )( )

(

(

which are multiples

13.

14.

15.

[Ans. C] (x ) , (x)V(x) Where V(x) is the variance of x, Since variance is σ and hence never negative, ≥

( t le st one he d)

TT )

So required prob bility

)

∴ equired rob bility 12.

)

( ∪ )

16. No. of divisors of of o of divisors of ( )( )

Mathematics

[Ans. B] Required Probability = P (getting 6 in the first time) + P (getting 1 in the first time and getting 5 or 6 in the second time) + P (getting 2 in the first time and getting 4 or 5 or 6 in the second time) + P (getting 3 in the first time and getting 3 or 4 or 5 or 6 in the second time) ( )

( )

( )

17.

[Ans. C] The p.d.f of the random variable is x +1 P(x) 0.5 0.5 The cumulative distribution function F(x) is the probability upto x as given below x +1 F(x) 0.5 1.0 So correct option is (C)

18.

[Ans. C] e (k)

[Ans. A] + The five cards are * Sample space ordered pairs st nd P (1 card = 2 card + 1) )( )( )( )+ *(

k P is no. of cars per minute travelling.

[Ans. D] 𝛔y = a 𝛔x is the correct expression Since variance of constant is zero.

For no cars. (i.e. k = 0) For no cars. P(0) e So P can be either 0,1,2. (i.e. k = 0,1,2) For k = 1, p(1)=e

[Ans. A] Let A be the event of head in one coin. B be the event of head in second coin. The required probability is * ) ( ∪ )+ ( )| ∪ ) ( ∪ ) ( ) ( ∪ ) ( ) (both coin he ds)

For k = 2 , P(2)= Hence ( ) e e th

( )

( )

e 4

th

e 5

th


GATE QUESTION BANK

e (

(

20.

(

)

∴ equired prob bility is ( ∪ ( )

∪ )

⁄ )

(

( ) (

)

e

[Ans. *] Range 0.24 to 0.27 The smaller sticks, therefore, will range in length from almost 0 meters upto a maximum of 0.5 meters, with each length equally possible. Thus, the average length will be about 0.25 meters, or about a quarter of the stick.

24.

(

)

[Ans. 10] 22 occurs in following ways 6 6 6 4 w ys 6 6 5 5 w ys

[Ans. 0.25] ( ∪ ) P(S) = 1 ( ) ( ) ( ) utu lly exclusive ( ) ( ) ( ) et ( ) x; ( ) x P(A) P(B) = x( x) Maximum value of y = x ( x) dy ( x) x dx = 2x = 1 x

equired prob bility

(max)

x 21.

( )

) e

19.

( )

Mathematics

ximum v lue of y

[Ans. *] Range 11.85 to 11.95 For functioning 3 need to be working (function)

ECE 1.

(

)

[Ans. D]

3 1  6 2 3 1 P(even number )   6 2 Since events are independent, therefore 1 1 1  P(odd/even)    2 2 4 P(Odd number) 

p 22.

[Ans. *] Range 3.8 to 3.9 Expected length = Average length of all words

2.

[Ans. A] I

∫ e

(

√ omp ring with (

23.

σ√ ut μ

[Ans. *] Range 0.259 to 0.261 Let A = divisible by 2, B = divisible by 3 and C = divisible by 5, then n(A) = 50, n(B) = 33, n(C) = 20 n( ) n( ) n( ) n( ) P( ∪ ∪ )

∫

∫ rom x σ σ th

)

th

)

e

dx

σ

√

dx

e

dx

…

nd x

th


)

GATE QUESTION BANK

Put σ ∫

3.

in

Probability of failing in paper 2, P (B) = 0.2 Probability of failing in paper 1, when

equ tion

e

√

A   0.6 B  A  P A  B We know that, P   PB  B

student has failed in paper 2, P

[Ans. C] 

 P x.dx  1





 Ke



Mathematics

(

∴

.dx  1

ax

)

e dx

∫

e

dx

 x x,for x  0      x for x  0 K K   1 a a 

( )

= 0.6 0.2 = 0.12



or ∫

( )

7.

[Ans. A] CDF: F x  

x

 PDF

dx



For x<0, F x  

x

  x  1

dx

1

4.

[Ans. D] . / ( )

P (Y/D) =

. / ( )

. / ( )

= 5.

. / ( )

=0.4

0

1

F0 

1 2

conc ve upw rds

For x>0, F x   F0 

x

 x  1

dx

0

[Ans. A] var[x]=σ =E[(x x)2] Where, x=E[x] x= expected or mean value of X defining

 1  x2    x  concave downwards 2  2  Hence the CDF is, shown in the figure (A).



E[X] =

 xf  xdx x



8.

[Ans. A]

Given: Px  x   Me2|x|  Ne3|x|



    x P xi   x  xi dx    i



 P  xdx  1 x

 xiP xi 

 



i

Variance σ is a measure of the spread of the values of X from its mean x. Using relation , E[X+Y]= E[X]+E[Y] And E[CX]=CE[X] On var[x]= σ =E[(x x)2] σ = ,Xx2 = E[X2] [ ,X-] 6.











Me2|x|  Ne3|x| dx  2 Me2|x|  Ne3|x| dx  1 0

By simplifying

2 3

 M N 1 9.

[Ans. B] x+y=2 x y=0 => x =1, y = 1 P(x=1,y=1) = ¼

[Ans. C] Probability of failing in paper 1, P (A) = 0.3 th

th

¼ = 1/16 th


GATE QUESTION BANK

10.

[Ans. C]

14.

Probability of getting head in first toss = Probability of getting head in second toss =

[Ans. C] P(no. of tosses is odd) (no of tosses is

…)

P(no. of toss is 1) = P(Head in 1st toss)

and in all other 8 tosses there should

P(no. of toss is 3) = P(tail in first toss, tail in second toss and head in third toss)

be tail always. So probability of getting head in first two tosses ( )( )( )…………… ( ) = (1/2)10

P(no. of toss is 5) = P(T,T,T,TH) . /

11.

Mathematics

[Ans. B] Both the teacher and student are wrong Mean = ∑ k = 0.1 + 0.4 + 1.2 + 0.8 + 0.5 = 3.0 E(x2) = ∑ k

… etc

So, P(no. of tosses in odd)

⁄ ⁄ ⁄ ⁄

Variance(x)= E(x2) – * (x)+ =10.2 9=1.2 12.

[Ans. D] P(H, H, H, T) +P (H, H, H, H ) =

13.

. /

. /

15.

[Ans. B] ( V ≥ V) ( V V≥ ) *z v v+ Linear combination of Gaussian random variable is Gaussian ∴ (z ≥ ) and not mean till zero because both random variables has mean zero hence ( ) Hence Option B is correct

16.

[Ans. D] F(x) = P{X x} (x) * X x+ x 2X 3

. / =

[Ans. C] Total number of cases = 36 Favorable cases: (1, 1) (2, 1) (3, 1) (4, 1) (5, 1) (6, 1) (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) (1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3) (1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4) (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5) (1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6) Total number of favorable cases

For positive value of x, (x) (x) is always greater than zero For negative value of x (x) (x)is ve ut , (x) (x)- x ≥

Then probability th

th

th


GATE QUESTION BANK

17.

18.

[Ans. *] Range 0.65 to 0.68 et ‘ ’ be different types of f milies nd ‘S’ be there siblings. S S S (siblings) Probability that child chosen at random having sibling is 2/3

(x)

et S

,

∑x f(x)

[Ans. C]

21.

[Ans. *] Range 2.9 to 3.1 Let the first toss be head. Let x denotes the number of tosses(after getting first head) to get first tail. We can summarize the even as Event(After x Probability getting first H) T 1 1/2 HT 2 1/2 1/2=1/4 HHT 3 1/8 nd so on…

II)gives

(

)S

……

(x) i.e. The expected number of tosses (after first head) to get first tail is 2 and same can be applicable if first toss results in tail. Hence the average number of tosses is

22.

-

20.

(I

(II)

S

∑x …

……

(I)

S

f(x) ∴ (x)

……

S

[Ans. *] Range 0.32 to 0.34 This is a tricky question, here, X X X independent and identically distributed random variables with the uniform distribution , -. So, they are equiprobable. So X X or X have chances being largest are equiprobable. So, [P {X is largest}] or [P {X is largest}] or [P {X is largest}] =1 and P {X is largest} = P {X is largest} = P {X is largest}

[Ans. *] Range 49.9 to 50.1 Set of positive odd number less than 100 is 50. As it is a uniform distribution

∑ x (x) …

∴ *X is l rgest + 19.

Mathematics

[Ans. *] Range 0.15 to 0.18 X X X X X X et z X X X (X ) X X (z ) Pdf of z we need to determine. It is the convolution of three pdf

(z

23.

th

)

∫

z

dz

z

|

[Ans. *] Range 0.79 to 0.81 |x| ,|x|- ∫ e dx √

th

th


GATE QUESTION BANK

√ √ √ √ √

∫ |x| exp 4

x x

∫ x exp 4

∴pr(P ∪ Q) pr(P) + pr(Q) (D) is true since P Q P n(P Q) n(P) pr(P Q) pr(P)

5 dx 5 dx

5 dx 2.

x

∫ x exp 4

x

∫ x exp 4

√ [ exp (

√ , 24.

x

∫ |x| exp 4

-

[Ans. B] P(A|B) =

5 dx 5 dx

x ) dx]

( he d in tosses nd first toss in he d) = P(HHT) + P(HTH)

4/5

Parcel is sent to R

∴ Required probability = R

3.

1/5

S

Parcel is lost Parcel is lost

parcel

is

=

[Ans. C] If two fair dices are rolles the probability distribution of r where r is the sum of the numbers on each die is given by r P(r)

4/5

that

) ( )

Also, P(first toss is head) =

√

Parcel is sent to

Probability

(

∴ P(2 heads in 3 tosses | first toss is head) ( he ds in tosses nd first toss in he d) (first toss is he d)

[Ans. *] Range 0.43 to 0.45 Pre flow diagram is

1/5

Mathematics

lost

2 Probability that parcel is lost by 3 Probability that parcel is lost by provided that the parcel is lost

4 5

EE 1.

6 [Ans. D] (A) is false since of P & Q are independent pr(P Q) = pr(P) pr(Q) which need not be zero. (B) is false since pr(P ∪ Q) = pr(P) + pr(Q) – pr(P Q) (C) is false since independence and mutually exclusion are unrelated properties.

7 8 9 10 11 12 th

th

th


GATE QUESTION BANK

The above table has been obtained by taking all different all different ways of obtaining a particular sum. For example, a sum of 5 can be obtained by (1, 4), (2, 3), (3, 2) and (4, 1). ∴ P(x = 5) = 4/36 Now let us consider choice (A) Pr(r > 6) = Pr(r ≥ 7)

P(1, 5, 6) =

=

P(3, 4, 5) =

=

P(1, 2, 5) =

=

∴ Choice (C) P(1, 5 and 6) = 5.

is correct.

[Ans. C] x is uniformly distributes in [0, 1] ∴ Probability density function

= =

Mathematics

=

f(x) =

∴ Choice (A) i.e. pr(r > 6) = 1/6 is wrong. Consider choice (B) pr(r/3 is an integer) = pr(r = 3) + pr (r = 6) + pr (r = 9) + pr (r = 1)

=

=1

∴ f(x) = 1 0 < x < 1 Now E(x ) = ∫ x f(x)dx ∫x

dx

= =

6.

=

[Ans. B] Let N people in room. So no. of events that at least two people in room born at same date

∴ Choice (B) i.e. pr (r/3) is an integer = 5/6 is wrong. Consider choice (C) Now, pr(r/4 is an integer) = pr(r = 4) + pr (r = 8) + pr (r = 12) = =

≥

N

Solving, we get N = 7

+

7.

[Ans. C] (II is red|I is white) (II is red nd I is white) (I is white) (I is white nd II is red) (I is white)

8.

[Ans. B]

=

pr(r = 8) = ∴ pr(r = 8 | r/4 is an integer) =

…

=

= ∴ Choice (C) is correct. 4.

[Ans. C] Dice value 1 2

Probability

and

is the entrie

3

rectangle The region in which maximum of {x, y} is

4

less than ⁄

is shown below as shaded

region inside this rectangle

5 6

th

th

th


GATE QUESTION BANK y (

)

(

12.

)

Mathematics

[Ans. *] Range 0.35 to 0.45 (

)

x ∫

dx

∫

x|

(

(

)

dx

∫

dx

x|

)

13. p .m x,x y-

∫ f(x)dx

[Ans. *] Range 0.4 to 0.5 ∫ f(x) dx

/

by property

∴ ∫ kx dx k 9.

14.

[Ans. A] (x

) ,e

10.

∫ e dx e -

, e -

e

[Ans. B] Let number of heads = x, ∴ Number of tails n x ∴ ifference x (n x)or (n x n or n x If x n n x n x

n

If n

x

n

IN 1.

x

x

|

[Ans. D]

=52 weeks and 2 days are extra. Out of x)

x

7, (SUN MON) (or) (SAT SUN) are favorable. So, Probability of this event= 2.

or x

[Ans. C] Since the reading taken by the instrument is normally distributed, hence P(x

x )

Where, [Ans. *] Range 0.13 to 0.15 Let proportionality constant = k ∴ ( dot) k ( dots) k ( dots) k ( dots) k ( dots) k ( dots) k ∴k k k k k k k ∴k ∴ rob bility of showing dots

∴k

[Ans. D] Since leap chosen will be random, so, we assume it being the case of uniform probability distribution function. Number of days in a leap year=366 days

As x and n are integers, this is not possible ∴ Probability 0 11.

k

Now

√

√

∫ e

(

)

.dx

μ e n of the distribution σ St nd rd devi tion of the distribution. ∫

exp(

)dx

where, n=x 10 (∴μ kg) and from the data given in question √

∫

e

dx

On equating, we get 0.05=0.84 σ k

σ

th

th

th


GATE QUESTION BANK

3.

[Ans. D] Mean=

8.

[Ans. B] By definition of Gaussian distribution, total area under the curve =1. Hence half of the area =0.5

9.

[Ans. A]

=5.9 V. (

√

S

̅)

(

̅)

(

̅)

V (closest answer is 0.2)

P(x)= 4.

[Ans. C] ( )

=

Mean = μ ( )

∫ x (x)dx = ∫

Var(x)= ∫ (x

1 2  3 3 5.

Mathematics

=∫

[Ans. A] ]

(x

(x)dx

μ)

) dx =

10.

[Ans. C] Probability of incorrect report

11.

[Ans. C] σ mm μ mm Then probability



 P(x)dx  1

x dx = 6



 Area under triangle =

c 1 2

α 6.

[Ans. A] Probability that the sum of digits of two dices is even is same either both dices shows even numbers or odd numbers on the top of the surface ( ) ∴ ( ) ( ) Where ( ) Probability of occurring even number of both the dices ( ) Probability of occurring odd number of both the dices (

(X where x

(

X

μ

σ mm

(

)

)

( )

e

√ e

√

So, number of measurement more than 10.15mm P Total number of measurement

)

nd (

)

) ≃

∴ ( ) 12. 7.

[Ans. A] ∫ f(x) dx=P or ∫

e

or e

|

.dx =P

[Ans. D] For the product to be even, the numbers from both the boxes should not turn out to be odd simultaneously. ∴ ( )

( )( )

or P = .

th

th

th


GATE QUESTION BANK

13.

[Ans. A] ∫ f(x)dx e |

14.

15.

∫ e dx

e

[Ans. 2] For valid pdf ∫ ∫

Mathematics

dx

pdf dx

;

;k

[Ans. *] Range 0.890 to 0.899 Probability that job is successfully processed = ( )( )

th

th

th


GATE QUESTION BANK

Mathematics

Numerical Methods ME – 2005 1. Starting from x = 1, one step of Newton – Raphson method in solving the equation x³ +3x 7=0 gives the next value (x₁) as (A) x₁=0.5 (C) x₁ = .5 (B) x₁= . 0 (D) x₁=2 2.

With a 1 unit change in b, what is the change in x in the solution of the system of equation = 2 .0 0. = (A) Zero (C) 50 units (B) 2 units (D) 100 units

ME – 2006 3. Match the items in columns I and II. Column I Column II (P) Gauss-Seidel (1) Interpolation method (Q) Forward (2) Non-linear Newton-Gauss differential method equations (R) Runge-Kutta (3) Numerical method integration (S) Trapezoidal (4) Linear algebraic Rule equation (A) 2 (B) 2 (C) 2 (D) 2 4.

Equation of the line normal to function ) f(x) = (x (A) y = x 5 (B) y = x 5

at (0 5) is (C) y = x (D) y = x

5 5

ME – 2007 5. A calculator has accuracy up to 8 digits 2

after decimal place. The value of

 sinxdx 0

when evaluated using this calculator by trapezoidal method with 8 equal intervals, to 5 significant digits is (A) 0.00000 (C) 0.00500 (B) 1.0000 (D) 0.00025

ME – 2010 6. Torque exerted on a flywheel over a cycle is listed in the table. Flywheel energy (in J per unit cycle) using impson’s rule is Angle (degree) Torque (N-m) 0 0 60 1066 120 323 180 0 240 323 300 55 360 0 (A) 542 (C) 1444 (B) 992.7 (D) 1986 ME – 2011 7.

The integral ∫

dx, when evaluated by

using impson’s / rule on two equal subintervals each of length 1, equals (A) 1.000 (C) 1.111 (B) 1.098 (D) 1.120 ME – 2013 8. Match the correct pairs. Numerical Order of Fitting Integration Scheme Polynomial . impson’s / 1. First Rule Q. Trapezoidal Rule 2. Second . impson’s / 3. Third Rule (A) P – 2 , Q – 1, R – 3 (B) P – 3, Q – 2 , R – 1 (C) P – 1, Q – 2 , R – 3 (D) P – 3, Q – 1 , R – 2 ME – 2014 9. Using the trapezoidal rule, and dividing the interval of integration into three equal sub intervals, the definite integral ∫ |x|dx is ____________

th

th

th


GATE QUESTION BANK

10.

The value of ∫ .

( )

“value approximate estimate?

calculated using

the Trapezoidal rule with five sub intervals is _______ 11.

The definite integral ∫

12.

The real root of the equation 5x 2cosx = 0 (up to two decimal accuracy) is _______

13.

Consider

an

equation

= t

.If x =x at t = 0 , the

CE – 2005 Linked Answer Question 1 and 2 Give a>0, we wish to calculate its reciprocal value 1/a by using Newton Raphson Method for f(x) = 0.

2.

The Newton Raphson algorithm for the function will be (A) x

= (x

)

(B) x

= (x

x )

(C) x

= 2x

ax

(D) x

=x

x

in

the

(C) – (D)

CE – 2007 4. The following equation needs to be numerically solved using the NewtonRaphson method x3 + 4x – 9 = 0 the iterative equation for the purpose is (k indicates the iteration level)

differential

increment in x calculated using RungeKutta fourth order multi-step method with a step size of Δt = 0.2 is (A) 0.22 (C) 0.66 (B) 0.44 (D) 0.88

1.

(A) – (B) 0

value”)

is evaluated

using Trapezoidal rule with a step size of 1. The correct answer is _______

ordinary

Mathematics

For a = 7 and starting with x = 0.2 the first two iteration will be (A) 0.11, 0.1299 (C) 0.12, 0.1416 (B) 0.12, 0.1392 (D) 0.13, 0.1428

CE – 2006 3. A 2nd degree polynomial f(x) has values of 1, 4 and 15 at x = 0, 1 and 2 respectively.

5.

(A) x

=

(B) x

=

(C) x

=x

(D) x

=

x

Given that one root of the equation x 10x + 31x – 30 = 0 is 5, the other two roots are (A) 2 and (C) and (B) 2 and (D) 2 and

CE – 2008 6. Three values of x and y are to be fitted in a straight line in the form y = a + bx by the method of least squares. Given ∑x = 6, ∑y = 2 ∑x = and ∑xy = the values of a and b are respectively (A) 2 and 3 (C) 2 and 1 (B) 1 and 2 (D) 3 and 2 CE – 2009 7. In the solution of the following set of linear equation by Gauss elimination using partial pivoting 5x + y + 2z = 34; 4y – 3z = 12; and 10x – 2y + z = 4; the pivots for elimination of x and y are (A) 10 and 4 (C) 5 and 4 (B) 10 and 2 (D) 5 and 4

The integral ∫ f(x) dx is to be estimated by applying the trapezoidal rule to this data. What is the error (define as true th

th

th


GATE QUESTION BANK

CE – 2010 8. The table below given values of a function F(x) obtained for values of x at intervals of 0.25. x 0 0.25 0.5 0.75 1.0 F(x) 1 0.9412 0.8 0.64 0.50 The value of the integral of the function between the limits 0 to using impson’s rule is (A) 0.7854 (C) 3.1416 (B) 2.3562 (D) 7.5000 CE 9.

2011 The square root of a number N is to be obtained by applying the Newton Raphson iterations to the equation x = 0. If i denotes the iteration index, the correct iteration scheme will be (A) x (B) x

= (x

)

= (x

CE – 2013 12. Find the magnitude of the error (correct to two decimal places) in the estimation of following integral using impson’s ⁄ Rule. Take the step length as 1.___________ ∫ (x

1.

Consider

)

(D) x

= (x

)

he error in

 xe dx

=

)

(

)

is 2

1 R xn1   xn   can be used to compute 2 xn 

.

the (A) square of R (B) reciprocal of R (C) square root of R (D) logarithm of R

0 .

CE – 2012 The estimate of ∫ .

1 3

to an accuracy of at least 106

The Newton-Raphson iteration

for a continuous

The values of and ( ) are 19.78 and 500.01, respectively. The corresponding error in the central difference estimate for h = 0.02 is approximately (A) . 0 (C) .5 0 (B) .0 0 (D) .0 0

11.

x

1

function estimated with h=0.03 using the central difference formula f(x)|

=

using the trapezoidal rule is (A) 1000e (C) 100e (B) 1000 (D) 100

f(x)|

(

series

CS – 2008 2. The minimum number of equal length subintervals needed to approximate

3. 10.

the

= 0.5 obtained from the NewtonRaphson method. The series converges to (A) 1.5 (C) 1.6 (D) 1.4 (B) √2

)

= (x

0) dx

CS – 2007

2

(C) x

Mathematics

obtained using

impson’s rule with three – point function evaluation exceeds the exact value by (A) 0.235 (C) 0.024 (B) 0.068 (D) 0.012

CS – 2010 4. Newton-Raphson method is used to compute a root of the equation x 13 = 0 with 3.5 as the initial value. The approximation after one iteration is (A) 3.575 (C) 3.667 (B) 3.677 (D) 3.607

th

th

th


GATE QUESTION BANK

CS – 2012 5. The bisection method is applied to compute a zero of the function f(x) = x x x in the interval [1,9]. The method converges to a solution after ___________ iterations. (A) 1 (C) 5 (B) 3 (D) 7 CS – 2013 6. Function f is known at the following points: x f(x) 0 0 0.3 0.09 0.6 0.36 0.9 0.81 1.2 1.44 1.5 2.25 1.8 3.24 2.1 4.41 2.4 5.76 2.7 7.29 3.0 9.00 he value of ∫ f(x)dx computed using the trapezpidal rule is (A) 8.983 (C) 9.017 (B) 9.003 (D) 9.045 CS – 2014 7. The function f(x) = x sin x satisfied the following equation: ( ) + f(x) + t cos x = 0. The value of t is _________. 8.

In the Newton-Raphson method, an initial guess of = 2 made and the sequence x x x .. is obtained for the function 0.75x 2x 2x =0 Consider the statements (I) x = 0. (II) The method converges to a solution in a finite number of iterations. Which of the following is TRUE? (A) Only I

Mathematics

(B) Only II (C) Both I and II (D) Neither I nor II 9.

With respect to the numerical evaluation of the definite integral,

= ∫ x dx where a and b are given, which of the following statements is/are TRUE? (I) The value of K obtained using the trapezoidal rule is always greater then or equal to the exact value of the defined integral (II) The value of K obtained using the impson’s rule is always equal to the exact value of the definite integral (A) I only (B) II only (C) Both I and II (D) Neither I nor II ECE– 2005 1. Match the following and choose the correct combination Group I Group II (A) Newton1. Solving nonRaphson linear equations method (B) Runge-Kutta 2. Solving linear method simultaneous equations (C) impson’s 3. Solving ordinary Rule differential equations (D) Gauss 4. Numerical elimination integration method 5. Interpolation 6. Calculation of Eigen values (A) A-6, B-1, C-5, D-3 (B) A-1, B-6, C-4, D-3 (C) A-1, B-3, C-4, D-2 (D) A-5, B-3, C-4, D-1

th

th

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GATE QUESTION BANK

ECE– 2007 2. The equation x3 x2+4x 4=0 is to be solved using the Newton-Raphson method. If x=2 is taken as the initial approximation of the solution, then the next approximation using this method will be (A) 2/3 (C) 1 (B) 4/3 (D) 3/2

(A) 2

x

x

.

sin x ..

2 cos x

x

.

..

(C) 2

x

x

.

..

(D) 2

x

x

.

..

8.

The series ∑

eXn 1  eXn X2  eXn 1  Xn   1 (D) Xn1  n Xn -eXn

ECE– 2014 6. The Taylor expansion of is

x

Match the application to appropriate numerical method. Application Numerical Method P1:Numerical M1:Newtonintegration Raphson Method P2:Solution to a M2:Runge-Kutta transcendental Method equation P3:Solution to a M : impson’s system of linear 1/3-rule equations P4:Solution to a M4:Gauss differential equation Elimination Method (A) P1—M3, P2—M2, P3—M4, P4—M1 (B) P1—M3, P2—M1, P3—M4, P4—M2 (C) P1—M4, P2—M1, P3—M3, P4—M2 (D) P1—M2, P2—M1, P3—M3, P4—M4

(C) Xn1  1  Xn 

ECE– 2013 5. A polynomial f(x) = a x a x a x a x a with all coefficients positive has (A) No real root (B) No negative real root (C) Odd number of real roots (D) At least one positive and one negative real root

(B) 2

7.

ECE– 2008 3. The recursion relation to solve x= using Newton-Raphson method is (A) =e (B) = e

ECE– 2011 4. A numerical solution of the equation f(x) = x √x = 0 can be obtained using Newton – Raphson method. If the starting value is x = 2 for the iteration, the value of x that is to be used in the next step is (A) 0.306 (C) 1.694 (B) 0.739 (D) 2.306

Mathematics

converges to

(A) 2 ln 2 (B) √2

(C) 2 (D) e

EE– 2007 1.

The differential equation

=

is

discretised using Euler’s numerical integration method with a time step T > 0. What is the maximum permissible value of T to ensure stability of the solution of the corresponding discrete time equation? (A) 1 (C) (B) /2 (D) 2 EE– 2008 2. Equation e = 0 is required to be solved using ewton’s method with an initial guess x = . Then, after one

th

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GATE QUESTION BANK

step of ewton’s method estimate x of the solution will be given by (A) 0.71828 (C) 0.20587 (B) 0.36784 (D) 0.00000 3.

A differential equation dx/dt = e u(t) has to be solved using trapezoidal rule of integration with a step size h = 0.01 sec. Function u(t) indicates a unit step function. If x(0)= 0, then value of x at t = 0.01 s will be given by (A) 0.00099 (C) 0.0099 (B) 0.00495 (D) 0.0198

EE– 2009 4. Let x 7 = 0. The iterative steps for the solution using Newton – aphson’s method is given by (A) x

= (x

(B) x

=x

(C) x

=x

(D) x

=x

)

EE– 2013 7. When the Newton – Raphson method is applied to solve the equation f(x) = x 2x = 0 the solution at the end of the first iteration with the initial guess value as x = .2 is (A) 0.82 (C) 0.705 (B) 0.49 (D) 1.69 EE– 2014 8. The function ( ) = is to be solved using Newton-Raphson method. If the initial value of is taken as 1.0, then the absolute error observed at 2nd iteration is ___________ IN– 2006 1. For k = 0 2 . the steps of Newton-Raphson method for solving a non-linear equation is given as

2 5 xk 1  xk  xK2 . 3 3 (x

Starting from a suitable initial choice as k tends to , the iterate tends to (A) 1.7099 (C) 3.1251 (B) 2.2361 (D) 5.0000

)

EE– 2011 5. Solution of the variables and for the following equations is to be obtained by employing the Newton-Raphson iterative method equation(i) 0x inx 0. = 0 equation(ii) 0x 0x cosx 0. = 0 Assuming the initial values = 0.0 and = .0 the jacobian matrix is 0 0. 0 0. (A) * (C) * + + 0 0. 0 0. 0 0 0 0 (B) * (D) * + + 0 0 0 0 6.

Mathematics

IN– 2007 2. Identify the Newton-Raphson iteration scheme for finding the square root of 2.

3.

(A) x

=

(x

)

(B) x

= (x

)

(C) x

=

(D) x

= √2

x

The polynomial p(x) = x + x + 2 has (A) all real roots (B) 3 real and 2 complex roots (C) 1 real and 4 complex roots (D) all complex roots

Roots of the algebraic equation x x x = 0 are ) (A) ( (C) (0 0 0) (B) ( j j) (D) ( j j)

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GATE QUESTION BANK

IN– 2008 4. It is known that two roots of the nonlinear equation x3 – 6x2 +11x 6 = 0 are 1 and 3. The third root will be (A) j (C) 2 (B) j (D) 4

IN– 2013 8. While numerically solving the differential equation


=

with

x(0) = 0 and the constant 0 is to be numerically integrated using the forward Euler method with a constant integration time step T. The maximum value of T such that the numerical solution of x converges is (C) (A) (B)

2xy = 0 y(0) =

using

Euler’s predictor – corrector (improved Euler – Cauchy )method with a step size of 0.2, the value of y after the first step is (A) 1.00 (C) 0.97 (B) 1.03 (D) 0.96

IN – 2009 5.

Mathematics

IN– 2014 9. The iteration step in order to solve for the cube roots of a given number N using the Newton- aphson’s method is

(D) 2

(A) x

=x

(B) x

= (2x

(C) x

=x

(D) x

= (2x

(

x ) )

(

x ) )

IN– 2010 6. The velocity v (in m/s) of a moving mass, starting from rest, is given as

=v

t.

Using Euler’s forward difference method (also known as Cauchy-Euler method) with a step size of 0.1s, the velocity at 0.2s evaluates to (A) 0.01 m/s (C) 0.2 m/s (B) 0.1 m/s (D) 1 m/s IN– 2011 7. The extremum (minimum or maximum) point of a function f(x) is to be determined by solving

( )

= 0 using the

Newton-Raphson method. Let f(x) = x x and x = 1 be the initial guess of x. The value of x after two iterations (x ) is (A) 0.0141 (C) 1.4167 (B) 1.4142 (D) 1.5000

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GATE QUESTION BANK

Mathematics


y = sin ( ) = 2

[Ans. C] By N-R method ,

=x –

f(x) = x  f( ) =

x

x =x

(

)

(

)

( ) ( )

y = sin (

x =

y = sin( ) = 0 5 y = sin ( ) =

7

y = sin (

 f (x) = x f ( )= , 

=1

(

)

) = 0.70 0

)=

7 y = sin ( ) =

[Ans. C] Given x y = 2 (i) .0 x 0.0 y = b (ii) Multiply 0.99 is equation (i) and subtract from equation (ii); we get ( .0 0. )x = b (2 0. ) 0.02x = b . 0.02Δx = Δb Δx =

0.02

[Ans. D]

4.

[Ans. B] Given f(x) = (x 2 ) f (x) = (x

f(x)dx = [(y

∫ y ∫

0.70 0

6.

y )

[(0

0)

0.70 0

0.70 0

[(0

y )]

7.

[Ans. C] x y= ∫

( 0

0)

2( 2.7 /unit cycle.

=

Slope of normal = 3 (∵ roduct of slopes = 1) Slope of normal at point (0, 5) y 5 = (x 0) y= x 5 [Ans. A] b a 2 0 h= = = n y = sin(0) = 0

0=0

[Ans. B] ower = ω = Area under the curve. h (y = [(y y ) y y )

/

Slope of tangent at point (0, 5) 2 ) / = m = (0

y

2(0.70 0

2(y

)

2(y

)]

sinx dx =

=

5.

)=0

Trapezoidal rule

= 50 units

3.

0.70 0

(0.5) = .5

=

y = sin ( 2.

0.70 0

1 1

x dx = (y

= (

55)

2 )]

3

2 h

x

2

2

0

y 2

y )

)

= . 8.

[Ans. D] By the definition only

y = sin ( ) = 0.70 0

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GATE QUESTION BANK

9.

[Ans. *] Range 1.10 to 1.12

By intermediate value theorem roots lie be between 0 and 1. et x = rad = 57. 2 By Newton Raphson method f(x ) x =x f (x ) 2x sin x 2 cos x x = 5 2 sin x x = 0.5 2 x = 0.5 25 x = 0.5 2

∫ |x|dx is h ∫ ydx = [y 2

2(y

y

y

x

1

y

1

0.33

2

y ]

y

0.33

2

∫ |x|dx =

.)

y

[

y

0.33

1

0.333

1

2(0.

0.

)]

13.

= . 0 10.

[Ans. *] Range 1.74 to 1.76 2.5 h= = 0. 5 y 2y 2y ∫ . ln (x)dx = [ 2y y =

.

[ln(2.5)

2ln( . ) = .75 11.

2(ln2. )

.

2y

]

2 ln( . )

CE 1.

[Ans. *]Range 1.1 to 1.2

t|

Δx = 0.0

0. = 0.

= 2t

t|

Set up the equation as x = i.e. = a

∫ f(x)dx = [y

y

2(y

iven in question 0 1 1 2 1 0.5

∫ dx = [y x 2

.

.

t Δx = 2

To calculate using N-R method

rapezoidal rule

x y

)dt

[Ans. C]

∫ dx by trapezoidal rule x

h=

[Ans. D] The variation in options are much, so it can be solved by integrating directly dx = t dt ∫ dx = ∫ ( t

ln( )]

2ln( .7)

Mathematics

y

y

..y

)]

a=0 i.e. f(x) =

2 3 0.33

a=0

Now f (x) = f(x ) =

a

f (x ) =

2(y )]

For N-R method = [ 2 = .

0.

2

0.5]

x

=x x

12.

[Ans. *] Range 0.53 to 0.56 Let f(x) = 5x 2 cos x f (x) = 5 2 sin x f(0) = f( ) = 2.

=x

(

)

(

) (

)

Simplifying which we get x = 2x ax

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GATE QUESTION BANK

2.

[Ans. B] For a = 7 iteration equation Becomes x = 2x 7x with x = 0.2 x = 2x 7x = 2× 0.2 – 7(0.2) = 0.12 and x = 2x 7x = 2× 0.12 7(0. 2) = 0.1392

x

[Ans. A] f(x) = 1, 4, 15 at x = 0, 1 and 2 respectively ∫ f(x)dx = (f

2f

=x x

=

5.

f )

∫ f(x)dx = (1 + 2(4) + 15) = 12

and α β

4.

αβ

(

βγ =

)

= 30

βγ

γα = 5β

βγ

5γ =

=

 5 (β γ) βγ = ince βγ = from (i) 5 (β γ) = β γ=5 βγ = olving for β and γ β (5 β) = β 5β =0  β = 2 and γ = Alternative method 5 1 0 31 0 0 5 25 30 1 5 6 0 (x 5)(x )=0 5x (x 5)(x 2)(x )=0 x=2 5

x )dx + =

2=

[Ans. A ] Given f(x) = x x =0 f (x) = x Newton – Raphson formula is

γα =

 βγ = (i) Also

Error = exact – Approximate value =

βγ

α βγ = 5

Now exact value ∫ f(x)dx

= *x

2x x

αβγ=

Approximate value by rapezoidal ule = 12 Since f(x) is second degree polynomial, let f(x) = a0 + a x + a x f(0) = 1 a 0 0= a = f(1) = 4 a a a =  1+ a a = a a =  f(2) = 15 a 2a a = 5  2a a = 5  2a a = Solving (i) and (ii) a = and a = f(x) = 1 – x + 4 x x

=

[Ans. A] Given x – 10 x + 31x 30 = 0 One root = 5 Let the roots be α β and γ of equation ax + bx + cx + d = 0

(3 points Trapezoidal Rule) Here h = 1

=∫ (

f(x ) f (x ) (x x ) ( x ) x x x ( x )

=x

x 3.

Mathematics

6.

[Ans. D] Y = a + bx Given n= ∑x = and ∑xy = th

th

∑y = 2 ∑x = 14

th


GATE QUESTION BANK

Normal equations are ∑y = na b∑x ∑xy = a∑x b∑x Substitute the values and simply a= b=2 7.

9.

5 [0 0

2

(

)

| 2] → 2

0 [0 5

2

=

f(x ) f(x )

=x

=x

(

x

)

2x

x

=

2x

2

[x

x

]

10.

[Ans. D] Error in central difference formula is (h) This means, error If error for h = 0.03 is 2 0 then Error for h = 0.02 is approximately (0.02) 2 0 0 (0.0 )

11.

[Ans. D] Exact value of ∫ .

| 2] 2

So the pivot for eliminating x is a = 10 Now we eliminate x using this pivot as follows : 0 2 [0 | 2] 5 2 5 0 2 0 2] → [0 0 2 /2 Now to eliminate y, we need to compare the elements in second column at and below the diagonal element Since a = 4 is already larger in absolute value compares to a = 2 The pivot element for eliminating y is a = 4 itself. The pivots for eliminating x and y are respectively 10 and 4 8.

[Ans. A] x

[Ans. A] The equation is 5x + y + 2z = 34 0x + 4y – 3z = 12 and 10x – 2y + z = The augmented matrix for gauss elimination is 5 2 [0 | 2] 0 2 Since in the first column maximum element in absolute value is 10 we need to exchange row 1 with row 3

Mathematics

.

dx = .0

Using impson’s rule in three – point form, b a .5 0.5 h= = = 0.5 n 2 So, x 0.5 1 1.5 y 2 1 0.67 ∫

= =

[

0.5

] [2

0. 7

]

= . So, the estimate exceeds the exact value by Approximate value – Exact value = 1.1116 1.0986 =0.012(approximately)

12.

[Ans. *](Range 0.52 to 0.55) Using impson’s ule X 0 1 2 3 Y 10 11 26 91

4 266

[Ans. A] I = h(f = 0. = 0.7 5

f

2f

f

0.25(

0.

0.

0.5)

∫ (x

f ) 2

= [( 0

2

0)dx 2

)

2(2 )

(

)]

= 2 5. The value of integral th

th

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GATE QUESTION BANK

∫ (x

0) dx = *

=

0

=2

x 5

=

0x+ 3.

.

x

=

[Ans. A] +

, x = 0.5

)

(α

)

2α =

=x =α

2α = α + R α =R α=√ So this iteration will compute the square root of R

α= + α= 8α = 4α +9 α = 4.

α = = 1.5 [Ans. A] Here, the function being integrated is f(x) = xe f (x) = xe + e = e (x + 1) f’ (x) = xe + e + e = e (x + 2) Since, both are increasing functions of x, maximum value of f ( ) in interval 1 2, occurs at = 2 so ( )| (2 max |f =e 2) = e Truncation Error for trapezoidal rule = TE (bound)

[Ans. D] y=x dy = 2x dx f(x)= x x

= .5

5.

=

(b – a) max |f ( )| 1

=

(2 – 1) [e (2 + 2)]

=

e

Now putting

(

)

=

= 57 7

)=5 f(x ) 2 oot lies between and

x =(

0

0

)=2 f(x ) 0 2 After ' ' interations we get the root

x =(

= max |f ( )|

.

[Ans. B] f( ) = 5 f( ) = 5 72 ) ) f( 0 f( 0

is number of subintervals

=

.

= . 07

max |f ( )|

Where

(x

2α=α+ =

when the series converges x

=

= 1000 e

At convergence x =x =α α=

Given x

2.

)/

[Ans. C]

5 Magnitude of error = 2 5. 2 . = 0.5 CS 1.

(

Mathematics

2

6.

[Ans. D] h ∫ f(x)dx = [f(0) 2 =

0

=

=

0 [

. .

[

0.

f( )

2(0.0 0. . . 7.2 )

2(∑f )] ]

5 . ]

= 9.045

h= Now, No. of intervals,

= th

th

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GATE QUESTION BANK

7.

8.

9.

ECE 1.

2.

[Ans. – ] Given (x) + f(x) + t cos x = 0 and f(x) = x sin x f (x) = x cos x + sin x f (x) = x ( sin x) + cos x + cos x = 2 cos x – x sin x = 2 cos x – f (x) 2 cos x – f (x) + f(x) +t cos x = 0 2 cos x = tcos x t = 2 [Ans. A] f(x) = 0.75x 2x 2x f (x) = 2.25x x 2 x =2 f = 2 f = f x =x =0 f f = f = 2 f x =x =2 f f = 2 f = f x =x =0 f Also, root does not lies between 0 and 1 So, the method diverges if x = 2 nly ( )is true.

x1  2  3.

 1  x n  4.

x1  x0 

e e

e xn 1  exn

[Ans. C] x

f(x ) f (x )

=x

f(2) = (2

x

) = √2

√2

f (2) =

√

=2

and

√

=

√ )

(√ √

= .

√

5.

[Ans. D] f(x) = a x a x a x a x a If the above equation have complex roots, then they must be in complex conjugate pair, because it’s given all co-efficients are positive ( they are real ) So if complex roots are even no. (in pair) then real roots will also be even. ption ( )is wrong From the equation ( 0) roduct of roots = As no. of roots = 4, Product of roots < 1 either one root 0 (or) Product of three roots < 0 ption ( )is rong. Now, take option (A), Let us take it is correct . Roots are in complex conjugate pairs = Product of roots 0 | | | | 0 which is not possible ption (A) is wrong orrect answer is option ( )

[Ans. C] By definition (& the application) of various methods

4=0

Next approximation x1  x0 

8 4  12 3

[Ans. C] Given : f(x)= x e By Newton Raphson method, f(x ) x x =x =x f (x )

[Ans. C] For value of K if trapezoidal rule is used then the value is either greater than actual value of definite integral and if impson’s rule is used then value is exact Hence both statements are TRUE

[Ans. B] y(t) =x3 x2 + 4x x0 = 2

Mathematics

f  x0  f ' x0 

x03  x02  4x0  4 3x02  2x0  4 th

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GATE QUESTION BANK

6.

[Ans. A] sin x 2 cos x x = (x – )

7.

[Ans. B]

8.

[Ans. D] ∑

n

2( –

x 2

.

x

)

Put x = as given, x = [e ( 2) ]/e = 0.71828

)

[Ans. C]

=e

.. = e

u(t)

x

x 2

.

. . x in t

[Ans. D] Here,

x = ∫ e u(t) dt = ∫ f(t) dt At t = 0.01, x = Area of trapezoidal

= x

f(x y) =

x

h

=(

)x h

4.

x

5.

=[

=*

( )

f(x ) = e f’(x ) = e

6.

(

]

)

(

)

+

0x cos x 0x sinx

20x

0sinx ] 0cosx

0 0

is

0 + 0

[Ans. D] x x x =0 (x )(x )=0 x =0 x =0 x= x= j

=x –(

(

The matrix at x = 0 x =

( )

=

.

[Ans. B] u(x x ) = 0x sin x 0. = 0 v(x x ) = 0x 0x cosx 0. = 0 The Jacobian matrix is u u x x v v [ x x ]

[Ans. A] Here f(x) = e f (x) = e The Newton Raphson iterative equation is

=

=x

= *x

|

Δ 2 o maximum permissible value of Δ is 2 .

i.e. x

e

[Ans. A]

since h = Δ here Δ

x

[

= x

h

=x

.

= 0.0099

h

or stability |

x

f(0.0 )] =

= [f(0)

Euler’s method equation is x = x h. f(x y ) x x = x h( )

2.

i=0

(

x =

2

as e = EE 1.

Now put

3.

=

Mathematics

)

)

th

th

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GATE QUESTION BANK

7.

[Ans. C] x =x =

.2

Hence, it will have atleast 5 (0+1)= 4 complex roots.

f(x ) f (x ) ( .2) 2( .2) ( .2) 2

4.

[Ans. C] Approach- 1 Given, x3 – 6x2 + 11x – 6 = 0 Or (x – 1)(x – 3)(x – 2) = 0 x= 1, 2, 3.

= 0.705 8.

[Ans. *] Range 0.05 to 0.07 Clearly, x = 0 is root of the equation f(x) = e =0 f (x) = e and x = .0 Using ewton raphson method (e . ) f(x ) x =x = = f (x ) e. e and x = x

f(x ) = f (x ) e =

(e

Approach- 2 For ax3 +bx2 + cx +d = 0 If the three roots are p,q,r then Sum of the roots= p+q+r= b/a Product of the roots= pqr= d/a pq+qr+rp=c/a

) e

5.

[Ans. D] dx x = dt f(x, y) =

e

e = 0. 7 0. = 0.0 Absolute error at 2nd itteration is |0 0.0 | = 0.0 IN 1.

x

=x h

=( [Ans. A] As k ∞ xk+1 ≈xk xk = x

h (x y ) = x )x

2.

3.

h

h(

x

)

)

h

|

h

x

/

(

or stability |

Δ

x = x x =5 x =5

Mathematics

Δ

= 1.70

[Ans. A] Assume x = √ f(x) = x =0 f(x ) x =x = [x f (x ) 2

6.

[Ans. A] dv =v t dt t v dv =v t dt 0 0 0 0+0 0. = 0 0.1 0 0+0.1 0. = 0.0

7.

[Ans. C] f(x) = x x f (x) = x = g(x) x = initial guess g (x) = x g (x ) x =x g (x )

2 ] x

[Ans. C] Given p(x) = x + x + 2 There is no sign change, hence at most 0 positive root ( rom escarte’s rule of signs) p( x) = x x+2 There is one sign change, hence at most 1 negative root ( rom escarte’s rule of signs)

2

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GATE QUESTION BANK

(

= x =x = .5 = .

)

= .5

g(x ) g (x ) 0.75 7

8.

[Ans. D] dy = 2xy x = 0 y = h = 0.2 dx y =y h. f(x y ) (0.2)f(0 ) = = and y = y [f(x y ) f(x y )] (0. )[f(0 ) f(0.2 )] = = 0. is the value of y after first step, using Euler’s predictor – corrector method

9.

[Ans. B] For convergence x

Mathematics

= x =x x=

x =

(2x

x

)

x= √

th

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GATE QUESTION BANK

Mathematics

Calculus ME – 2005 1.

ME – 2006

The line integral ∫ ⃗ ⃗⃗⃗⃗ of the vector function ⃗ ( ) 2xyz ̂+ x²z + ̂ k²y ̂ from the origin to the point P (1,1,1) (A) is 1 (B) is Zero (C) is 1 (D) cannot be determined without specifying the path

2.

be (A) (B) 8.

(A)

(C)

(B)

(D)

Changing the order of the integration in the double integral I = ∫ ∫

(

∫ (

What is q?

(A)

(C) X (D) 8 )

(A)

√

(C)

√

(B)

√

(D)

.

(

10.

/

)

(A) 0 (B) ⁄

is equal to 11.

∫

(C) (D) 1

The area of a triangle formed by the tips of vectors a , b and c is (A)

(

)(

(D) Zero

(B)

|(

)

Stoke’ theorem connects (A) A line integral and a surface integral (B) Surface integral and a volume integral (C) A line integral and a volume integral (D) Gradient of function and its surface integral

(C)

|

(D)

(

(C) 2∫ (

√

ME – 2007

(B) 2∫

6.

1 and t is a real number,

Let x denote a real number. Find out the INCORRECT statement + represents the set if all (A) S * real numbers greater then 3 + represents the empty (B) S * set + represents the (C) S * union of set A and set B + represents the set (D) S * of all real umbers between a and b, where and b are real number

)

leads to (A) 4y (B) 16y²

(C) 0 (D) ⁄

dt is:

∫

9.

4.

)

⁄ ⁄

√

By a change of variables x(u,v) = uv, y(u,v) = v/u is double integral, the integral f(x,y) changes to f(uv, u/v) ( ). Then, ( ) (A) 2 v/u (C) v² (B) 2 u v (D) 1

I =∫ ∫ (

2x2  7x  3 , then limf(x) will x 3 5x2  12x  9

Assuming i =

The right circular cone of largest volume that can be enclosed by a sphere of 1 m radius has a height of

3.

5.

If f( x ) =

7.

)

th

th

) (

)|

| )

th


GATE QUESTION BANK

12.

√

If

√

y (2) = (A) 4 or 1 (B) 4 only 13.

, then

√

14.

20.

The divergence of the vector field )̂ ( ) ̂ is (x y) ̂ ( (A) 0 (C) 2 (B) 1 (D) 3

21.

Let

(C) 1 (D) 1/ln2

y2  4x and x2  4y is ⁄ (A) (B) 8 23.

⁄ (C) (D) 16

The distance between the origin and the point nearest to it on the surface

The directional derivative of the scalar

z2  1  xy is

function f(x, y, z) = x2  2y2  z at the

(A) 1

point P = (1, 1, 2) in the direction of the vector ⃗ ̂ ̂ is (A) 4 (C) 1 (B) 2 (D) 1

at x=2, y=1?

ME – 2009 22. The area enclosed between the curves

Which of the following integrals is unbounded? (C) ∫ (A) ∫ (D) ∫

What is

(A) 0 (B) ln2

The length of the curve

(B) ∫ 16.

In the Taylor series expansion of ex about x = 2, the coefficient of (x 2)4 is ⁄ (A) ⁄ (C) ⁄ (B) (D) ⁄

The minimum value of function y = x2 in the interval [1, 5] is (A) 0 (C) 25 (B) 1 (D) Undefined

between x = 0 and x = 1 is (A) 0.27 (C) 1 (B) 0.67 (D) 1.22 15.

19.

(C) 1 only (D) Undefined

ME – 2008

Mathematics

(C) √ (D) 2

(B) √ ⁄ 24.

A path AB in the form of one quarter of a circle of unit radius is shown in the figure. Integration of  x  y  on path AB 2

17.

Consider the shaded triangular region P shown in the figure. What is∬ xydxdy? y

P

18.

B

X

x

A

2

⁄ ⁄

The value of (A) (B)

Y

x+2y=2

1

0 (A) (B)

traversed in a counter-clockwise sense is

⁄ ⁄

(C) ⁄ (D) 1

(

)

(C) (D)

25.

is ⁄ ⁄

th

(A)

(C)

(B)

(D) 1

The divergence of the vector field ̂ at a point (1,1,1) is ̂ ̂ equal to (A) 7 (C) 3 (B) 4 (D) 0 th

th


GATE QUESTION BANK

ME – 2010 26. Velocity vector of a flow field is given as ⃗ ̂ .̂ The vorticity vector at (1, 1, 1) is (A) 4 ̂ ̂ (B) 4 ̂ ̂ 27.

The function (A) o o

(C) ̂ (D) ̂

∀ ∀ (B) o o ∀ ∀ except at x = 3/2 (C) o o ∀ ∀ except at x = 2/3 (D) o o ∀

28.

29.

ME – 2012 33. Consider the function ( ) in the interval . At the point x = 0, f(x) is (A) Continuous and differentiable. (B) Non – continuous and differentiable. (C) Continuous and non – differentiable. (D) Neither continuous nor differentiable.

̂ ̂

R R R R

34.

R R

ME – 2011 30. If f(x) is an even function and is a positive real number, then ∫ ( )dx equals

31.

What is (A) (B)

32.

36.

For the spherical surface the unit outward normal vector at the point

is

(C) π (D) π

(C)

(B) (C)

is

√

(A) (B)

has

/

√

̂

√

̂

√

̂

√

̂

√

(C) ̂ (D) 37.

equal to?

A series expansion for the function (A)

.

∫ ( )

(C) 0 (D) 1

(C) 1 (D) 2

At x = 0, the function f(x) = (A) A maximum value (B) A minimum value (C) A singularity (D) A point of inflection

R except at x = 3 ∀ R

(D)

/ is

35.

The parabolic arc is √ revolved around the x-axis. The volume of the solid of revolution is (A) π (C) π (B) π (D) π

(A) 0 (B)

. (A) 1/4 (B) 1/2

The value of the integral ∫ (A) –π (B) –π

Mathematics

√

̂

√

̂

√

̂

The area enclosed between the straight line y = x and the parabola y = in the x – y plane is (A) 1/6 (C) 1/3 (B) 1/4 (D) 1/2

ME – 2013 38. The following surface integral is to be evaluated over a sphere for the given steady velocity vector field defined with respect to a cartesian coordinate system having i, j and k as unit base vectors. ∫∫ (

(D) th

th

) th


GATE QUESTION BANK

Where S is the sphere, and n is the outward unit normal vector to the sphere. The value of the surface integral is (A) π (C) π⁄ (B) π (D) π 39.

45.

If a function is continuous at a point, (A) the limit of the function may not exist at the point (B) the function must be derivable at the point (C) the limit of the function at the point tends to infinity (D) the limit must exist at the point and the value of limit should be same as the value of the function at that point

The value of the definite integral ( )

∫ √

is

(A)

√

(C)

√

(B)

√

(D)

√

46.

Divergence of the vector field ̂ ( ̂ ̂ ) is (A) 0 (C) 5 (B) 3 (D) 6

(C) 3 (D)Not defined

47.

The value of the integral

ME – 2014 40.

is (A) 0 (B) 1

∫ 41.

Which one of the following describes the relationship among the three vectors ̂ ̂ ̂ ̂ + ̂ + ̂ ̂ ̂ ̂ (A) The vectors are mutually perpendicular (B) The vectors are linearly dependent (C) The vectors are linearly independent (D) The vectors are unit vectors

42.

.

(

)

/ is equal to

(A) 0 (B) 0.5 43.

̂

̂

)̂ )̂ ̂ ̂

̂ ̂ ̂ ̂

)

(

)

) (

)

(A) 3 (B) 0 48.

(C) 1 (D) 2

The value of the integral ∫ ∫ is (

)

(C) (

(B) (

)

(D) .

(A)

) /

).

Where, c is the square cut from the first quadrant by the lines x = 1 and y = 1 will ( G ’ h o o h h line integral into double integral) (A) ⁄ (C) ⁄ (B) 1 (D) ⁄

̂ ̂

( (

CE – 2005 1. Value of the integral ∮ (

(C) 1 (D) 2

Curl of vector ⃗ ̂ (A) ( (B) ( (C) (D)

44.

Mathematics

̂ ̂ 2.

A rail engine accelerates from its stationary position for 8 seconds and travels a distance of 280 m. According to the Mean Value theorem, the speedometer at a certain time during acceleration must read exactly. (A) 0 kmph (C) 75 kmph (B) 8 kmph (D) 126 kmph

The best approximation of the minimum value attained by (100x) for ≥ is _______

th

th

th


GATE QUESTION BANK

CE – 2006 3. What is the area common to the circles o 2 (A) 0.524 a (C) 1.014 a2 (B) 0.614 a2 (D) 1.228 a2 4.

The directional derivative of f(x, y, z) = 2 + 3 + at the point P (2, 1, 3) in the direction of the vector a= k is (A) (C) (B) (D)

CE – 2007 5. Potential function  is given as = . When will be the stream function () with the condition  = 0 at x = y = 0? (A) 2xy (C) (B) + (D) 2 6.

Evaluate ∫ (C) π⁄ (D) π⁄

(A) π (B) π⁄ 7.

10.

12.

transformed to (A) (B)

9.

(C) √ (D) 18

parabola is y = 4h

(A) ∫ √ √

√

(D)

√

√

(C) ∫

= 0 by substituting (C)

where x is the

horizontal coordinate and y is the vertical coordinate with the origin at the centre of the cable. The expression for the total length of the cable is

= 0 can be

(D)

14.

√

∫

.

The



/

is

(A) 2/3 (B) 1

The inner (dot) product of two vectors ⃗ and ⃗ is zero. The angle (degrees) between in two vectors is (A) 0 (C) 90 (B) 30 (D) 120

(C) 40.5 (D) 54.0

√

(B) 2∫

+

is

CE – 2010 13. A parabolic cable is held between two supports at the same level. The horizontal span between the supports is L. The sag at the mid-span is h. The equation of the

CE – 2008 +

)

For a scalar function f(x, y, z) = the directional derivative at the point P(1, 2, 1) in the ⃗ is direction of a vector (A) (B)

A velocity is given as ̅ = 5xy + 2 y2 + 3yz2⃗ . The divergence

The equation

The value of ∫ ∫ ( (A) 13.5 (B) 27.0

CE – 2009 11. For a scalar function f(x, y, z) = + 3 + 2 the gradient at the point P (1, 2, 1) is (A) 2 + 6 + 4⃗ (C) 2 + 12 + 4⃗ (D) √ (B) 2 + 12 – 4⃗

of this velocity vector at (1 1 1) is (A) 9 (C) 14 (B) 10 (D) 15

8.

Mathematics

15.

th

(C) 3/2 (D)

Given a function ( ) The optimal value of f(x, y) th

th


GATE QUESTION BANK

(A) Is a minimum equal to 10/3 (B) Is a maximum equal to 10/3 (C) Is a minimum equal to 8/3 (D) Is a maximum equal to 8/3

CE – 2013 21.

CE – 2011 16.

∫

√ √

√

?

22.

(C) a (D) 2a

/

o

magnitudes a and b respectively. |⃗ ⃗ | will be equal to (A) – (⃗ ⃗ ) (C) + (⃗ ⃗ ) (B) ab ⃗ ⃗ (D) ab + ⃗ ⃗ CE – 2012 19. For the parallelogram OPQR shown in the sketch, ̅̅̅̅ ̂ ̂ and ̅̅̅̅ R ̂ .̂ The area of the parallelogram is Q

(C) 1 (D)

24.

A particle moves along a curve whose parametric equation are : and z = 2 sin (5t), where x, y and z show variations of the distance covered by the particle (in cm) with time t (in s). The magnitude of the acceleration of the particle (in cm ) at t = 0 is ___________

25.

If {x} is a continuous, real valued random variable defined over the interval ( ) and its occurrence is defined by the density function given as:

(C) 1 (D) π

If ⃗ and ⃗ are two arbitrary vectors with

R P

.

( )

/

√

wh

‘ ’

‘ ’

the statistical attributes of the random variable {x}. The value of the integral

O

(A) ad –bc (B) ac+bd

.

With reference to the conventional Cartesian (x, y) coordinate system, the vertices of a triangle have the following coordinates: ( ) ( ) ( ) ( ) ( ) ( ). The area of the triangle is equal to (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄

π

(A) 0 (B) π

(C) ad + bc (D) ab – cd

∫

o

o

. √

/

dx is

(A) 1 (B) 0.5

The infinity series

(A) sec (B)

(C) 1 (D) 8/3

23.

π

20.

o

(A) (B)

Wh ho h o λ such that the function defined below is continuous π ? f(x)={

18.

The value of ∫ (A) 0 (B) 1/15

CE – 2014

(A) 0 (B) a/2 17.

Mathematics

o

26.

(C) o (D)

(C) π (D) π⁄

The expression

o

(A) log x (B) 0 th

th

(C) x log x (D) th


GATE QUESTION BANK

CS – 2005 1.

Let G(x) 

CS – 2010

1   g(i)xi where |x|<1. 2 (1  x) i0 

What is g(i)? (A) i (B) i+1

(C) 2i (D) 2i

CS – 2007 2. Consider the following two statements about the function f(x) =|x|: P: f(x) is continuous for all real values of x Q: f(x) is differentiable for all real values of x Which of the following is true? (A) P is true Q is false (B) P is false Q is true (C) Both P and Q are true (D) Both P and Q are false CS – 2008 3.

4.

x sinx equals Lim x   x cosx (A) 1 (B) 1

(C) (D)

Let P=∑

∑

7.

What is the value of (A) 0 (B)

∫ (A) 0 (B) 1

(C) (D) 1

(A) 0 (B) 2

(C) –i (D) i

CS – 2012 9. Consider the function f(x)= sin(x) in the interval x ,π⁄ π⁄ -. The number and location(s) of the local minima of this function are (A) One , at π⁄ (B) One , at π⁄ (C) Two , at π⁄ and π (D) Two , at π⁄ and π CS – 2013 10. Which one the following function is continuous at x =3? (A) ( )

{

(B) ( )

2

(C) ( )

2

(D) ( ) CS – 2014 11. Let the function ( )

CS – 2009 6.

/ ?

∫

A point on a curve is said to be extreme if it is a local minimum or a local maximum. The number of distinct extrema for the 4 3 2 curve 3x  16x  24x  37 is (A) 0 (C) 2 (B) 1 (D) 3

.

CS – 2011 8. Given i = √ , what will be the evaluation of the definite integral

where k is a positive integer. Then (A) (C) (B) (D) 5.

Mathematics

(π (π

|

Where evaluates to

o o (π o (π

) )

) )

(π (π

)| )

1 and ( ) denote the

0

derivation of f with respect to . Which of the following statements is/are TRUE? (I) There exists

(C) ln2 (D) ln 2

. th

th

/

h h th

( )


GATE QUESTION BANK

(II) There exists . (A) (B) (C) (D) 12.

13.

14.

/

(A) h h

( )

I only II only Both I and II Neither I nor II

(B)

A function f (x) is continuous in the interval [0, 2]. It is known that f(0) = f(2) = 1 and f(x) = 1. Which one of the following statements must be true? (A) There exists a y in the interval (0,1) such that f(y) = f(y + 1) (B) For every y in the interval (0, 1), f(y) = f(2 y) (C) The maximum value of the function in the interval (0, 2) is 1 (D) There exists a y in the interval (0, 1) such that f(y) ( )

(C)

(D)

ECE – 2006 2. As x is increased from function f  x  

If and are 4 – dimensional subspace of a 6 – dimensional vector space V, then the smallest possible dimension of is ____________. If ∫

dx = π, then the value of k

e 1  ex



3 The integral sin  d is given by

3.

0

1 2 2 (B ) 3

o

(A) π (B) π

, the



The value of the integral given below is ∫

to

x

(A) monotonically increases (B) monotonically decreases (C) increases to a maximum value and then decreases (D) decreases to a minimum value and then increases

is equal to_______. 15.

Mathematics

4 3 8 (D) 3

(A) (C) – π (D) π

ECE – 2005 1. The derivative of the symmetric function drawn in given figure will look like

(C)

  P  ds , where P is a vector, is

4.

equal to (A) ∮ P  dl

(C) ∮  P  dl

(B) ∮  P dl

(D)

  Pdv

 P , where P is a vector, is equal to

5.

th

2 (A) P P  P

(C) 2P  P

2 (B)  P P

(D)  P 2P

th

th


GATE QUESTION BANK

ECE – 2007 6. For the function , the linear approximation around = 2 is (A) (3 x) (B) 1 x





ECE – 2008 12. Consider points P and Q in the x –y plane, with P=(1,0) and Q=(0,1). The line Q

integral 2  xdx  ydy  along the

 P

(C) 3  2 2  1  2 x e





2

semicircle with the line segment PQ as its diameter (A) is 1 (B) is 0 (C) is 1 (D) depends on the direction (clockwise or anticlockwise) of the semicircle

(D) 7.

8.

9.

10.

For x <<1, coth(x) can be approximated as (A) x (C) (B) x2 (D)

In the Taylor series expansion of exp(x)+sin(x) about the point x=π the coefficient of (x π)2 is (π) (A) (π) (C) (π) (π) (B) (D)

14.

Which of the following functions would have only odd powers of x in its Taylor series expansion about the point x=0? (A) sin(x3) (C) cos(x3) 2 (B) sin(x ) (D) cos(x2)

15.

The value of the integral of the function g(x, y)=4x3+10y4 along the straight line segment from the point (0,0) to the point (1,2) in the x y plane is (A) 33 (C) 40 (B) 35 (D) 56

16.

For real values of x, the minimum value of the function f(x)=exp(x)+ exp( x) is (A) 2 (C) 0.5 (B) 1 (D) 00

17.

Consider points P and Q in the x-y plane, with P=(1, 0) and Q= (0, 1).

Which one of the following function is strictly bounded? 2 (A) (C) x x (B) e (D)

(A) 0.5 (B) 1 11.

13. Consider the function f(x) = x – 2. The maximum value of f(x) in the closed interval [ 4,4] is (A) 18 (C) 2.25 (B) 10 (D) Indeterminate

sin   /2 lim is 0  (C) 2 (D) not defined

The following Plot shows a function y which varies linearly with x. The value of 2



the integral I  ydx is 1

Y 3 2

) along ∫ ( the semicircle with the line segment PQ as its diameter (A) Is (B) Is 0 (C) Is 1 h

1 1

Mathematics

1

(A) 1.0 (B) 2.5

2

3

X

(C) 4.0 (D) 5.0

th

th

th


GATE QUESTION BANK

(D) Depends on the direction (clockwise or anti-clockwise) of the semicircle

21.

ECE – 2009 18. The Taylor series expansion of

sinx at x   is given by x  (A) 1 

 x  2  ..... 3!

(B)

2 x    1   .....

(C)

2 x    1  .....

the value of the integral ∯ (A) 3V (B) 5V

3!

 x  

2

19.

3!

 .....

If a vector field ⃗ is related to another ⃗ , which vector field ⃗ through ⃗ = of the following is true? Note: C and refer to any closed contour and any surface whose boundary is C. (A) ∮ ⃗ ⃗ = ∬ ⃗ ⃗ (B) ∮ ⃗ ⃗ =

∬⃗ ⃗

(C) ∮

⃗ ⃗ =

∬⃗ ⃗

(D) ∮

⃗ ⃗ =

∬⃗ ⃗

ECE – 2010 20. If ⃗

̂ ̂ , then ∮ ⃗ ⃗⃗⃗ over the path shown in the figure is

(A) 0 (B) ⁄√

̂⃗

is

(C) 10V (D) 15V

ECE\IN – 2012 23. The direction of vector A is radially outward from the origin, with where and K is constant. The value of n for which . A = 0 is (A) 2 (C) 1 (B) 2 (D) 0 ECE\EE – 2012 24. The maximum value of ( ) in the interval [1,6] is (A) 21 (C) 41 (B) 25 (D) 46 ECE – 2013 25. The maximum value of unit which the approximation holds to within 10% error is (A) (C) (B) (D) 26.

√

, then has a maximum at minimum at maximum at minimum at

ECE – 2011 22. Consider a closed surface S surrounding a volume V. If is the position vector of a point inside S, with ̂ the unit normal of S,

3!

(D) 1 

If (A) (B) (C) (D)

Mathematics

The divergence of the vector field ⃗ ̂ ̂ ̂ is (A) 0 (B) 1/3

√

(C) 1 (D) 3

(C) 1 (D) 2√

th

th

th


GATE QUESTION BANK

27.

Consider a vector field ⃗ ( ) The closed loop line integral ∮ ⃗  can be expressed as ⃗ )  over the closed (A) ∯( surface boundary by the loop (B) ∰(  ⃗ )dv over the closed volume bounded by the loop (C) ∭(  ⃗ )dv over the open volume bounded by the loop ⃗ ) over the closed surface (D) ∬( bounded by the loop

Mathematics

34.

The magnitude of the gradient for the function ( ) at the point (1,1,1) is_______.

35.

The directional derivative of ( ) ( ) ( )in the direction √

of the unit vector at an angle of with y axis, is given by ________________. EE – 2005 1.

For the scalar field u =

, magnitude

of the gradient at the point (1, 3) is ECE – 2014 28. The volume under the surface z(x, y) = x+y and above the triangle in the x-y plane defined by {0 y x and 0 x 12} is______ 29.

30.

For function ( ) (A) o (B) o The value of

the maximum value of the occurs at (C) (D) o .

(A) (B)

(B) √ ⁄ 2.

For the function f(x) = , the maximum occurs when x is equal to (A) 2 (C) 0 (B) 1 (D) 1

3.

If S = ∫

/ is

31.

The maximum value of the function ( ) ( ) (wh ) occurs at x =____.

32.

The maximum value of ( ) 0 x 3 is ______.

in the interval

EE – 2006 4. A surface S(x, y) = 2x + 5y – 3 is integrated once over a path consisting of the points that satisfy (x+1)2+ (y 1)2 =√ . The integral evaluates to (A) 17√ (C) √ /17 (D) 0 (B) 17/√ 5.

33.

, then S has the value (C) ⁄ (D) 1

⁄ ⁄

(A) (B)

(C) (D)

(C) √ (D) ⁄

⁄

(A) √

The expression V = ∫ πR (

h

)

h

for the volume of a cone is equal to

For a right angled triangle, if the sum of the lengths of the hypotenuse and a side is kept constant, in order to have maximum area of the triangle, the angle between the hypotenuse and the side is (A) 12 (C) 60 (B) 36 (D) 45

th

(A) ∫ πR (

h

)

(B) ∫ πR (

h

)

(C) ∫

π

(

(D) ∫

π

(

th

h

R) h ⁄R )

th


GATE QUESTION BANK

EE – 2007 6.

EE – 2010

The integral equals (A) (B) 0

(

∫

o

) o

EE – 2009 8. f(x, y) is a continuous function defined over (x, y) [0, 1] [0, 1]. Given the two constraints, x > and y > , the volume under f(x, y) is (A) ∫

∫

(B) ∫

∫

(C) ∫

∫

9.

10.

√

11.

∫

√

(

)

(

)

( √

a minimum a discontinuity a point of inflection a maximum

Divergence of the three-dimensional radial vector field is (A) 3 (C) ̂ ̂ ̂ (B) 1/r (D) ̂ ( ̂ ̂)

13.

The value of the quantity P, where ∫ (A) 0 (B) 1

, is equal to (C) e (D) 1/e

EE – 2011 14. The two vectors [1, 1, 1] and [1, a, where a = . (A) (B) (C) (D)

)

A cubic polynomial with real coefficients (A) can possibly have no extrema and no zero crossings (B) may have up to three extrema and upto 2 zero crossings (C) cannot have more than two extrema and more than three zero crossings (D) will always have an equal number of extrema and zero crossings

has

12.

) (

At t = 0, the function ( ) (A) (B) (C) (D)

(C) (1/2) o (D) (1/2)

EE – 2008 7. Consider function f(x)= ( ) where x is a real number. Then the function has (A) only one minimum (B) only two minima (C) three minima (D) three maxima

(D) ∫

Mathematics

15.

√

],

/, are

orthonormal orthogonal parallel collinear

The function f(x) = 2x – has (A) a maxima at x = 1 and a minima at x=5 (B) a maxima at x = 1 and a minima at x= 5 (C) only a maxima at x = 1 (D) only a minima at x = 1

EE – 2013 16. Given a vector field , the line integral

F(x, y) = ( )̂ ( )̂ ’ line integral over the straight line from ( ) = (0, 2) to ( ) = (2, 0) evaluates to (A) –8 (C) 8 (B) 4 (D) 0

evaluated along a segment on the x∫ axis from x = 1 to x = 2 is (A) 2.33 (C) 2.33 (B) 0 (D) 7 17.

th

The curl of the gradient of the scalar field defined by (A) th

th


GATE QUESTION BANK

(B) (C) ( ( (D)

)

(

19.

20.

21.

23.

( ) Where f and v are scalar and vector fields respectively. If h is (A) (B) (C) (D)

24.

The minimum value of the function ( ) 0 in the interval , - is (A) 20 (C) 16 (B) 28 (D) 32

)

)

EE – 2014 18. Let ( ) . The maximum value of the function in the interval ( ) is (A) (C) (B) (D) The line integral of function , in the counterclockwise direction, along the circle is (A) π (C) π (B) π (D) π Minimum of the real valued function ( ) occurs at x equal to ( ) (A) (C) 1 (B) (D)

IN – 2005 1. A scalar field is given by f = x2/3 + y2/3, where x and y are the Cartesian coordinates. The derivative of f along the line y = x directed away from the origin, at the point (8, 8) is

To evaluate the double integral ∫ .∫

( ⁄ )

.

/

substitution u = (

/ dy, we make the ) and

2.

)

( ) ∫ (∫

)

( ) ∫ (∫

)

( ) ∫ (∫

)

(A)

√

(B)

√

(A) (B) 0 3.

√

∫ ( )

is

(C) f(1) (D) f(0)

The value of the integral ∫ (A) 2 (B) does not exist

(C) (D)

is 2

̅(t) has a constant magnitude, If a vector R then

4.

A particle, starting from origin at t = 0 s, is traveling along x-axis with velocity π π o . /

(D)

√

f(t) defined over [0,1],

̅ (A) R 22.

(C)

Given a real-valued continuous function

. The

integral will reduce to ( ) ∫ (∫

Mathematics

̅ (B) R 5.

̅

̅

̅ R ̅ (C) R ̅

̅ (D) R

̅ R

̅

If f = + …… + where ai (i = 0 to n) are constants,

At t = 3 s, the difference between the distance covered by the particle and the magnitude of displacement from the origin is _________

then

is

(A) ⁄ (B) ⁄

th

th

(C) nf (D) n√

th


GATE QUESTION BANK

6.

The plot of a function f(x) is shown in the following figure. A possible expression for the function f(x) is f(x)

(B) once differentiable but not twice (C) twice differentiable but not thrice (D) thrice differentiable 11.

x

0 (A)

(

)

(C)

(

(B)

. /

(D)

. /

)

IN – 2006 7. The function ( ) is approximated as where is in radian. The maximum value of for which the error due to the approximation is within (A) 0.1 rad (C) 0.3 rad (B) 0.2 rad (D) 0.4 rad

()

( )∫

(A) (

()

(

))

(B) (

()

(

))

(

()

(

))

(

()

(

))

()

(

(D) (

(

)

(

) ( )

( )) ( ))

13.

The expression (A) – (B) x

14.

Given y =

(A) √π⁄ (B) √π 10.

(C) Π (D) π⁄

Consider the function f(x) = , where x is real. Then the function f(x) at x = 0 is (A) continuous but not differentiable

+ 2x + 10, the value of

|

(C) 12 (D) 13

15. (A) Indeterminate (B) 0

(C) 1 (D)

IN – 2009 16. A sphere of unit radius is centered at the origin. The unit normal at a point (x, y, z) on the surface of the sphere is (A) (x, y, z) (C) . /

IN – 2007 9. The value of the integral dx dy is.

√

for x > 0 is equal to (C) (D)

is equal to (A) 0 (B) 4

(B) .

∫ ∫

(C) (D)

is.

IN – 2008 12. Consider the function y = x2 6x + 9. The maximum value of y obtained when x varies over the interval 2 to 5 is (A) 1 (C) 4 (B) 3 (D) 9

The solution of the integral equation ()

(C)

For real x, the maximum value of (A) 1 (B) e 1

1

8.

Mathematics

√

√

√

/

(D) .

√

√

√

√

√

√

/

IN – 2010 17. The electric charge density in the region R: is given as σ( ) , where x and y are in meters. The total charge (in coulomb) contained in the region R is (A) π (C) π⁄ (B) π (D) 0 th

th

th


GATE QUESTION BANK

18.

The integral ∫

.

evaluates to (A) 6 (B) 3 19.

/ sin(t) dt

23.

A scalar valued function is defined as ( ) , where A is a symmetric positive definite matrix with dimension n× n ; b and x are vectors of dimension n×1. The minimum value of ( ) will occur when x equals ) (A) ( (C) . / ) (B) – ( (D)

24.

Given ()

(

()

o .

(C) 1.5 (D) 0

The infinite series ( )

…………

converges to (A) cos (x) (B) sin(x)

(C) sinh(x) (D)

IN – 2011 20.

The series ∑ for (A) (B)

(

)

Mathematics

π) π

π /

The o w (A) A circle (B) A multi-loop closed curve (C) Hyperbola (D) An ellipse

converges

(C) (D)

IN – 2013 21. For a vector E, which one the following statement is NOT TRUE? (A) E E o o (B) If E E is called conservative (C) If E E is called irrotational (D) E E -rotational IN – 2014 22. A vector is defined as ̂ ̂ ̂ ̂ are unit vectors in Where ̂ ̂ Cartesian ( ) coordinate system. The surface integral ∯ f.ds over the closed surface S of a cube with vertices having the following coordinates: (0,0,0),(1,0,0),(1,1,0),(0,1,0),(0,0,1), (1,0,1),(1,1,1),(0,1,1) is________

th

th

th


GATE QUESTION BANK

Mathematics


= =

Since, potential function of ⃗ is x²yz ( ) ( ) ( ) 2.

0 o

1

[Ans. A]

[Ans. D] o

h

π h

π o

(

√

0

9.

[Ans. B]

10.

[Ans. B] (

)

For V to be max

)

This is of the form . /

Hence, h

Applying L hospital rule (

3.

√

1

)

[Ans. A]

. /

= (

)

|

|

|

|

= = 11.

4.

[Ans. A] (

After changing order ∫ ∫ 5.

[Ans. A] I= ∫ (

)

=2∫

[ ∫

[Ans. B] Let the vectors be

) ( )(⃗ )(⃗ )

]

= 2∫ 6.

[Ans. A] A Line integral and a surface integral is connected by stokes theorem

7.

[Ans. B]

Now Area vector will be perpendicular to plane of i.e. will be the required unit vector. And option (A), (D) cannot give a vector product )| |(⃗ ⃗ ) (⃗ 12.

[Ans. B]

Applying ’ Hospital rule, we get I= 8.

[Ans. A] ∫

√

Given:

I=

(

)

(

)

√

√

√

For 0 1

[

] th

th

th


GATE QUESTION BANK

13.

14.

But y is always greater than x. Hence y= 4 only.

= ∫

[Ans. B] Since interval given is bounded, so minimum value of functions is 1.

= 0

(

)

)

)

Now by partial fractions, (a3 8) = (a 2)(a2+2a+4)

) |

⇒L=

[Ans. D] To see whether the integrals are bounded or unbounded, we need to see that the o ’ h h interval of integration. Let us write down the range of the integrands in the 4 options, Thus, (D) , i.e., ∫ [Ans. B] h

19.

o

o

o Φ (

Φ)

̂

̂ ). (

̂

( ̂

( )

(

( )

)

( )

( )

Coefficient of (x- )⁴ Now f(x)= ex ⇒ (x)= ex ⇒ (a)= ea ( )

Hence for a=2, 20.

⃗⃗

[Ans. D] div {( (

Hence directional derivative is (grad (x2+2y2+z)).

=

[Ans. C] Taylor series expansion of f(x) about a is given by ( )

dx is unbounded.

along a vector ⃗

(2x ̂

( (

A (0,1); B (0,1); C (0, ); D (0, )

16.

[Ans. B]

Let x= a3 ⇒ a=2

=1.22 15.

1

).dx

= (

)

L=

∫√

L =∫ (√

)

= 18.

h

(

= ∫ (

[Ans. D] h

Mathematics

)̂

(

)

)̂ (

)̂}

( )

(

)

=3

̂)

√ ̂)

21.

[Ans. C]

= Hence at (1,1,2), ⇒

Directional derivative = 17.

[Ans. A] I = ∬ .dx dy The limit of y is form 0 to

and limit

of x from 0 to 2 I =∫ ∫

⇒

( )

⇒

( (

∫

.

)(

) )

/ th

th

th


GATE QUESTION BANK

22.

[Ans. A] Given:

23.

y2  4x x2  4y

Mathematics

[Ans. A] Short method: Take a point on the curve z = 1, x = 0, y=0 Length between origin and this point ) ( ) ( ) =1 √( This is minimum length because all options have length greater than 1.

(4,4) (0,0)

24.

x4   4x 16

[Ans. B] Y

or x4  64x

B

or x(x  64)  0 3

or x3  64 or x  4

x = cos  y=sin 

 y 4  Required area = ∫ .√

Path is x2  y2  1

/

R  e (x  y)2  1  2sin  cos 

4

 2 x3   2 x3 2    3 120  4 64  (4)3 2  3 12 32 16 16    3 3 3

2

2

 cos2    (1  sin2)d   2 0  0 =

Alternately For point where both parabolas cut each other

 1 1     1 2 2 2 2

Alternately Given: x2  y2  1 Put x=cos  , and y=sin 

y2  4x, x2  4y

 x  y 2  cos2  sin2  2sincos

 x  4 4x 2

= 1  sin2

or x2  8 x or x4  64x

∫ (

or x3  64

 x  4,0 ,(4,0)

2

4

x2 dx 0 4

     1 2 

  4x   0

)

 cos2    1 1        2 0  2 2 2  

 Required area 4

X

A

4

 2 x3  16  2 x3 2     3 120 3 

25.

[Ans. C]

F  3xzi  2xyj  yz2k ⃗ ⃗

th

th

th


GATE QUESTION BANK

(

)

(

)

(

√

)

 3z  2x  2yz

π ∫

At point (1, 1, 1), divergence =3+2 2=3 26.

⃗

o

30.

⃗ ̂

̂

||

31. ( ̂

27.

)̂

(

∫ ( )

[Ans. D] Standard limit formulae

) ̂

32.

[Ans. B]

33.

[Ans. C] The function is continuous in [ 1, 1] It is also differentiable in [ 1, 1] except at x = 0. Since Left derivative = 1 and Right derivative = 1 at x = 0

34.

[Ans. B]

[Ans. C]

1

1

2

y is continuous for all x differentiable for all x since at

o

o

R, and R, except at

o

,

Using this standard limit, here a = 1 then = ( ) /2 =1/2

’ h

value towards the left and right side of

35.

[Ans. D] ( ) ( ) ( ) ( ) f(x) has a point of inflection at x =0.

36.

[Ans. A]

[Ans. D] ,

∫

29.

( )

̂

⇒

28.

π

)

[Ans. D] If f(x) even function ∫

||

o

π* +

π(

[Ans. D] ⃗ ̂

Mathematics

-

[Ans. D]

π

∫ π

(

) ̂

Volume from x = 1 to x = 2,

̂

∫π ( √ th

th

√

̂ ̂ ̂

̂

) th


GATE QUESTION BANK

̂

√ √ ̂

√ ̂

̂ ̂

√ ̂

38.

(√ √

37.

̂

̂ √

[Ans. A] By Gauss Divergence theorem, ∬( ̅ ̅)

√ √ The unit outward normal vector at point P is (

Mathematics

)

∭(

(Surface Integral is transformed to volume Integral)

)

( )

( )

( )

)̂

√ ̂

)

∭(

∭

[Ans. A] The area enclosed is shown below as shaded

π π

(

∬( ̅ ̂)

)

)

∭(

( π) (

)

The coordinates of point P and Q is obtained by solving y = x and y = simultaneously, i.e. x = ) ⇒ ( ⇒ Now, x = 0 ⇒ which is point Q(0,0) and x = 1 ⇒ which is point P(1,1) So required area is

π 39.

[Ans. C] ∫(√ ) ( ) Using Integration by parts ∫

∫

Here, f=ln(x) and dg=√ and g=

∫

* +

∫

o ∫(√ ) ( )

* + [

]

∫

[

]

[

(

th

th

( ) ] )

th


GATE QUESTION BANK

40.

π

[Ans. A] o

0

1

So the minimum value is

[Differentiating both o o Hospital method]

o w

.

=

’

. /

45.

o

41.

Mathematics

/

[Ans. D] o o o ( ) ( )

o

( )

o

o

otherwise it is said to be discontinuous. So the most appropriate option is D.

[Ans. B] G

o 46.

|

|

[Ans. C] ̂ ̂ Div

̂

(

)

Vectors are linearly dependent 42.

[Ans. B] (

) ) ( ) , o ( )( ) ( ) o ( )( ) 43.

(

47. -

ho

[Ans. B] Let ∫

(

)

(

( o (

)

[Ans. A] ̂

(

⃗ [ ̂[

(

] )

(

) ∫

̂ ̂

)

48.

)]

)

()

o

̂[ ,̂ ( 44.

(

) (

( ) )̂

∫

|

,

-

)] (

,̂ (

o

[Ans. B] ∫ ∫

̂[

o

∫

)] ̂,

)̂

(

-

∫ (

)

|

)̂ [

[Ans. *] Range 1.00 to 0.94 h o π

,

th

th

] -

,

th

-


GATE QUESTION BANK

CE 1.

a =2a cos i.e, cos [Ans C] G ’ theorem is 

∮(

)

∬(

∮ ((



(

( )

)

= y and

=∫

∫

=∫

,

(

)-

=

×

( )

π . /

|

π . /

π (

π *

[Ans. D] Since the position of rail engine S(t) is continuous and differentiable function according to Lagrange’s mean value theorem more )

(

o ) o

∫(

w

∫

(t) = v(t) =

(

π * (

( )

π

π π

√

| π

√

√

)

√

)

)+

+

)

m/sec kmph

4.

[Ans. C] f = 2 +3

= 126 kmph Where v(t) is the velocity of the rail engine. 3.

∫

∫

(

-

= 2y

=∫

=

∫

)

’ h o I= ∫

)

, ∫

 = xy 

)

(

)

)

⇒

R

Here I = ∮ (

2.

Mathematics

[Ans. D] h ’ o h r=2acos (i) r = a represents a circle with centre ( ) ‘ ’ (ii) r = 2acos represents a circle symmetric about OX with centre at ( ) ‘ ’ The circles are shown in figure below. At h o o o ‘ ’ P y Q π 3

O

A

(

)⃗

= 4xi + 6yj + 2zk At P (2, 1, 3) Directional derivation ̂ ( ) ( ) ( ) √ ( ) ( ) ( ) √ √ 5.

[Ans. A] Potential function,

x

th

th

th


GATE QUESTION BANK

8. Integrating ∫

[Ans. D] Put

∫

wh

Mathematics

o

⇒

√



=√



=√

…… ( )

Now given equation is ……….. (ii) 6.

[Ans. B] Let I(α) =∫ (

h ∫ )

.

/

h

(

dx …( ) =

(

)

(

h

)

√ ) [ from eqn(i)]

=

∫ Then Integrating by parts we get, =

0

=

.

( α

√

h

(

)

o )1

/

√

(

√

(

(

h

)

h

)

= dI = Integrating, I = ( )

α o

h

√

)

h

() ( )

+C=0 C= (α)  ( )

α

π

Now substitute in eqn (ii) we get h h

π

I(0) = But from equation (i), I(0) = ∫ ∫

h

⇒ dx

h

⇒

dx =

h h

Which is the desired form 7.

[Ans. D] ̅=5 +2

√

+ 3y ⃗

(⃗ )

9.

[Ans. C] ̅ ̅=0 ̅ ̅ If ̅ ̅ = 0

= 5y + 4y + 6yz At(1, 1, 1) div ( ) = 5.1 + 4.1 + 6.1.1 = 15

is the correct transformation.

o

 o Since P and Q are non-zero vectors  o 0  th

th

th


GATE QUESTION BANK

10.

[Ans. A] Since the limit is a function of x. We first integrate w.r.t. y and then w.r.t. x )

∫ ∫(

∫

√ √

√

)

[Ans. D] Length of curve f(x) between x = a and x = b is given by ∫√

∫

*

√

√ 13.

∫(

Mathematics

(

)

+ Here,

∫ (

4h … … ( )

= 8h

)

Since ∫ (

and y = h at x =

)

* ( ) *

(As can be seen from equation (i), by substituting x = 0 and x = L/2)

( )+

(Length of cable)

+

√

=∫

.

/ ∫ √

ho 11.

[Ans. B] f = + 3 +2 f = grad f = i

+j

[Ans. A]

15.

[Ans. A] ( )

+k

= i(2x) + j(6y) + k(4z) The gradient at P(1, 2, 1) is = i(2×1) + j(6×2) + k(4 ( )) = 2i + 12j – 4k 12.

14.

[Ans. B] (

)

⃗

⃗ ⃗

̂

⃗

h

Putting,

√ o (

Given,

̂ )

.

/ is the only stationary point.

√ *

+ .

√

th

th

/

th


GATE QUESTION BANK

*

Since the limit is in form of

+ .

*

’ ho and get λ

/

+ .

Since, We have either a maxima or minima at

o ()

⇒λ 18.

/

Also since, r=0 )

o

1

.

/

[Ans. A]

= 8 > 0, the point

(

o

)

,

)

19.

o -

[Ans. A] Area = |̅ ̅ | ̅̅̅̅ ̅̅̅̅ R (

So the optimal value of f(x, y) is a

|

) o

o

(

The minimum value is (

o

π

⇒λ

(

o

, we can use

/

Since,

.

Mathematics

)

(

)

|

minimum equal to 16.

̅̅̅̅ ̅̅̅̅ R ̅̅̅̅ ̅̅̅̅ R

[Ans. B] Let I = ∫

√ √

Since ∫ ( ) I=∫

√ √

…( )

√

√

∫ (

20.

[Ans. B]

21.

[Ans. B]

( )

( )

̅(

)

)

…( )

(i) + (ii)  2I = ∫

√

√

√

√

 2I = ∫

∫

2I = |

o ∫

o

o

 I = a/2 17.

⇒

[Ans. C] For a function f(x) to be continuous, at x=a ( ) ( )

o ∫ (

⇒

)

∫ (

)

If f(x) is continuous at x= π . /

*

λ o

+

[

th

th

]

th


GATE QUESTION BANK

[

]

|

(

Mathematics

)

(

)

(

)

|

Substituting the values we get ( ) ( ) ( ) | | 24. π

π

∫

o

∫

o

∫

o

(

[Ans. 12]

o ) ( )

o

o ( ) ∫

o

( )

[ 22.

⇒ Magnitude of acceleration

]

=√

[Ans. C] ( ⇒

) (

25.

(

)

[Ans. B] We have

) ∫ ( )

⇒ , ow

-

∫ ( )

=1+0=1 Hence correct option is (C) 23.

∫ ( )

∫ ( )

[Ans. A] (4, 3) a (2, 2) b

c

x

( )μ

0.5

(1, 0)

0.5

o ∆ wh o –ordinate points are given is given by

th

th

th


GATE QUESTION BANK

26.

[Ans. A]

=

α Use L – hospital Rule

= 4.

α

=1

[Ans. A] P=∑

o

…

‘ ’

= log x

 1 n 1 r  Cr xr  n 1  x  r0





r 0

r 0

‘ ’



r 0

i 0

5.

= 12

g(i) =i+1

– 24 48

)

+ 37

– 48 x = 0

√

x= =2

[Ans. A] f(x)= |x| Continuity: In other words, f(x) = x o ≥ x for x< 0 Since, = =0 , f (x) is continuous for all real values of x Differentiability:

=

√

96x

48

=

√

√ = 36

Now at x = 0 =

48

0

At 2 ± √ also

0 (using

calculator) There are 3 extrema in this function

( )

)

6.

( )

[Ans. D] Since ∫ ( )

R h So |x| is continuous but not differentiable at x=0 3.

(

x (12 – 48x 48 ) = 0 x = 0 or 12 – 48x – 48 = 0 4x – 4 = 0

∑ ()

(

k

)

[Ans. D] y = 3 – 16

(since r is a dummy variable, r can be replaced by i)

)

k

⇒

  r  1 xr   i  1 xi

(

–1)

)

w h (

  r1 Cr xr   r1 C1xr 

( …

 1 21 r  Cr xr  1  x 2 r0

Putting n=2,

2.

)

= Q=∑

[Ans. B]

(

w h a =1, l=2k 1

P= ( CS 1.

Mathematics

I =∫ =∫



=∫ (

)

 (

) (

)

Since tan (A B) =

[Ans. A] =

⁄ ⁄

th

th

th


GATE QUESTION BANK 

[

]

[

]

Mathematics

∫ 

∫

0

1

0

1

9.

[Ans. B] f(x) = sin x ⇒ ( ) o ( ) ⇒ o π π π [



( (

∫

) )

( (

) )

π ]

( )





∫

∫

At

. /

gives maximum

value =,

)-

(

At

= ln ( sec ) – ln (sec 0) = ln (√ ) = ln (

. /

value

( ) 10.

)–0=

[Ans. A] For x =

7.

[Ans. B] (

8.

)

*

(

(

) [

*(

) +

) + .

/

]

11.

[Ans. C] By Mean value theorem

12.

[Ans. A] Define g(x) = f(x) – f(x + 1) in [0, 1]. g(0) is negative and g(1) is positive. By intermediate value theorem there is €( ) h h g(y) = 0 That is f(y) = f(y + 1) Thus Answer is (A)

13.

[Ans. 2] * w + * w + For min maximum non – common elements must be there ⇒ * + must be common to any 2 elements of V1 ( )minimum value = 2

o o ∫

∫

*

+ [

]

,

-(

o π

, f(x) =

For x = , f(x) = 3 – 1 = 2 For x = 3, f(x) = 2 ( ) ( ) = f(3)

[Ans. D] ∫

gives minimum

π

)

th

th

th


GATE QUESTION BANK

14.

Mathematics

[Ans. 4] ∫ ∫

|

∫∫

π ( ) o π o (π) π Hence option (A) is correct

oπ π o π

∫

∫(

) ECE 1.

[Ans. C]

∫ ∫ (∫

∫ o

)

dy  0 for x< 0 dx dy  0 for x> 0 dx

∫ o

o Substituting the limits π o (π) o ( ) π

2.

[Ans. A] Given,

f x 

∫

 f ' x  

1  e .e  e 1  e 

|

∫∫

3.

= x cos

∫(

x 2

2x





ex

1  ex



2

0

o

)

Let cos = t ⇒ At o π o π o

o

∫ o

x

∫

[Ans. A] ∫

x

[Ans. C]

= π o ( π) π o π = π LHS = I + II = π π π⇒ 15.

ex 1  ex

∫

|

∫

∫ o

∫(

)

∫(

∫

∫

th

th

th

)

|

|

(

)

(

)


GATE QUESTION BANK

8.

Mathematics

[Ans. A]

Given, f  x   x2  x  2

df  x  0 dx 4.

 2x 1=0

[Ans. A] o ’ h o )

∬(

 x

∮

1 2

d2f  x  = 2 ve dx2 So it shows only minima for interval [ 4, 4], it contains a maximum value that will be at x= 4 or x=4  f( 4)=18 and f(+4)=10 

5.

6.

[Ans. D] From vector triple product ( ) ( ) ( ) Here, ( ) ( ) ⇒ ( ) ( ) ( )

[Ans. D]

y  f  x ; x  0, 

[Ans. A] ( )

 f  x0 

For strictly bounded, 0  limy  

2 x  x0  f' x0   x  x0  f'' x0      ......

1

 e  (x  2)(e 2

x0

or 0  lim y  

2

 x  2 )

2

2

2

x 2

So, y  e x is strictly bounded

e ...... 2

  x  22  ......   e2  3  x    2   (Neglecting higher power of x)

7.

9.

10.

lim 0

=

ex  e x ex  ex

x x2 x3 e  1     .......... 1 2 3

11.

x

ex  1 

[Ans. B] Two points on line are ( 1, 0) and (0, 1) Hence line equation is,

 y y  y   2 1 x  c  x2  x1  y  x c y  x 1 … ( )

x x2 x3    .......... 1 2 3

x2 x4   .......... ex  ex 2 4  x x  e e x3 x5 x    .......... 3 5 1

or cot h (x)=

sin  /2 1  sin  /2  lim   0 2    /2 

1  sin  /2 1 = lim  2  0  /2  2

[Ans. C] coth (x)=

[Ans. A]

2 2 5 I   ydx    x  1dx   2.5 2 1 1

1 x

(Since at x=1,y=2)

(Neglecting x2 and higher order)

th

th

th


GATE QUESTION BANK

12.

[Ans. B] Taking f(x, y)= xy, we can show that, xdx+ydy, is exact. So, the value of the integral is independent of path

15.

Mathematics

[Ans. A]

Given : g  x,y   4x3  10y 4 The straight line can be expressed as y=2x Then g(x,y)=4x3+ 10 (2x)4

(0, 1)

 1

1





4 I   4x3  10 2x   dx   4x3  160x4 dx   0 0 1

 4x4 160 5  =   x  33 5 0  4

(1, 0)

)

∫(

∫

[Ans. A] f(x)= + (x)= =0 x=0 (x)= + >0  x R. Hence minimum at x=0 f(0)=1+1=2 Alternatively: For any even function the maxima & minima can be found by A.M. >= GM => exp(x) + exp( x) ≥ 2 Hence minimum value = 2

17.

[Ans. B]

∫

[ |

13.

16.

| ]

[Ans. B] Let f(x)  ex  sinx o ’ 2 x  a  f  x   f a    x  a f'a   f''a  2!

Q

where, a= 2 x    f  x   f     x   f'   f''  2!

Coefficient of (x )2 is

f ''  2

P

f''   ex  sinx |at x   e

 Coefficient of (x )2=0.5 exp () 14.

∫(

)

[Ans. A]

o Thus, ( ( )w o ( )w o ( )w

)w h h h

h

∫

∫

[ |

| ]

o ow ow ow ow th

th

th


GATE QUESTION BANK

18.

21.

[Ans. D] sinx = x = (x – π ) –

y=

(

or

19.

(

)

(

)

sin x = (x – π ) –

or

)

(

)

=1 –

(

) (

= 1

(

) (

)

o

.... ( )

)



...

o

....

(

)

( )



....

=

Therefore, at

22.

∬⃗ ⃗

̂

̂

̂

̂

∯

⃗

∭ ( )∭ and is the position vector)

(

23.

⃗⃗⃗ ⃗⃗⃗

has a maximum.

[Ans. D] Apply the divergence theorem

[Ans. C]

[Ans. A] ̂

Y

S

3

R

1

̅

Q

P

⇒

√

√

⇒

∮ ⃗ ⃗⃗⃗ ∫ ⃗ ⃗⃗⃗

∫ ⃗ ⃗⃗⃗ √

∫ ⃗ ⃗⃗⃗

∫

√

∫ ⃗ ⃗⃗⃗

∫ .√ /

∫ ⃗ ⃗⃗⃗

∫√ √

[ ∫ ⃗ ⃗⃗⃗

√

* +

[

) )

⇒

25.

[Ans. B]

, √ √

( )

)]

( (

)]

[Ans. C] ( ) , ( ) ( ) ( ) ⇒ are the stationary points ( ) ( ) ( ) and f(2) = 25 and f(4)=21 M o ( ) , f(6)=41

√

. /

(

(

24. ]

] ∫ .√ /

[

[

̂

⇒

∫ ⃗ ⃗⃗⃗

along PQ y =1 dy =0]

∫ ⃗ ⃗⃗⃗

⇒

⇒

X

= [

o

Since

[Ans. D] o ’ h o ⃗ ⃗ = ∮

⃗⃗⃗

 o

o

According Stokes Theorem ⃗⃗⃗ ⃗ ∮ ⃗ ⃗ =∮

20.

[Ans. A]

....

sin (x –π )

or

Mathematics

∮ ⃗ ⃗⃗⃗ ⇒ th

th

th


GATE QUESTION BANK

o

30.

[Ans. C]

E o

(

E o

31.

o

[Ans. *] Range ( ) ( )

̂

̂

⇒ ⇒ ( )

=1+1+1 =3 [Ans. D] o ’ h o “ h integral of a ⃗ vector around a closed path L is equal to the integral of curl of ⃗ over the open ∮⃗

h

⃗

∬(

o

∫

∫

∫

∫

(

∫

*

32.

h ”

33.

)

) ) ) )

[Ans. C] Let x (opposite side), y (adjacent side) and z (hypotenuse side) of a right angled triangle

+

∫

29.

[Ans. *] Range 5.9 to 6.1 Maximum value is 6 ( ) ( ) ( ( ( (

⃗ )⃗

[Ans. *] Range 862 to 866 Volume under the surface ∫

( ) ( ) h

o

o

28.

to 0.01

( )

[Ans. D] ̅ ̂

=

27.

)

π

⇒

26.

Mathematics

Given

o

… )(

(

[Ans. A] o ( ) ̇( )

o

⇒ o

⇒ ( ) Since ( ) is negative, maximum value of f(x) will be where ( )

⇒

o 0(

⇒ ⇒

)

o ⇒ ( )

⇒

o

( )

( )

o

oh

(

)

)

1

( ( (

th

th

)(

))

) th


GATE QUESTION BANK

By trial and error method using options π

34.

Now at x = 2 (2) = ( ) = ( ) = 2 <0 At x = 2 we have a maxima.

[Ans. *] Range 6.8 to 7.2 ⃗ ̂ ̂ ̂ ̂

̂

=

̂

̂

̂

3.

=∫

[Ans. *] Range 2.99 to 3.01 √

⃗

√

)

√

(

(

)

)̂

At (1, 1), ⃗

√

(̂

Given unit vector, ̂

(

√

√

)̂

(̂

√

√

.

-

,

-

/

[Ans. D]

5.

[Ans. D] We consider options (A) and (D) only because which contains variable r. By integrating (D), we get π , which is volume of cone.

6.

[Ans. D] By property of definite integral

̂ ) ̂ )

=3

) ∫ ( ) ∫ ( π On simplification we get option (D)

[Ans. C] Grad u = ̂

⁄

At (1, 3) Grad u = √

,( ⁄ )

7.

[Ans. B] f(x) = ( ) (x) = 2( ) =4x( ) =0 x = 0, x = 2 and x = 2 are the stationary points. (x) = 4[x(2x) +( ) ] = 4[2 = 4 [3 = 12 (0) = < 0, maxima at x = 0 (2) =(12) = 32 > 0, minima at x = ( 2) =12( ) = 32 > 0; minima at x = There is only one maxima and only two minima for this function.

-

=√ 2.

,

4.

̂ )

So, directional derivative ⇒ ⃗ ̂ (̂ ̂ ) (̂

EE 1.

1

[Ans. C]

=

(

0

̂

At (1, 1, 1) ⃗ |⃗ | √

35.

Mathematics

[Ans. A] f(x) = (x) = ( ) = ( ) Putting ( (x) = 0 ( )=0 ( )=0 x = 0 or x = 2 are the stationary points. Now, ( ) ( (x) = )( ) = ( ( )) = ( ) ( )=2 At x = 0, (0) = Since (x) = 2 is > 0 at x = 0 we have a minima. th

th

th


GATE QUESTION BANK

8.

Mathematics

= , =, ( = (

[Ans. A]

))

(

) =1

1

14.

[Ans. B] Dot product of two vectors =1+a+ =0 So orthogonal

15.

[Ans. C] f(x) = ( ) ( ) So the equation f(x) having only maxima at x = 1

16.

[Ans. B]

0

√

∫

9.

10.

(

∫

)

[Ans. C] ( ) ( ) ( ) ( ) ( )

̂

[Ans. D] ̅=( )̂ ( ( ) = (0, 2) ( ) = (2, 0) Equation of starting line

̂ ̂ ∫

11.

) y = 2 – x and dy = – dx

17.

C

)

o ( )

o

.(

)̂

̂

(

)̂

̂

̂ ||

( ̂ ( =0

⃗ ̂

̂

||

̂ ̂

)̂ /

̂

is undefined

[Ans. A] ̂ Div ( ) =.

‘ ’

(

Discontinuous

/(

̂

̂

̂)

18.

= 1+1+1= 3 13.

[Ans. D]

)

But at

12.

∫

[Ans. B] (

̂

Along x axis ,y=0,z=0 The integral reduces to zero.

=

∫ (

̂ ̂

̂

)̂

⇒ y = 2 – x , dy = – dx ̅ ̅ =( ( ) Putting ∫̅ ̅ ∫

̇̂

[Ans. B] P=∫ th

[Ans. A] ( ) o ⇒ M th

) ( ) ̂ (

( ( )

th

) ( ) ̂ (

)

(

) (

)


) )

GATE QUESTION BANK

19.

Mathematics

[Ans. B]

.

0

π

/1

∫⃗ ∫

[

(

o

o π

)( )(

∫ ( ∫

⇒

o

23.

[Ans. A] ( ⃗) ⃗ ⃗ ( ) ( ) ⃗

24.

[Ans. B] ( ) ( ) ⇒ ⇒ )( ⇒( ⇒ ( ) ( ) ( )

) )

[

]

π 20.

]

[Ans. C] ( )

(

)

( )

( ) For number of values of ) o ( ) ( ( ) ( ( )

( ) ⁄

,

w

-

) ( ) ( )

M 21.

)

[Ans. B]

IN 1.

G o

h

[Ans. A]

o

o

⇒ o

⇒

(

o

)

Unit vector along y = x is G

∫ (∫ ∫ (∫

22.

[Ans. 2] ( ∫

)

̂

)

̂

π

√ o

√ ̂

.

) ∫

π

o

o .

π

/

.

π

/.

√

/

√

/

√

√

√ √

2.

[Ans. D] Using L Hospital Rule., numerator becomes =

From the graph, distance at

th

th

()

= ( )

th


GATE QUESTION BANK

3.

[Ans. B]

Mathematics

When

Given integral is, I=∫

( )

Let f(x) = so curve of 1/

will be

(

)

(

)

And when

f(x)

( )

The possible expression for f(x) is 1

. 7.

-1

0

1

/

[Ans. B]

x

Error,

This curve will be discontinuous at x=0 o ’ w o

For error to be minimum (

4.

[Ans. A] ̅ (t) =x (t) ̂+y (t) + Let R ̂ z (t) ̂ ̅( ) =K (constant) |R i.e., (t) + (t) + (t) = constant. On analyzing the given (A) option, we find ̅(t) that R

̅( )

⇒ ⇒

[Ans. C] Given : f= + where,

(

)

⇒

√

will give constant magnitude, √

1

G …… + (i=0 to n) are constant.

=

+(n 1)

o

…… ⇒

+ and

)

o

so first differentiation of the integration will be zero. 5.

o

=0+

+

(n 1)

√

…… ⇒

+n

√

+ = , = nf 6.

+

+

⇒

-

(

)

⇒ ⇒

[Ans. B] ( )

(

)

8.

[Ans. B] ()

When ( )

(

)

(

)

( )

∫

…( )

Differentiating the above equation

When ( )

th

th

th


GATE QUESTION BANK

()

,

14.

() ∫

Mathematics

[Ans. B] Given y = x2 + 2x + 10 = 2x + 2

( ) -

| From equation (i) () ()

⇒

()

()

This is Leibnitz linear equation Integrating factor I.F = ∫ the solution is ()

15.

[Ans. C] By definition

16.

[Ans. A]

()

Unit vector=

=xi+yj+zk

and 17.

∫

[Ans. C] R: Y

( ) 1 1

, [Ans. D]

10.

[Ans. A] This is a standard question of differentiability & continuity

Area =

[Ans. C] y= =(

X

- o

9.

11.

+1

( )

Total charge = σ = = 18.

).(cos x + sin x) = 0

⇒ tan x = 1 Or x =

coulomb.

[Ans. B] We know that ∫

() (

∫

.

) π

( )wh π . /

/

y will be maximum at x = y=

19.

= 12.

13.

[Ans. C] y(2) = y(5) =

=

√

[Ans. B] Expansion of sin x ........

( ) ( )

20.

[Ans. B] In a G.P

(

)

For a G.P to converge

[Ans. C] y= y=

(

⇒

)

⇒

(

)

⇒ th

th

th


GATE QUESTION BANK

21.

[Ans. D] .E=0 is not irrational (it is solenoidal)

22.

[Ans. 1] From Gauss divergence theorem, we have ∫ ̅ ̅

̅

∫

/dxdydz

∫ [ ⇒

̅

∫

∫.

Mathematics

∫ ̅

[Ans. C]

24.

[Ans. D]

̂

) ̂

̂

]

o .

π

π /

23.

(

∫

(

π )

th

th

th


GATE QUESTION BANK

Mathematics

Differential Equations ME – 2005 1.

If x

xy

n y

what is y (A) e (B) 1

then

⁄ ⁄

(C) (D)

(B) degree 1 order 1 (C) degree 2 order 1 (D) degree 2 order 2 ME – 2007 7.

2.

3.

Statement for Linked Answer Questions 2 and 3. The complete solution of the ordinary differential equation y y p qy s x x y Then, p and q are (A) p =3, q = 3 (C) p =4, q = 3 (B) p =3, q = 4 (D) p =4, q = 4 Which of the following is a solution of the differential equation y y p q y x x (A) (C) x (B) x (D) x

The solution of

For

ME – 2008 8. It is given that + 2y + y = 0, y (0) = 0, y(1) = 0. What is y (0.5)? (A) 0 (C) 0.62 (B) 0.37 (D) 1.13 9.

Given that ẍ + 3x = 0, and x(0) = 1, ẋ (0) = 0, what is x(1)? (A) 0.99 (C) 0.16 (B) 0.16 (D) 0.99

ME – 2009 10.

+ 3y =

, the particular

integral is: (A) (B) (C) (D) 5.

The solution of x

y

x

with the

s

(A) y

(C) y

(B) y

(D) y

ME – 2010 11. +

The solution of the differential equation

(A) (1+ x)

(C) (1 x)

(B) (1+ x)

(D) (1

The Blasius equation,

, is a

(A) second order non-linear ordinary differential equation (B) third order non-linear ordinary differential equation (C) third order linear ordinary differential equation (D) mixed order non-linear ordinary differential equation

2 dy  2xy  ex with y (0) = 1 is: dx

6.

(D) 2  x  2

(B)   x  1

condition y +4

with initial value

y (0) = 1 is bounded in the interval (C) x  1,x  1 (A)  x 

ME – 2006 4.

y

x)

The partial differential equation (

)

(

)= 0 has

(A) degree 1 order 2 th

th

th


GATE QUESTION BANK

ME – 2011 12. Consider

the

differential

x ,yt x ,yt

equation

y x. The general solution with constant c is (A) y

t n

(B) y

t n (

(C) y

t n ( )

(D) y

t n(

17.

)

18.

differential y

equation

with the boundary

(B) s n ( )

(D)

partial u

15.

19.

s n( )

differential

equation

is a

linear equation of order 2 non – linear equation of order 1 linear equation of order 1 non – linear equation of order 2

subjected to the boundary conditions u(0) = 0 and u(L) = U, is (A) u (C) u ( ) (

)

(D) u

(

x

y and

x

x ,y-

*

x + ,y-

*

x + ,y-

y is

20.

with t __________

The general solution of the differential os x

constant, is (A) y s n x

y

(B) t n (

)

y

(C)

os (

)

x

(D) t n (

)

x

y with c as a x

Consider two solution x(t) = x t and x t x t of the differential equation x t x t t su t t t x t x | t x t | t

t s (A) 1 (B) 1

x t

x t

| t

(C) 0 (D)

The solution of the initial value problem xy y

)

ME – 2014 16. The matrix form of the linear system

x ,yt

at x

The wronskian W(t) =|

where k is a constant,

t

x + ,y-

If y = f(x) is the solution of

x

The solution to the differential equation

(B) u

*

x

conditions of y(0) =0 and y(1) = 1. The complete solution of the differential equation is (A) x (C) s n( )

(A) (B) (C) (D)

x + ,y-

the boundary conditions y

)

the

x

ME – 2013 14. The

*

equation

ME – 2012 13. Consider x

t n

Mathematics

is

(A)

(C)

(B)

(D)

CE – 2005 1. Transformation to substituting v = y

linear form by of the equation

+ p(t)y = q(t)y ; n > 0 will be (A) th

+ (1 n)pv = (1 n)q th

th


GATE QUESTION BANK

2.

(B)

+ (1 n)pv = (1+n)q

(C)

+ (1+n)pv = (1 n)q

(D)

+ (1+n)pv = (1+n)q

CE – 2007 6. The degree of the differential equation + 2x = 0 is (A) 0 (B) 1

in the range (A) (B)

y x

( os x ( os x

(C)

( os x

(D)

( os x

,

( )

7.

The solution for the differential equation = x y with the condition that y = 1 at

is given by

x = 0 is

s n x)

(B) In(y) =

s n x) s n x)

8.

xy

x

+4

(D) y =

A body originally at 600C cools down to C in 15 minutes when kept in air at a temperature of 250C. What will be the temperature of the body at the end of 30 minutes? (A) 35.20C (C) 28.70C 0 (B) 31.5 C (D) 150C

CE – 2008 9.

The general solution of (A) (B) (C) (D)

The solution of the differential equation x

(C) In(y) =

(A) y =

s n x)

CE – 2006 3. A spherical naphthalene ball expanded to the atmosphere losses volume at a rate proportional to its instantaneous surface area due to evaporation. If the initial diameter of the ball is 2 cm and the diameter reduces to 1 cm after 3 months, the ball completely evaporates in (A) 6 months (C) 12 months (B) 9 months (D) Infinite time

+ y = 0 is

y = P cos x + Q sin x y = P cos x y = P sin x y=Psn x

given that at x = 1, 10.

y = 0 is

5.

(C) 2 (D) 3

The solution of y

4.

Mathematics

(A)

(C)

(B)

(D)


= 0.25 y is to be

solv us ng t b w r mpl t Eul r’s method with the boundary condition y = 1 at x = 0 and with a step size of 1. What would be the value of y at x = 1? (A) 1.33 (C) 2.00 (B) 1.67 (D) 2.33

Solution of (A) x (B) x

=

at x = 1 and y = √ is

y y

(C) x (D) x

y y

CE – 2009 11. Solution of the differential equation 3y

+ 2x = 0 represents a family of

(A) Ellipses (B) Parabolas

(C) circles (D) hyperbolas

CE – 2010 12. The order and degree of the differential equation

+ 4 √( )

respectively (A) 3 and 2 (B) 2 and 3 th

th

y

= 0 are

(C) 3 and 3 (D) 3 and 1 th


GATE QUESTION BANK

13.

The solution to the ordinary differential equation (A) (B) (C) (D)

14.

+

y= y= y= y=

The following differential equation has 3

d2y  dy   4   y2  2  x dt2  dt  (A) degree=2, order=1 (B) degree=1, order=2 (C) degree=4, order=3 (D) degree=2, order=3

6y = 0 is

3

+ + + +

The partial differential equation that can be formed from z = ax + by + ab has the form (w t p (A) (B) (C) (D)

2.

n q

Mathematics

)

ECE – 2006

3.

For the differential equation

z = px + qy z = px + pq z = px + qy + pq z = qx + pq

the boundary conditions are (i) y=0 for x=0 and (ii) y=0 for x=a The form of non-zero solutions of y (where m varies over all integers) are m x y ∑ sn

CE – 2011 15. The solution of the differential equation + = x, with the condition that y = 1 at x = 1, is (A) y =

+

(D) y =

+

CE – 2012 16. The solution of the ordinary differential y=0 for the boundary

condition, y=5 at x = 1 is (A) y (C) y (B) y (D) y CE – 2014 17. The

y

∑

y

∑

y

∑

os

m x

(C) y = +

(B) y = +

equation

d2y  k2y  0 2 dx

integrating

for

the

equation (A) (B)

x

ECE – 2007 4. The solution of the differential equation

d2y  y  y 2 under the boundary dx2 conditions (i) y=y1 at x=0 and (ii) y=y2 at x=, where k, y1 and y2 are constants, is (A) y y y xp( x⁄ ) y (B) y y y xp x⁄ y (C) y y y s n x⁄ y (D) y y y xp x⁄ y k2

differential s

(C) (D)

ECE – 2005 1. A solution of the following differential equation is given by

d2y dy  5  6y  0 dx dx2

ECE – 2008 5. Which of the following is a solution to the differential equation

2x 3x (A) y  e  e

2x 3x (C) y  e  e

2x 3x (B) y  e  e

2x 3x (D) y  e  e

(A) t (B) x t

th

th

x t

x t

(C) x t (D) x t

th

t t


GATE QUESTION BANK

ECE – 2009 6. The order of the differential equation

ECE– 2011 10. The solution of the differential equation

3

d2y  dy    y4  et is dt 2  dt  (A) 1 (C) 3 (B) 2 (D) 4

7.

Match each differential equation in Group I to its family of solution curves from Group II. Group I Group II 1. Circles dy y  P.

dx x dy y  Q. dx x

(A) (B) (C) (D)

3. Hyperbolas

dy x  dx y

y x

(C) y (D) y

ECE\EE\IN – 2012 11. With initial condition x(1) = 0.5, the solution of the differential equation, t

x

t is

(A) x

t

(C) xt

(B) x

t

(D) x

ECE\IN – 2012 12. Consider the differential equation y t y t y t t t t wt y t | n | num r

l v lu o

(A) (B)

x with the initial condition

y s ng Eul r’s rst or r m t o with a step size of 0.1, the value of y is (A) 0.01 (C) 0.0631 (B) 0.031 (D) 0.1 A function n x satisfies the differential equation

is

(A) x (B) x

P-2, Q-3, R-3, S-1 P-1, Q-3, R-2, S-1 P-2, Q-1, R-3, S-3 P-3, Q-2, R-1, S-2

ECE – 2010 8. Consider a differential equation

9.

y y

2. Straight Lines

dy x R.  dx y S.

Mathematics

where L is a

constant. The boundary conditions are: n and n . The solution to this equation is (A) n x xp x (B) n x xp x √ (C) n x xp x (D) n x xp x

y | t (C) (D) 1

s

ECE – 2013 13. A system described by a linear, constant coefficient, ordinary, first order differential equation has an exact solution given by y t or t when the forcing function is x(t) and the initial condition is y(0). If one wishes to modify the system so that the solution becomes – 2y(t) for t > 0, we need to (A) Change the initial condition to – y(0) and the forcing function to 2x(t) (B) Change the initial condition to 2y(0) and the forcing function to –x(t) (C) Change the initial condition to j√ y(0) and the forcing function to j√ x(t) (D) Change the initial condition to – 2y (0) and the forcing function to – 2x(t)

th

th

th


GATE QUESTION BANK

ECE – 2014 14. If the characteristic equation of the differential equation y t

15.

has two equal roots,

n t v lu s o (A) ±1 (B) 0,0

r (C) ±j (D) ±1/2

xy

(C)

(B)

xy

(D)

(B) x t (C) x t (D) x t EE – 2011 3. With K as a constant, the possible solution for the first order differential equation is

Which ONE of the following is a linear non-homogeneous differential equation, where x and y are the independent and dependent variables respectively? (A)

Mathematics

xy

(C) (D)

(A) (B)

EE – 2013 4. A function y x x is defined over an open interval x = (1,2). At least at one point in this interval ,

16.

17.

18.

If z

xy ln xy then

(A) x

y

(C) x

y

(B) y

x

(D) y

x

If a and b are constants, the most general solution of the differential equation x x x s t t (A) (C) bt (B) bt (D)

(A) 20 (B) 25

tx

s

6.

x x (A) x t

n

with initial conditions | t

, the solution is

(B) s n t

os t

(C) s n t

os t

(D) os t

t

Consider

the

x

EE – 2005 1. The solution of the first order differential qu t on x’ t 3x(t), x (0) = x is (A) x (t) = x (C) x (t) = x (B) x (t) = x (D) x (t) = x EE – 2010 2. For the differential equation

(C) 30 (D) 35

EE – 2014 5. The solution for the differential equation x x w t n t l on t ons x t x n | s t (A) t t

With initial values y(0) = y (0) = 1, the solution of the differential equation y

is exactly

x

differential

equation

y

Which of the following is a solution to this differential equation for x > 0? (A) (C) x (D) ln x (B) x IN– 2005 1. The general solution of the differential equation (D2 4D +4)y = 0, is of the form (given D = d/dx), and C1 and C2 are constants (A) C1e2x (C) C1e2x + C2 e2x 2x (B) C1e + C2 (D) C1e2x + C2x th

th

th


GATE QUESTION BANK

2.

urv s or w t urv tur ρ t 3 any point is equal to cos θ w r θ s t angle made by the tangent at that point with the positive direction of the x-axis, r

gv nρ

⁄

, where y and y

are the first and second derivatives of y with respect to x) (A) circles (C) ellipses (B) parabolas (D) hyperbolas IN– 2006 3. For an initial value problem ÿ ẏ y y n ẏ various solutions are written in the following groups. Match the type of solution with the correct expression. Group 1 Group 2 P. General solution 1. 0.1ex of homogeneous equations Q. Particular integral 2. (A cos 10 x + B sin 10 x) R. Total solution 3. cos 10 x + x satisfying boundary 0.1e conditions (A) P-2, Q-1, R-3 (C) P-1, Q-2, R-3 (B) P-1, Q-3, R-2 (D) P-3, Q-2, R-1 4.

A linear ordinary differential equation is given as

d2y dy  3  2y  δ(t) 2 dt dt Where  (t) is an impulse input. The solut on s oun by Eul r’s orw r difference method that uses an integration step h. What is a suitable value of h? (A) 2.0 (C) 1.0 (B) 1.5 (D) 0.2

Mathematics

IN– 2007 5. The boundary-value problem y λy y y w ll v non-zero solut on n only t v lu s o λ r (A) ± ± … (B) … (C) … (D) … IN– 2008 6. Consider the differential equation = 1 + y2. Which one of the following can be a particular solution of this differential equation? (A) y = tan (x + 3) (C) x = tan (y + 3) (B) y = tan x + 3 (D) x = tan y + 3 IN– 2010 7. Consider y

the

differential

equation

with y(0)=1. The value of

y(1) is (A)

(C)

(B)

(D)

IN – 2011 8. Consider the differential equation ÿ ẏ y with boundary conditions y(0) = 1, y(1) = 0. The value of y(2) is (A) 1 (C) – (B) (D) IN– 2013 9. The type of the partial differential equation

is

(A) Parabolic (B) Elliptic 10.

th

(C) Hyperbolic (D) Nonlinear

The maximum value of the solution y(t) of the differential equation y t ÿ t with initial conditions ẏ and y , for t is (A) 1 (C) (B) 2 (D) √

th

th


GATE QUESTION BANK

Mathematics

IN– 2014 11. The figure shows the plot of y as a function of x

y

y

x

x

The function shown is the solution of the differential equation (assuming all initial conditions to be zero) is : (A) (B)

x

(C)

x

(D)

|x|

th

th

th


GATE QUESTION BANK

Mathematics


4. [Ans. D] x y y

lnx x

xy x

lnx x

y

omp r ng w g t ow y(I.F.) = ∫ x y

∫

olv ng bov utt ng x x 2.

3.

[Ans. B] The given differential equation may be written as y y y ux l ry qu t on s

w

lnx x

x ∫

∫

x

Substituting D=2, we get

x x

(

x

n t v lu o t n n t v lu o y t

5.

[Ans. B] First order equation,

sy

y

dy  Py  Q, dx

Where P = 2x and Q = Since P and Q are functions of x, then Integrating factor,

[Ans. C] Given equation is y p qy x x p q y p q ts solut on s y um o roots p p ro u t o roots q q [Ans. C] Given equation is y y p q x x p q ut p n q y

)

2

I.F. = e  Pdx  e x Solution is y

∫

y

x

∫

x

2

yex  x  c ,c=1

Since, y

 x2

(1 + x) e

y 6.

[Ans. A] Order: The order of a differential equation is the order of the highest derivative appears in the equation Degree: The degree of a differential equation is the degree of the highest order differential coefficient or derivative, when the differential coefficients are free from radicals and fraction. The general solution of differential qu t on o or r ‘n’ must nvolv ‘n’ arbitrary constant.

y

x

th

th

th


GATE QUESTION BANK

7.

y ( )y x … x x Standard form y y … x Where P and Q function of x only and solution is given by

[Ans. C] y x

v n y

y x

y

nt gr t ng y

y

x

∫

x nx

y

n

x

x

x

olut on y x x

and x

Given condition y m ns t x

y

₂ r or yx

y [Ans. D] ẍ x Auxiliary equation is m2 + 3 = 0 i.e. m = ±√ x os√ t sn√ t ẋ os√ t s n√ t √ At t = 0 1=A 0=B x = os √ t x

11.

x

[Ans. B] is third order ( is linear, since the product

) and it is not

allowed in linear differential equation 12.

os √ t

[Ans. D] y x y ∫ y t n

10.

x

x

y 9.

∫x x x

x

yx [Ans. A] y y y A.E is, D2+2D+1 =0 2=0 m 1 The C.F. is (C1+C2x)e-x P.I. = 0 ow y ₁ n y ₂

∫

x

∫

x x

8.

x

Where, integrating factor (I.F) r

y

Mathematics

[Ans. A] Given differential equation is y x y x x

y

th

y t n.

th

y x ∫ x x x

/

th


GATE QUESTION BANK

13.

[Ans. A] 17. y y x x y x x y n y Choice (A) satisfies the initial condition as well as equation as shown below y x y n y y lso x x y x 18. y y x x y x x x x x x x x x o y x is the solution to this equation with given boundary conditions

14.

[Ans. D]

15.

[Ans. B] m m u u At x=0, At x=L, (

[Ans. *] Range 34 to 36 y x y x y x tx y y tx x y x tx y [Ans. D] y os x y x Let x y z y z x x z os z x z os z x z s ( ) z x

z os ( )

Integrating z t n( ) x z t n( ) x x t n(

) n

19.

u x

Solving we get u = U( 16.

Mathematics

)

[Ans. A] x x y t y x y t So by observation it is understood that, x x ,y- * + ,yt

y )

[Ans. A] Since the determinant of wronskian matrix is constant values for, therefore it is same for both t=0 and t= t

20.

x

x t

x t t

[Ans. B] y ∫ ∫ x x y y ln ( ) x y

x t t

x t

ln y

x

ln

v ny n th

y th

th


GATE QUESTION BANK

CE 1.

y tx x

[Ans. A] Given

+ p(t)y = q(t) y

n y

3.

Multiplying by (1 n) y we get v p t n y q t t Now since y = v we get v n pv n q t Where p is p (t) and q is q(t)

t Where, V =

n

4 r

=

sn x os x sn x os x sn x

os x

r )

 r

t utt ng r

n

y

r

r t  dr = kdt Integration we get r = kt + C At t = 0, r = 1 1= k×0+C C=1 r = kt + 1 Now at t = 3 months r = 0.5 cm 0.5 = k × 3 + 1

)

t

…

r r t t Substituting in (i) we get

±

os x

sn x)

r

y (

( os x

[Ans. A]

[Ans. A] y y y x x y y ( ) x This is a linear differential equation

n

s

+ p(t) y = q(t) y ; n > 0

Given, v = y v y n y t t y v t n y t Substitution in the differential equation we get

2.

Mathematics

n solv ng g v s t

sn x

 t = 6 months

y os

sn

4. sn x os x

[Ans. A] Given y x xy – x x y x xy x x Dividing by x

os x

sn x sn x os x

th

th

x

th

y


GATE QUESTION BANK

y x ( )y ( ) x x x Which is a linear first order differential of the form y y x Integrating factor = I.F = ∫ = ∫ y × I.F = ∫ .(I.F)dx x yx ∫( )x x x Now at x = 1, y = 0

Hence, here the degree is 1, which is power of

7.

[Ans. D] y x y x This is variable separable form

x

= x dx

∫ x

∫

x

y y

∫x

tx log y

C x y

5.

–x x

[Ans. B] =

x

y

0C

= Now at t = 30 minutes Θ

±√

y =2

6.

t

 ln θ θ = kt + θ θ C. θ θ C. Given θ = 250C Now t t θ 60 = 25 + C.e0 C = 35 θ At t m nut s θ 40 = 25 + 35

y

y +1=0

t =∫

∫

y y y  0.25hy y +y =0 Putting k = 0 in above equation 0.25h y y +y =0 Since, y = 1 and h = 1 0.25 y

θ θ0) (Newton’s law of cooling)

θ θ θ

[Ans. C] y y y tx x h=1 Iterative equation for backward (implicit) Euler methods for above equation would be y

y x

y

8. y

x

x

log y

i.e. 0 ×

x

Mathematics

= 25 + 35 (

)

= 25 + 35 × ( ) (s n

[Ans. B] Degree of a differential equation is the power of its highest order derivative after the differential equation is made free of radicals and fractions if any, in derivative power.

)

= 31. ≈ C

th

th

th


GATE QUESTION BANK

9.

12.

[Ans. A] +y=0

10.

y

.

13.

∫ x x

√ C=2 Solution is y x

[Ans. D] y y x n y x x This is a linear differential equation of the form y y wt n x x x IF = Integrations factor

x x ∫

y x

y

( )

( )

x

x (

x x

x

y

(

∫

∫

x

Solution is y (IF) = ∫ x  y. x = ∫ xx x  yx = ∫ x x

y )

+

15.

x x y

[Ans. C] y y x x Auxiliary equation is +D–6=0 (D 2) = 0 D = 3 or D = 2 Solution is y =

[Ans. C] Z = ax + by + ab … z p x z q b y Substituting a and b in (i) in terms of p and q we get z = px + qy + pq

[Ans. A]

∫ y y

y y / 0( ) y 1 x x The order is 3 since highest differential

14.

x +y =4

y x y y

y

is

x

3y

y ) x

Removing radicals we get

At x = 1, y = √

11.

√(

The degree is 2 since power of highest differential is 2

[Ans. D] y x x y y dy = x dx ∫y y

[Ans. A] y x

+1=0 E sm m ± General solution is y= [ cos (1 × x) + sin (1 × x)] = cosx + sinx = P cosx + Q sinx Where P and Q are some constants

Mathematics

)

Which is the equation of a family of ellipses

 yx =

th

th

+C

th


GATE QUESTION BANK

y=

 sin ka=0 m x 

+

Now y(1) = 1  ot

 solut on s y

y

17.

m x

[Ans. D] k2D2y= y

y2

y2  2 1  D  k2  y  k2   1 m1 =  k x/k x/k C.F. = C1e  C2 e

[Ans. D]

x/k x/k  y2 y= C1e  C2 e

∫

At y=y1, x=0  y1 = C1+ C2+y2 … At y=y2 , x=  Hence C1 must be zero  y1 = C2+y2  C2 =y1 - y2

[Ans. B]

d2y dy  5  6y  0 dx dx2 A.E. is D2 5D  6  0  D=2,3 2x 3x Hence, solution is y  e  e

2.

sn

x

[Ans. D] y y x y y y

Particular integral (P.I) = = ECE 1.

∑

x 4.

16.

Mathematics

x  y=(y1 – y2) exp    + y2  k 5.

[Ans. B] x t x t t (D +3) x(t) = 0

[Ans. B] 3

d2y  dy   4   y2  2  x dt2  dt  Order of highest derivative=2 Hence, most appropriate answer is (B) 3

3.

[Ans. A] Given, Differential equation,

d2y  k2y  0 dx2 Auxilary equation is y ± Let y os x sn x At x=0, y = 0  A=0 y sn x At x=a, y=0  B sin ka=0 B0 otherwise y=0 always

So, x  t   ke3t , Hence x  t   2e3t is one solution (for some boundary / initial condition) 6.

[Ans. B] The order of a differential equation is the order of the highest derivative involving in equation, so answer is 2. The degree of a differential equation is the degree of the highest derivative involving in equation, so answer is 1.

7.

[Ans. A] P.

∫

∫

log y log x log y xw s qu t on o str g t l n th

th

th


GATE QUESTION BANK

Q.

∫ log y y

x

log x

∫y y y

x

8.

yp rbol

y

∫x x

9.

t ∫t t

xt

t

xt 12.

x

y

[Ans. D] Approach 1: y t y t y t t t t Converting to s-domain s y s sy y sy s y s s s y s s s y s s s n nv rs pl tr ns orm y t t u t y t t t y t | t

y

[Ans. D] Approach 2: y t y t y t t t t Applying Laplace Transform on both sides y s y s sy | t (sy s y ) y s s y s s sy s y s s s y s s s s s

n x m

Auxiliary equation m olut on n x Since, n

±



Since, n  must be zero) Therefore

(hence

The solution is, n x 10.

∫

r l

y old y +0.1 ( ) new x y x y 0 0 0+0 0+0.1×0=0 =0 0. 0 0.1+0 0+0.1×0.1=0.01 1 =0.1 0. 0.0 0.2+0.01 0.01+0.21×0.1 2 =0.21 1 =0.031 The value of y at x= 0.3 is 0.031. x

x=1

Using initial condition, at t = 1, x = 0.5

ypr bol

… Equ t on o

[Ans. B] y x y x x  x y

t

ol s xt

qu t on o

∫y y x

x

∫x x

…

S.

[Ans. C] t

log

… qu t on o

R.

11.

∫

Mathematics

[Ans. C] Given y ln y When y y

and x

y

y t

x

t

t

t

y

y t y t t

y

t

t t t t

th

th

th


GATE QUESTION BANK

z y

x ln xy

ox

z x

y t | t 13.

[Ans. D] Let the differential equation be y t y t x t t Apply Laplace transform on both sides y t {x t } 2 y t 3 t sy s y y s x s s y s x s y x s y y s s s Taking inverse Laplace on both sides x s {y s } 2 3 y { } s s y t x t y So if we want y t as a solution both x(t) and y(0) has to be multiplied by . Hence change x(t) by x t and y(0) by y

14.

[Ans. A] y y y x x The auxiliary equation is m m ± then either m or m i.e., roots of the equation are equal to or

y

z y

18.

EE 1.

xy ln xy

xy

[Ans. C] z xy ln xy z y ln xy x

y

y ln xy

…

x (t) x

x ∫ t x lnx = t x Putting x Now putting initial condition x(0) = x x x Solution is x = x i.e. x(t) = x

0 is a first order linear

xy

.

∫

omog n ous

xy

xy

t

equation (homogeneous) r non l n r qu t ons 16.

x

xy

[Ans. *] Range 0.53 to 0.55 E m m m olut ons s y bx y bx b … s ng y y n gv s n b y x tx y

[Ans. A] v n x’ t

is a first order linear

equation non

x ln xy

[Ans. B] x x x t t Pre auxiliary equation is m m Pre roots of AE are m Repeated roots are present. So, most general solution in n t bt

[Ans. A] xy

x

xy

z y

y

i.e. 15.

xy

xy ln xy

z x

ox 17.

Mathematics

y

th

th

th


GATE QUESTION BANK

2.

[Ans. B] x x x t t Auxiliary equation m m m m (m+4)(m+2)=0 m= 2, 4 x(t) =

6.

[Ans. C] x y xy y y x

m

t

nx

 … On solving (1) & (2), we get

IN 1.

y

x

x( s ts

x

)

x

s

[Ans. C] y

and

m

x(t)= 2 3.

x

x

… (1)



|

y y x

w subst tut y

n x(0) = 1 1=

Mathematics

Since there is double root at 2, so general solution of the given differential equation would be +

[Ans. A] y

m

x

Integrate on both sides 2.

y 4.

5.

[Ans. B] v n ρ

os θ … y n ρ … y now y’ t nθ … Equating equations (i) and (ii) and using equation (iii) in equation (ii), we get y os θ= os θ

[Ans. B] y x x p n nt rv l x y x x y x x Value is in between 20 and 30 So it is 25 [Ans. C] x x gv n t x os t sn t x n x sn t os t t x | t x

os t

y= .x Which is equation of a parabola. 3.

[Ans. A] A.E. D= 1+ 10i C.F = (A cos10 x + B sin 10 x) x

4.

[Ans. C]

5.

[Ans. C]

x

sn t

th

th

th


GATE QUESTION BANK

6.

[Ans. A] Given

Hyperbolic if El ps Compare the given differential equation with standard from A = 1, B = 0, C = 0

= 1 + y2

Integrating ∫

=∫ x

Or t n y = x + c Or y = t n x 7.

Parabolic

[Ans. C] y y x Auxiliary equation, m + 1 = 0 m= 1 C.F =

10.

y

[Ans. C]

The solution for the differential equation is y x Now, y and y , placing these values We get, and y

s nx

s nx

os x

So, y os x s n x or m x m y s nx os x s nx os x x y os x s n x y or x m xm y m x os sn

y

√ 11.

[Ans. A] Given partial differential equation is x

± os x

ẏ ẏ

y

9.

[Ans. D] y t ÿ t

y y

y=

8.

Mathematics

√

√

√

[Ans. D] By back tracking, from option (D) y |x| x or x x = x or x Integrating y ∫ ∫ x x or x x

t

∫ x x or x

x t We know that

x

y (x y

or x

)

or x

is said to be Parabolic if

th

th

th


GATE QUESTION BANK

Mathematics

Complex Variables ME – 2007 1. If x y and (x, y) are functions with continuous second derivatives, then x y + i (x, y) can be expressed as an analytic function of x + i (i = √ ), when (A)

ME – 2014 6.

The argument of the complex number where i = √ π π 2

7. +

(D)

+

(A) 2πi (B) 4πi

An analytic function of a complex variable z x + i y is expressed as z u x y +iv x y where i √ f u(x,y)= 2xy, then v(x,y) must be (A) x + y + onst nt (B) x y + onst nt (C) x + y + onst nt (D) x y + onst nt

9.

An analytic function of a complex variable z = x + i y is expressed as f(z) = u(x, y) + i v(x, y), where i = √ . If u (x, y) = x – y , then expression for v(x, y) in terms of x, y and a general constant c would be (A) xy + (C) 2xy +

ME – 2009 3. An analytic function of a complex variable z = x + iy is expressed as f(z) = u(x, y) +iv(x, y) where i = 1 . If u = xy, the expression of v should be

x

2

2

y 2

2

 k

y (C) (D)

2

 x2 2

 k

 x  y 2  k 2

ME – 2010 4. The modulus of the complex number

(B)

) is

(A) 5 (B) √

(C) 1/√ (D) 1/5

traversed in

8.

(C) 2πi (D) 0

 x  y 2  k

is evaluate

counter clock wise direction. The integral is equal to π (A) 0 – 2 π π – 4 4

+

ME – 2008 2. The integral ∮ z z evaluated around the unit circle on the complex plane for z is

(

x y

along the circle x + y

(C)

(B)

π 2 π

The integral ∮ y x

(B)

(A)

, is

10.

+

(D)

+

If z is a complex variable, the value of is

∫

(A) i (B) 0.511+1.57i (C) i (D) 0.511+1.57i

ME – 2011 5. The product of two complex numbers 1 + i and 2 – 5i is (A) 7 – 3i (C) 3 – 4i (B) 3 – 4i (D) 7 + 3i th

th

th


GATE QUESTION BANK

CE – 2005 1. Which one of the following is NOT true for complex number and ? (A) (B) | (C) | (D) |

2.

̅̅̅̅

=|

|

|≤| |+| | |≤| | | | | +| | 2| | + 2| |

+ +

CE – 2011 6. For an analytic function, f(x + iy) = u(x, y)+iv(x, y), u is given by u = 3x 3y . The expression for v considering K to be a constant is (C) 6x 6y+k (A) 3y 3x + k (D) 6xy +k (B) 6y – 6x + k CE – 2014

Consider likely applicable of u hy’s integral theorem to evaluate the following integral counter clockwise around the unit circle c. ∮s

z z

7.

z

πn

2

+

i i

ECE – 2006 1. The value

of

∮|

2.

-

π/2: singul riti s s t { nπ n 2 } (D) None of the above

∮

dz is

(A)

4πi

(C)

(B)

πi

(D) 1

πi

(C) (D)

the

+

contour

i i

integral

z in positive sense is

|

(A)

(C)

(B)

(D)

For the function of a complex variable W = In Z (where, W = u + jv and Z = x + jy), the u = constant lines get mapped in Z-plane as (A) set of radial straight line (B) set of concentric circles (C) set of confocal hyperbolas (D) set of confocal ellipses

(C) I

CE – 2006 3. Using Cauchy’s is integral theorem, the value of the integral (integration being taken in counter clockwise direction)

can be expressed as

(A) (B)

z being a complex variable. The value of I will be (A) I = 0: singularities set = ϕ (B) I = 0: singularities set =,

Mathematics

ECE – 2007 3. If the semi-circular contour D of radius 2 is as shown in the figure, then the value of the integral ∮

is

j

CE – 2009 4.

The analytic function f(z) = singularities at (A) 1 and 1 (B) 1 and i

5.

has

j2

(C) 1 and i (D) i and i

The value of the integral ∫

j2

2

dz (A) jπ (B) jπ

(where C is a closed curve given by |z| = 1) is (A) –πi (C) (B) (D) πi th

th

(C) π (D) π

th


GATE QUESTION BANK

ECE – 2008 4. The residue of the function

f z 

5.

1

z  2 z  22 2

(A) (B)

at z=2 is

(A)

(C)

(B)

(D)

(C) (D) 2

∮(

) z is

(A) 4π (B) 4π

The equation sin(z)=10 has (A) no real or complex solution (B) exactly two distinct complex solutions (C) a unique solution (D) an infinite number of complex solutions

12.

+ j2 j2

(C) 4π (D) 4π

+ j2 j2

The real part of an analytic function z where z x + jy is given by cos(𝑥). The imaginary part of z is (A) os x (C) sin x (B) sin x (D) sin x

EE – 2007

If f(z) =

+

is given by (A) 2π (B) 2π +

z

, then ∮

z

(C) 2πj (D) 2πj

z and 1 and

(C) (D)

The value of

+

at its poles are

(A)

1.

where C is the

∮

contour |z-i/2| = 1 is (A) 2πi (C) t n z (B) π (D) πi t n z

ECE – 2010 7. The residues of a complex function

(B)

2

ECE – 2014 11. C is a closed path in the z-plane given by |z|=3. The value of the integral

ECE – 2009 6.

Mathematics

EE – 2011 2. A point z has been plotted in the complex plane, as shown in figure below. nit ir l

and z

and

ECE – 2011 8.

The value of the integral ∮

z

ECE\EE\IN – 2012 9. If x = √ then the value of x is ⁄ (C) x (A) ⁄ (D) 1 (B) 10.

Given f (z)

nit ir l

lm

where is the circle |z| is given by (A) 0 (C) 4/5 (B) 1/10 (D) 1

lm

nit ir l

lm

nit ir l

y y lmlm

. If C is a

nit ir l

y

counterclockwise path in the z – plane such that |z+1| =1, the value of ∮

y

z z is th

th

th


GATE QUESTION BANK

EE – 2013 3.

z evaluated anticlockwise around

∮

the circle |z (A) 4π (B) 4.

i|

2 where i √ (C) 2 + π (D) 2 +2i

Square roots of – i, where i = √ (A) i, i (B)

os (

) + i sin (

, is

, are

)

IN – 2005 1. Consider the circle | | 2 in the complex plane (x, y) with z = x + iy. The minimum distant form the origin to the circle is (C) √ 4 (A) √2 2 (B) √ 4 (D) √2 2.

Let ̅, where z is a complex number not equal to zero. The z is a solution of (C) z (A) z (D) z (B) z

os ( ) + i sin ( ) (C)

os ( ) + i sin ( ) os ( ) + i sin ( )

(D) os ( ) + i sin ( os (

)

) + i sin ( )

EE – 2014 5. Let S be the set of points in the complex plane corresponding to the unit circle. {z: |z| } . Consider the (That is, function f(z)=zz* where z* denotes the complex conjugate of z. The f(z) maps S to which one of the following in the complex plane (A) unit circle (B) horizontal axis line segment from origin to (1, 0) (C) the point (1, 0) (D) the entire horizontal axis

IN – 2006 3. The value of the integral of the complex function 3s  4 f(s) (s  1)(s  2) Along the path s  3is (A) 2j (B) 4j

7.

All the values of the multi-valued complex function , where i √ are (A) purely imaginary. (B) real and non-negative. (C) on the unit circle. (D) equal in real and imaginary parts. Integration of the complex function z

, in the counter clockwise

(C) 6j (D) 8j

IN – 2007 4.

For the function

of a complex variable

z, the point z=0 is (A) a pole of order 3 (B) a pole of order 2 (C) a pole of order 1 (D) not a singularity 5.

6.

Mathematics

Let j = √ (A) √j (B) 1

.Then one value of (C)

is

(D)

IN – 2008 6. A complex variable x+j has its real part x varying in the range to + . Which one of the following is the locus (shown in thick lines) of 1/Z in the complex plane?

direction, around |z 1| = 1, is (A) πi (C) πi (B) (D) 2πi

th

th

th


l xis

(Note:

j m gin ry xis

m gin ry xis

j

l xis

IN – 2009 The value of ∮

where the contour

of integration is a simple closed curve around the origin, is (A) 0 (C) (D) (B) 2πj 8.

If z = x+jy, where x and y are real. The value of | | is (A) 1 (C) (D) (B) √

9.

One of the roots of the equation 𝑥 =j, where j is positive square root of 1, is √ (A) j (C) j +j

(D)

√

)

x

j

√

√ y

pl n

l xis

(B)

z is.

∮

l xis

j

7.

Mathematics

IN – 2010 10. The contour C in the adjoining figure is described by x + y . The value of

m gin ry xis

m gin ry xis

GATE QUESTION BANK

(A) 2πj (B) 2πj

(C) 4πj (D) 4πj

IN – 2011 11. The contour integral ∮ / with C as the counter-clockwise unit circle in the zplane is equal to (A) 0 (C) 2π√ (B) 2π (D)

j

th

th

th


GATE QUESTION BANK

Mathematics


2.

4. [Ans. B] By definition C-R equation holds [Ans. A] f(z)=

has simple pole at z = 0

Residue of f(z) at z = 0 lim z z lim os z ∫ z z 2πi (residue at z = 0) 2πi 2πi 3.

[Ans. B] + 4i + 2i 2i + 2i + i + 4i +4 Modulus = √

[Ans. C] Given u=xy For analytic function u v  x y and

[Ans. A] +i 2 2 i + 2i

6.

dw u v  i dz x x 

7.

∫y x

∮

Replacing x by z and y by 0, we get

∮ 8.

+ 2i + i i

r os x

r sin

r sin

r os

r

r

2π

π 2

[Ans. C] u v x y v 2y y 2y + x v 2 y + x v u v y x 2x x 2x + x 2 x x

z2 C 2 Where C is a constant, z v m0 i + 1 2 Integrating, w  i

 (x2  y2  2ixy)   mi  2   or v 

i

x y

y = r sin x y r os

dw  y  ix dz

Where, z = x + iy dw = izdz

i

[Ans. C]

u u i x y

dw  0  iz dz

i

[Ans. C] +i +i i i +i 2i i + i 2 +i rg ( ) t n ( ) i π⁄ 2

u v  y x

By Milne Thomson method Let w = u + iv

or

5.

+ 2i

y 2  x2 2

th

th

th


GATE QUESTION BANK

v

y y

9.

10.

[Ans. B] z ∫ ln z| z

[Ans. B] ∫s

z os z The poles are at z = n + /2 π = π/2 π/2 + π/2 None of these poles lie inside the unit circle |z| =1 Hence, sum of residues at poles = 0 Singularities set = ϕ and 2πi [sum o r si u s o t z t th poles] 2 πi

[Ans. C] iv n u x y v v v x+ y x y v u v u y x x y u u v x+ y y x 2y x + 2x y rm ont ing y t rms only llow v 2 xy +

3.

z z

ln i

ln

ln + ln i ln ln z os z i z i ln i ln z π i ( 2

ln + ln i + i sin i sin π/2

=

=| ̅̅̅̅

z

z= ∮

pplying z z

(

)

u hy’s int gr l th or m, using i .2πi ( )/

/

i

2πi

Now, ∮

/

ln

z

∮

i.e. ∮

)

[Ans. C] (A) is true since ̅̅̅̅

∫

[Ans. A] u hy’s int gr l th or m is f(a) =

+

CE 1.

2.

x x + onst nt

Mathematics

i .2πi 0( )

1/

i .2πi 0( )

1/

2π

̅̅̅̅

4πi

2π

|

(B) is true by triangle inequality of complex number (C) is not true since | |≥| | | | (D) is true since | + |2 = ( + ) ̅̅̅̅̅̅̅̅̅̅̅̅ + = ( + ) (z̅ + z̅ ) = z̅ + z̅ + z̅ + z̅ i ̅̅̅̅̅̅̅̅̅̅̅̅ And | |2 = ( + )

i

4.

4πi

[Ans. D] z z z z + z z i z+i The singularities are at z = i and –i

z

5.

[Ans. C]

= ( + ) (z̅ z̅ ) = z̅ + z̅ z̅ + z̅ ii Adding (i) and (ii) we get | + |2 + | |2 = 2 z̅ + 2 z̅ = 2| | + 2| |

r

∫

os 2πz 2z z *

2

th

th

∫

+ *z

+

th


GATE QUESTION BANK

in

z

is point with in |z|=1(the

los urv w n us integral theorem and say that

7.

[Ans. B] 2 i z +i Multiplying by conjugates 2 i i +i i 2i + i 2 + + i 2 + i

u hy’s

os 2πz [2πi ( )] wh r z 2 2 z [Notice that f(z) is analytic on all points inside |z| ] 2

[2πi

os 2

π

(

/2 )

Mathematics

]

2πi

6.

[Ans. D] f = u + iv u = 3x2 – 3y2 For f to be analysis, we have CauchyRiemann conditions, u v i x y u v ii y x From (i) we have u v x x y ∫ v

ECE 1.

Given ,

∮

I

2.

x +

x

z

z +4 j| 2

|z

2j   2j  2j 2

[Ans. B] iv n

log

1 y  u  iv  loge  x  iy   log x2  y2  i tan1   2 x Since, u is constant, therefore



x v + x 2 i.e. v = 3x2 + f(x) iii Now applying equation (iii) we get u v y x [ x+

1 1  z  4 z  2jz  2j 2

Pole (0, 2) lies inside the circle |z j|=2 y u hy’s nt gr l ormul

∫ x y

y

[Ans. D]



1 log x2  y2  c 2





x +y Which is represented set of concentric circles.

x

y

3.

[Ans. A] s

∮

y x x By integrating, f(x) = 6yx – 3x2 + K Substitute in equation (iii) v= 3x2 + 6yx – 3x2 + K v yx + K

2πj sum o r si u

Singular points are s = Only s= +1 lies inside the given contour lims  1 f s Residues at s= +1 = S1

lims  1 S1

n

th

∮

th



1 1  S 1 2

s

2



s

th

2πj ( ) 2

πj


GATE QUESTION BANK

4.

[Ans. A] Residue of z=2 is

Mathematics

z

d 2 z  2 f z   z2 dz  d 1  2 1   lim lim   2 3 z2 dz  z  2  z  2 32   z  2 

z+

)

z

+

lim

5.

+

(z

+ F(z 2πj 7.

+

)

[Ans. C] X(z) =

[Ans. D] sin z

Poles are Z= 0, Z =1, Z=2 Residue at Z=0 is lim

2i

Residue at Z =1 is lim

2 i

Residue z =2 is lim

2 i (

)

2 i

ut m m

8.

[Ans. A] z+4 ∮ z + 4z +

2 im 2 i

m 2 i

m

i

iz

log

4

2 2 i + i√ 2

√ 2 √

i

2 i

√

z

i

9.

√

i

i

i

√

log i + log( √ π iz log + i ( 2nπ) 2 +log √ π iz i ( 2nπ) + log 2 π z ( 2nπ) ilog( 2

i

)

[Ans. D] f(z) = + + z z z

∮ ∮(

+

log y i log i π i i 2

x log x i log π 2

⟹y √ √

)

10.

z

z+

z F z 2 π j r si u o Residue at z = 0 ( 2- order )

[Ans. C] z z

∫ 2πj

∫

z

*∫

∫

z+

z

z+

where f (z) =1

11. + z

x

log y

z +

x

√ ty

⟹ log y

( 2 infinite number of complex solutions sin z has infinite no. of complex solutions 6.

z + 4z + z+2 + 2 j will be outside the unit circle o th t int gr tion v lu is ‘z ro’

[Ans. A]

i√

i

iz

z

z

[Ans. C] s z lim

2j z + 2j

4+ j

2πj[ 4 + j

)

th

th

4π

th

+ 2j


GATE QUESTION BANK

12.

EE 1.

[Ans. B] Suppose that z u x y + iv x y is analytic then, u and v satisfy the Cauchy Riemann equation u v u v n x y y x r u xy os x u v sin x x y u v os x y x v sin x

z

z +

∮

z

∮ 2 πi 2.

√ + o / is outside the unit circle is IV quadrant 3.

[Ans. A] z 4 ∮ z +4 |z i| 2 z +4 z 4 z 2i For z 2i Residue at z +2i 4 4 +2i z + 2i +4i t z 2i li insi tz 2i li outsi z 4 o∮ 2πi sum o r si u z +4 2πi 2i 4π

4.

[Ans. B] Let + i √ i Squaring both sides we get +2 i i Equating real and imaginary parts

[Ans. B] Pole (z=i) lies inside the circle. |z-i/2|=1. Hence ∮

z+i z

i

2 πi i , wh r

z z

-

π

2i

[Ans. D] Let + i Since Z is shown inside the unit circle in I quadrant, a and B are both +ve and +

√ ow

2

+ i i

+ Since

Mathematics

+

+

wh n

i

2

2

i √2

+

√

wh n

+ o

+i in

qu

r nt wh n

| | √ in

√(

) +(

+

+

√ √

+

i

i

√2 √2 i i +i ( ) √2 √2 i i

√2

+

i √2

√2 √2 i i i +i + i( ) + √2 √2 √2 √2 i +( ) √2 √2 π π os ( ) + i sin ( ) 4 4

)

+

+

th

th

th


GATE QUESTION BANK

or

x π π os ( ) + i sin ( ) 4 4

5.

[Ans. C] z zz

} n s {z: |z| z All point of s will be mapped on the point (1, 0) 6.

[Ans. B] z log z i log z r l n Non-negative

7.

[Ans. C] ∫ x x

)

lim z

IN 1.

2.

int gr tion

2πi

z+

2

2 √2

o |z | king z |z| |z| z

2

uis o th ir l y 4

2

√2

[Ans. C] z z̅ Multiply both the side by z, we get z z̅ z |z| z |z| |z| wh r is ngl o z |z| since is a real quantity so in order to satisfy above equation has to be real quantity = 1 and , (where n = +2 )

z z

r +

√2 x

|

z lim z+ quir

x

2πi r s (f(a)) where a is a

singularity in contour c |z | r n pol s o z z nly z li s insi |z s(

y x x

|z|

Mathematics

π/2 ⁄

z 3.

[Ans. C]

πi

X X -2 -1 Cx y y (Cx ( -3

[Ans. A] | + i | 2 Radius of the circle is 2 and centre is at + i

3 Cx

y(n) n n y(n) )y(n)) 3s  4 1 2 C3 = F(s)  C3  . CC3 (sC 1)(s  2) s  1 s  2 y(n) 3 3 y(n) dz y(n)

2 + i

By Formula, y y ( ( Since, both n n contour, ) )

xy

 z  a  2j

the poles are enclosed by

therefore Value of integral=2πj + 2 2πj For distance to be min. The point P will be on the line passes through origin and centre of the circle. Slope of line OP = Slope of line OC

4.

πj

[Ans. B] Expand by Laurent series

𝑥 th

th

th


GATE QUESTION BANK

5.

[Ans. D]

10. ⁄

tx

j

⁄

(

⁄

log

log

)

2πj 11.

2πj

2

+

[sum o r si

2 o pol

4πj

j

x +

x +

j

∮

⁄

z

∮( + + + + ) z z 2z z The only pole of z is at z , which lies within |z| ∫ z z 2πi (residue) Note: Residue of z at z is coefficient ⁄ of z i.e. 1, here.

x j x + x

j j x +

lim { x +

j ption

⟹z j j ⟹ 2[ j

[Ans. C] z

x+j

|

z

⁄

[Ans. B] x+j

|

z

∮

)

/

x

7.

⁄

log (

π j 2 π j j 2

log

6.

⁄

log (

z=∮

Pole z j Residue at z

⁄

log x

[Ans. D] ∮

)

Mathematics

s tis y th

ov

}

on itions

[Ans. A] u hy’s int gr l ormul is ∫ Here a = 0, then f(0) = sin 0 = 0

8.

[Ans. D] z x + iy p | |= | = |

9.

| |=

|

|=

[Ans. B] Given x3 = j = e+jπ/2 x

⁄

x

os

π

+ j sin

π

√ +j 2 2

th

th

th


GATE QUESTION BANK

Mathematics

Laplace Transform ME – 2007 1. If F (s) is the Laplace transform of function f (t), then Laplace transform of t

 f (t) dt is 0

(A)

F (s)

(C) sF (s) – f (0)

(B)

F (s) f (0)

(D) CE – 2009 1. Laplace transfrm of the function f(x) = cosh(ax) is (A) (C)

ME – 2009 2.

The inverse Laplace transform of is (A) (B)



1 s s 2



(B) (C) 1 – (D)

ME – 2010 3. The Laplace transform of a function . The function

is

is

(A) (B)

(C) (D)

ME – 2012 4. The inverse Laplace transform of the function F(s)

is given by

(A) (B)

(C) (D)

(D)

ECE - 2005 1. In what range should Re(s) remain so that the Laplace transform of the function exists. (A) (C) (B) (D) ECE – 2006 2. A solution for the differential equation x’(t)+2x(t)= (t) with initial condition x( )=0 is (C) (A) (D) (B) ECE – 2008

ME – 2013 5. The function equation

3.

Consider the matrix P = *

satisfies the differential

value of eP is

and the auxiliary

conditions,

+ . The

(A) *

+

(B) [

]

. The

Laplace transform of

is given by

(A)

(C)

(B)

(D)

ME – 2014 6. Laplace transform of The Laplace transform of

(C) [

]

(D) [ is

.

]

ECE - 2010 4. The trigonometric Fourier series for the waveform f(t) shown below contains th

th

th


GATE QUESTION BANK

Mathematics

ECE – 2013 9. A system is described by the differential equation

=x(t). Let x(t)

be a rectangular pulse given by , Assuming that y(0) = 0 and (A) Only cosine terms and zero value for the dc component (B) Only cosine terms and a positive value for the dc component (C) Only cosine terms and a negative value for the dc component (D) Only sine terms and a negative value for the dc component. 5.

the Laplace transform of y(t) is

Given [

10.

]

then the value of K is (A) 1 (C) 3 (B) 2 (D) 4 ECE– 2011 6.

[

If

]

then the initial

and final values of f(t) are respectively (A) 0, 2 (C) 0, 2/7 (B) 2, 0 (D) 2/7, 0

The maximum value of the solution y(t) of the differential equation y(t) + ̈ with initial condition ̇ and ≥ (A) 1 (C) (B) 2 (D) √

ECE – 2014 11. The unilateral Laplace transform of . Which one of the following is the unilateral Laplace transform of ?

ECE/EE/IN – 2012 7. The unilateral Laplace transform of f(t) is . The unilateral Laplace transform

8.

of t f(t) is (A) –

(C)

(B)

(D)

Consider the differential equation

|

|

The numerical value of (A) (B)

EE – 2005 12. For the equation (t) + 3 (t) + 2x(t) = 5, the solution x(t) approaches which of the following values as t ? (A) 0 (C) 5 (D) 10 (B) EE – 2014

|

is

13.

(C) (D)

Let

be

the

transform of signal x(t). Then, (A) 0 (C) 5 (B) 3 (D) 21 th

th

th

Laplace is


GATE QUESTION BANK

14.

Mathematics

[ Let g: [ be a function [ ] where [x] defined by g(x) represents the integer part of x. (That is, it is the largest integer which is less than or equal to x). The value of the constant term in the Fourier series expansion of g(x) is_______


5.

[Ans. C]

[Ans. A] From definition, We know ∫

2.

Taking Laplace transformation on both sides [ ] [ ] ( ) ( )

[Ans. C]

1 1 1 1    (s  s) s(s  1) s (s  1) 2

(

3.

)

( )

(

[

)

(

) (

[

[Ans. A] [

]

6.

4.

)

]

s and constant

]

[Ans. B] It is the standard result that L (cosh at) =

ECE 1.

[Ans. A] [

[Ans. D] {

]

[Ans. D]

CE 1.

[

)

]

[

Matching coefficient of in numerator we get,

(

]

} 2. {

[Ans. A] ̇ (t) + 2x (t) = (t) Taking Laplace transform of both sides , we get

}

th

th

th


GATE QUESTION BANK

sX(s) X(0) + 2X(s) = 1 1 X(s) = s2 From Inverse Laplace transform gives, we get [ ] 3.

[Ans. D ] eP= [

Mathematics

cosine terms and a negative value of the dc component. 5.

[Ans. D]

[

]

] [

0 1  and P=   2  3 s 1  Where =   2 s+3 s  3 1  1  s  1s  2 2 s s 3    s  1 s  2 =   2  s  1 s  2  

]

[

1

1  s  1 s  2      s s  1s  2 

6.

]

[Ans. B]

Using initial value theorem:

 eP

 2 1  s  1  s  2   2  2  s  1 s  2

=

1 1   s  1 s  2    2 1   s  2 s  1  

=[

⁄

] =2

4.

[Ans. C] Since f(t) is an even function, its trigonometric Fourier series contains only cosine terms

∫

7.

∫

*∫

∫

[Ans. D]

+ t

[

(

)] 8.

[

[Ans. D]

]

Therefore, the trigonometric Fourier series for the waveform f(t) contains only

Taking Laplace transform on both the sides. We have, th

th

th


GATE QUESTION BANK

11.

Mathematics

[Ans. D]

By Laplace transform property, [

]

[

]

(

[

] [

|

] 12.

9.

[Ans. B] =5 By taking Laplace transform

[Ans. B] Writing in terms of Laplace transform

( ⁄

X(s) = (

) 13.

[Ans. B]

(

)

( ( 10.

)

[

)

]

(

)

)

[Ans. D] 14.

[Ans. 0.5]

∫ For t =

∫

|

Value of constant term = 0.5 + sin

√

th

th

th


. GATE QUESTION BANK

EM

Statics ME 1.

ME 2.

2005 Two books of mass 1 kg each are kept on a table, one over the other. The coefficient of friction on every pair of contacting surfaces is 0.3. The lower book is pulled with a horizontal force F. The minimum value of F for which slip occurs between the two books is (A) zero (C) 5.74 N (B) 1.06 N (D) 8.83 N 2006 If point A is in equilibrium under the action of the applied forces, the values of tensions and are respectively

P

F

Q

R

(A) 0.5 F (B) 0.63 F ME 5.

A

(C) 0.73 F (D) 0.87 F

2009 A block weighing 981N is resting on a horizontal surface. The coefficient of friction between the block and the horizontal surface is = 0.2. A vertical cable attached to the block provides partial support as shown. A man can pull horizontally with a force of 100N. What will be the tension, T (in N) in the cable if the man is just able to move the block to the right? T

(A) (B) (C) (D) 3.

ME 4.

600 N 520 N and 300 N 300 N and 520 N 450 N and 150 N 150 N and 450 N

If a system is in equilibrium and the position of the system depends upon many independent variables, the principle of virtual work states that the partial derivatives of its total potential energy with respect to each of the independent variable must be (A) (C) 1.0 (B) 0 (D) ∞

100N

G

(A) 176.2 (B) 196.0 ME 6.

2011 A 1 kg block is resting on a surface with coefficient of friction . A force of 0.8 N is applied to the block as shown in the figure. The friction force is 0.8 N

2008 Consider a truss PQR loaded at P with a force F as shown in the figure. The tension in the member QR is

(C) 481.0 (D) 981.0

1 kg

(A) 0 (B) 0.8 N th

th

(C) 0.98 N (D) 1.2N th



ME

2012 Common Data for Q 7 and Q8: Two steel truss member, AC and BC, each having cross sectional area of 100 mm , are subject to a horizontal force F as shown in figure. All the joints are hinged.

10.

A rigid link PQ is 2 m long and oriented at to the horizontal as shown in the figure. The magnitude and direction of velocity and the direction of velocity are given. The magnitude of (in m/s) at this instant is

A

m s

C

F

(A) 2.14 (B) 1.89

B

7.

If F =1kN, magnitude of the vertical reaction force developed at the point B in kN is (A) 0.63 (C) 1.26 (B) 0.32 (D) 1.46

8.

The maximum force F in kN that can be applied at C such that the axial stress in any of the truss members DOES NOT exceed 100 MPa is (A) 8.17 (C) 14.14 (B) 11.15 (D) 22.30

ME 9.

2014 A block R of mass 100 kg is placed on a block S of mass 150 kg as shown in the figure. Block R is tied to the wall by a massless and inextensible string PQ. If the coefficient of static friction for all surface is 0.4 the minimum force F(in kN) needed to move the block S is

(A) 0.69 (B) 0.88

EM

11.

(C) 1.21 (D) 0.96

A two member truss ABC is shown in the figure. The force (in kN) transmitted in member AB is _______

k

m

12.

A body of mass (M) 10 kg is initially stationary on a 45° inclined plane as shown in figure. The coefficient of dynamic friction between the body and the plane is 0.5. The body slides down the plane and attains a velocity of 20 m/s. The distance travelled (in meter) by the body along the plane is _______

(C) 0.98 (D) 1.37 th

th

th



13.

(A) (B) (C) (D) 15.

friction between the floor and the wardrobe, respectively? y m m

For the truss shown in the figure, the forces and are 9 kN and 3 kN, respectively. The force (in kN) in the member QS is All dimensions are in m

14.

In a statically determinate plane truss, the number of joints (j) and the number of members (m) are related by (A) j m (C) m j (B) m j (D) m j

EM

y

(A) (B) (C) (D) 16.

11.25 tension 11.25 compression 13.5 tension 13.5 compression

490.5 and 0.5 981 and 0.5 1000.5 and 0.15 1000.5 and 0.25

A ladder AB of length 5 m and weight (W) 600 N is resting against a wall. Assuming frictionless contact at the floor (B) and the wall (A), the magnitude of the force P (in Newton) required to maintain equilibrium of the ladder is _______ m

A wardrobe (mass 100 kg, height 4 m, width 2 m, depth 1 m), symmetric about the Y-Y axis, stands on a rough level floor as shown in the figure. A force P is applied at mid-height on the wardrobe so as to tip it about point Q without slipping. What are the minimum values of the force (in newton) and the static coefficient of

m

m

m

th

th

th



EM

Answer Keys & Explanations 1.

[Ans. D] for each surface is same Its free body diagram(m

olving equation (i and (ii , we get and m

g

1 2

3.

[Ans. B] Total potential energy = f(independentvariable) Hence for a system in equilibrium, Total potential energy =Constant Thus, partial derivatives of its total potential energy with respect to each of independent variable must be zero

4.

[Ans. B]

F

N = mg= g N mg g Equation of motion of 1 ×a (i a g towards right Equation of motion of 2 F ( ×a F–( g a (ii towards right Now for relative motion between two blocks N N 1

’ ’

F P

x R

Q x

x b

tan

F

1.732 x

x tan Taking moment about Q f×x × x b

’ mg

mg

a F

g g F > 3×0.3×9.8 F > 8.82 N F = 8.83N 2.

Let force in the member PQ is sin sin Force in member QR cos = 0.634 F

[Ans. A] 5.

[Ans. C] T

A

100 N 600 N

cos sin

cos sin

(i (ii

W=981 N th

th

th



(

Net normal force on block N = W T= 981 – T Frictional force, f Under equilibrium, i.e. when man is just able to move the block

–

=

[Ans. B] ×g × As theoretical frictional force is more than applied force P, Hence, F = P = 0.8N.

10.

sin

[Ans. D] ⃗⃗⃗⃗

cos ( )

( ( cos (r sin

(

sin(

(

(

cos(

cos(

11.

sin( * sin(

+

[Ans. *] Range 18 to 22 100

cos

kN ( sin

g ×

cos

sin

sin sin For F = 1kN ( cos

kN sin(

8.

(r

)

sin y and

× cos

g g

⃗⃗⃗⃗⃗⃗⃗⃗ (

(

(

k

(

[Ans. A] × cos

(

g Using 2 & 3 in 1, ( g = g g

(

7.

(

g Body R

(

6.

EM

sin(

[Ans. B] sin

k

cos

sin( tan

× ×

9.

12.

k

[Ans. *]Range 56 to 59 FBD of body of mass M : -

[Ans. D] g

g ( ody ( ody s

Body S

th

th

th



EM

gsin a

gcos

Body is in equilibrium along perpendicular to the Inclined plane g cos acceleration of body along the plane (a g sin a gsin

a

At joint S

g cos

a gsin g cos a × sin × × cos a m sec nitial velocity of body u final velocity of body m s a= 3.47 m/ sec We know:u as ( ( × ×s × s m × 13.

14.

[Ans. C] m j Perfectly frame truss or statically determinate plane truss m j Deficient frame m j Redundant frame

cos

cos k

k

But direction and are different Hence force in member QS = 11.25 tension 15.

[Ans. A] m m

[Ans. A] At joint P k

g

g ree body diagram

g sin

Now it can be solved by option checking i.e, and for ×

sin × sin

×

k

th

th

th



16.

EM

[Ans. *] Range 399 to 401

(

×

( (moment at point × × × from equation

(

th

th

th



EM

Dynamics ME 1.

2005 A 1 kg mass of clay, moving with a velocity of 10 m/s, strikes a stationary wheel and sticks to it. The solid wheel has a mass of 20 kg and a radius of 1 m. assuming that the wheel and the ground are both rigid and that the wheel is set into pure rolling motion, the angular velocity of the wheel immediately after the impact is approximately. 20kg 10m/s

A shell is fired from cannon. At the instant the shell is just about to leave the barrel, its velocity relative to the barrel is 3m/s, while the barrel is swinging upwards with a constant angular velocity of 2 rad/s. The magnitude of the absolute velocity of the shell is

2rad/s 2m

+

1kg

1m

(A) Zero (B) ⁄ rad/s 2.

3.

(A) 3m/s (B) 4m/s

(C) √ ⁄ rad / s ⁄ rad / s (D)

A elevator (lift) consists of the elevator cage and a counter weight, of mass m each. The cage and the counterweight are connected by a chain that passes over a pulley. The pulley is coupled to a motor. It is desired that the elevator should have a maximum stopping time of t seconds from a peak speed v. If the inertias of the pulley and the chain are neglected, the minimum power that the motor must have is

4.

A simple pendulum of length 5m, with a bob of mass 1 kg, is in simple harmonic motion. As it passes through its mean position, the bob has a speed of 5 m/s. The net force on the bob at the mean position is (A) zero (C) 5 N (B) 2.5 N (D) 25 N

5.

The time variation of the position of a particle in rectilinear motion is given by x t t t. If v is the velocity and a the acceleration of the particle in consistent units, the motion started with (A) v = 0, a = 0 (C) v = 2, a = 0 (B) v = 0, a = 2 (D) v = 2, a = 2

ME 6.

2007 A block of mass M is released from point P on a rough inclined plane with inclination angle shown in the figure below he co – efficient of friction is f tan then the time taken by the block to reach another point Q on the inclined plane, where PQ = s, is

pulley

v

chain

m cage

m

v

Counter weight

(A) (B)

mv

(C) 5m/s (D) 7m/s

(C) (D)

th

th

th



EM

mass of the counterweight, for static balance, is

P

g P

Q

R

Q

(A) √ (B) √

ME 8.

10.

(

(D) √

(

During inelastic collision of two particles, which one of the following is conserved? (A) total linear momentum only (B) total kinetic energy only (C) both linear momentum and kinetic energy (D) neither linear momentum nor kinetic energy

A straight rod of length L(t), hinged at one end and freely extensible at the other end, rotates through an angle (t about the hinge. At time t, L(t)=1m, ̇ (t =1m/s, (t = rad and ̇ (t =1 rad/s.The

ME 11.

2009 A uniform rigid rod of mass M and length L is hinged at one end as shown in the adjacent figure. A force P is applied at a distance of 2L/3 from the hinge so that the rod swings to the right. The reaction at the hinge is

P

(A) √ (B) √ 9.

30°

2L/3

P

V

(A) –P (B) 0

(C) V/2 (D) 2V/√

A cantilever type gate hinged at Q is shown in the figure. P and R are the centers of gravity of the cantilever part and the counterweight respectively. The mass of the cantilever part is 75 kg. The

(C) 225 kg (D) 300 kg

magnitude of the velocity at the other end of the rod is (A) 1 m/s (C) √ m/s (D) 2m/s (B) √ m/s

2008 A circular disc of radius R rolls without slipping at a velocity v. The magnitude of the velocity at point P (see figure) is

R

2.0m

(A) 75 kg (B) 150 kg

(

(C) √

7.

0.5m

(

ME 12.

th

L

(C) P/3 (D) 2P/3

2010 There are two points P and Q on a planar rigid body. The relative velocity between the two points (A) should always be along PQ (B) can be oriented along any direction th

th



(C) should always be perpendicular to PQ (D) should be along QP when the body undergoes pure translation ME 13.

2011 A stone with mass of 0.1 kg is catapulted as shown in the figure. The total force (in N) exerted by the rubber band as a function of distance x (in m) is given by x . If the stone is displaced by 0.1 m from the un-stretched position (x ) of the rubber band, the energy stored in the rubber band is

16.

A circular solid disc of uniform thickness 20mm, radius 200mm and mass 20 kg is used as a flywheel. If it rotates at 600 rpm, the kinetic energy of the flywheel, in Joules is (A) 395 (C) 1580 (B) 790 (D) 3160

ME 17.

2013 A link OB is rotating with a constant angular velocity of 2 rad/s in counter clockwisre direction and a block is sliding radially outward on it with an uniform velocity of 0.75 m/s respect to the rod , as shown in the figure below. If OA = 1 m , the magnitude of the absolute acceleration of the block at location A in m/s is B

Stone of mass 0.1 kg

(A) 0.01 J (B) 0.1 J 14.

ME 15.

A

O

(C) 1 J (D) 10 J

The coefficient of restitution of a perfectly plastic impact is (A) 0 (C) 2 (B) 1 (D) ∞

(A) 3 (B) 4 18.

2012 A solid disk of radius r rolls without slipping on a horizontal floor with angular velocity and angular acceleration . The magnitude of the acceleration of the point of contact on a disc is (A) zero (B) (C) √( (D) r

EM

(C) 5 (D) 6

A pin jointed uniform rigid rod of weight W and length L is supported horizontally by an external force F as shown in the figure below. The force F is suddenly removed. At the instant of force removal, the magnitude of vertical reaction developed at the support is F L

( (A) Zero (B) W/4

th

th

(C) W/2 (D) W

th



ME 19.

20.

2014 A circular object of radius r rolls without slipping on a horizontal level floor with the center having velocity V. The velocity at the point of contact between the object and the floor is (A) Zero (B) V in the direction of motion (C) V opposite to the direction of motion (D) V vertically upward from the floor

EM

21.

A truck accelerates up a 10° incline with a crate of 100 kg. Value of static coefficient of friction between the crate and the truck surface is 0.3. The maximum value unit of acceleration is (m/ ) of the truck such that the crate does not slide down is _______

22.

A mass m of 100 kg travelling with a uniform velocity of 5 m/s along a line collides with astationary mass m of 1000 kg. After the collision, both the masses travel together with the same velocity. The coefficient of restitution is (A) 0.6 (C) 0.01 (B) 0.1 (D) 0

A block weighing 200 N is in contact with a level plane whose coefficients of static and kinetic friction are 0.4 and 0.2, respectively. The block is acted upon by a horizontal force (in Newton) P=10t, where t denotes the time in seconds. The velocity (in m/s ) of the block attained after 10 seconds is _______


2.

[Ans. B] By conservation of linear momentum, mu = (m+M) v v = 10/21 m/s  = v/r = 10/21 1/3 rad/s

= mv

 Power = [Ans. C] r = 2×2 = 4m/sec vnet =√

4.

= 5 m/s

[Ans. A] Force at mean position is zero. rm s

v

mv

Final Kinetic energy = 0 Time duration it occurs = t

3.

[Ans. C]

mv

5m/s

m

v m

Power =

(rate of doing work or rate of

change of K.E. energy of the system) Initial Kinetic energy of the system th

th

th



5.

[Ans. D] Given, x = t v

t t



[Ans. D] For balancing the moments about the hinge Q, 75 × m × m = 300kg

10.

[Ans. D] L(T) = 1m, ̇ (t) = 1m/s, ̇ (t) = rad

,

m s

and a a

6.

9. t t t

m s

EM

[Ans. A]

t=

sec

Increase in length of the rod = L (T) × t = m Mg sin

New length = * Tangential velocity =* + ̇ *

Mg cos

Mg sin a g sin Now s

ut

g cos g cos

a

√

7.

8.

+m s

Radial velocity = 1 m/s Resultant velocity = √

at

*

+

= 2 m/s

But u t

+m

√

11.

(

[Ans. A] In any collision, linear momentum is conserved. Kinetic energy is conserved only in elastic collision whereas in inelastic collision, kinetic energy is converted into heat, sound or other forms of energy.

[Ans. B] A 2L/3

L G

P

a

P

Assume reaction at hinge is R, orque

[Ans. A]

( P

R

)

(

)

V

Linear acceleration,

V

a Magnitude of velocity at point P √ √ √

a

cos

(

)

Governing Equation, = Ma = M (P/M)

cos cos

√

th

th

th



12.

[Ans. C]

13.

[Ans. B]

16.

[Ans. B] ×(

Energy stored = work done ∫

EM

x dx

[

dx

∫

× ×

x ]

rad sec ×(

×

14.

15.

[Ans. A] For, perfectly plastic collision, co – efficient of restitution = 0

17.

Total acceleration f √f f f radial acceleration = r × (m s f tangential acceleration = 2v × × (m s

[Ans. A]

r

18.

(m s

[Ans. B] It undergoes fixed body rotation about 0

W O

Pure translation

Pure rotational

√

f

Motion of a body can be expressed as sum of pure rotation and pure translation. For no slip condition, translation acceleration of centre of mass a r Velocity of centre of mass v r

a v

r r

= O

CG

R a v

a v

[Ans. C]

ma

orque

r r

( )

m

r r

________1 ma

Now superimposing the two motions.

m(

Io

m(r

)

2(W-R) = mL __________2 By 1 and 2, ( a

a

Linear acceleration of bottom point r r Velocity of bottom point Centripetal acceleration of bottom point =0 Net acceleration of bottom point = 0

19.

[Ans. A] No slipping Zero Relative velocity

20.

[Ans. * ] (Range 4.8 to 5.0) Maximum static friction, f

th

th

th



( mg f it starts sliding

EM mg

x

rate

t it starts sliding (just in equilibrium t f f ( Using momentum principal ∫

dt

∫ (

f f

dt t

t

*

f

m(

∫ (

m( dt

t+

[(

]

g

g

m ⁄s [Ans. *] Range 1.0 to 1.3

22.

m g rate

(

(

f f Using 1&2, m a m g sin ( m g cos ( a g cos ( g sin ( [ ] a cos ( sin ( a a m s

g (

m a m gsin( m a m a m g sin ( m gcos ( m cos ( f

(

21.

y

a

f

th

[Ans. D] oefficient of restitution elative velocity of separation elative velocity of approach After collision, both the masses travel together with same velocity. So relative velocity of separation will be zero. Hence, coefficient of restitution will be zero

th

th



SOM

Simple Stress & Strain ME 1.

2005 A uniform, slender cylindrical rod is made of a homogeneous and isotropic material. The rod rest on frictionless surface. The rod is heated uniformly. If longitudinal and thermal stress are represented by nd respectively, then (A) (C) (B) (D)

ME 2.

2006 A bar having a cross-sectional area of 700 mm2 is subjected to axial loads at the positions indicated. The value of stress in the segment QR is: 63 kN

(A) 40 MPa (B) 50 MPa

4.

6.

A steel rod of length L and diameter D, fixed at both ends, is uniformly heated to a temperature rise of  T. The Young’s modulus is E and the co efficient of linear expansion is α. The therm l stress in the rod is (A) 0 (C) E   T (B) ∝∆T (D) E   TL

R

S

(C) 70 MPa (D) 120 MPa

Linked Answer Questions: Q. 7 – Q. 8 A machine frame shown in the figure below is subjected to a horizontal force of 600 N parallel to z – direction.

A steel bar of 40 mm 40 mm square cross-section is subjected to an axial compressive load of 200 kN. If the length of the bar is 2m and E = 200 GPa, the elongation of the bar will be: (A) 1.25 mm (C) 4.05 mm (B) 2.70 mm (D) 5.40 mm

y 500 mm 300 mm

3.

Q

2007 A 200 × 100 × 50 mm steel block is subjected to a hydrostatic pressure of 15 MPa. The Young’s modulus nd Poisson’s r tio of the material are 200 GPa and 0.3 respectively. The change in the volume of the block in mm³ is (A) 85 (C) 100 (B) 90 (D) 110

21 kN

35 kN 49 kN

P

ME 5.

According to Von-Mises distortion energy theory, the distortion energy under three dimensional state of stress is represented by

∅30 mm P

(A) (B)

600N N

Z

7.

X

The normal and shear stresses in MPa at point P are respectively (A) 67.9 and 56.6 (C) 67.9 and 0.0 (B) 56.6and 67.9 (D) 0.0and 56.6

(C) (D)

th

th

th



8.

The maximum principle stress in MPa and the orientation of the corresponding principle plane in degrees are respectively (A) 32.0 and 29.52 (B) 100.0and 60.48 (C) 32.0and 60.48 (D) 100.0and 29.52

ME

2008 Statement for Linked Answer Questions 9 and 10 A cylindrical container of radius R = 1 m, wall thickness 1 mm is filled with water up to a depth of 2 m and suspended along its upper rim. The density of water is 1000kg/m3 and acceleration due to gravity is 10 m/s2. The self-weight of the cylinder is negligible. The formula for hoop stress in a thin – walled cylinder can be used at all points along the height of the cylindrical container.

SOM

11.

A rod of Length L and diameter D is subjected to a tensile load P. Which of the following is sufficient to calculate the resulting change in diameter? (A) Young’s modulus (B) Shear modulus (C) Poisson’s r tio (D) Both Young’s modulus nd she r modulus

ME 12.

2011 A thin cylinder of inner radius 500 mm and thickness 10 mm subjected to an internal pressure of 5 MPa. The average circumferential (hoop) stress in MPa is (A) 100 (C) 500 (B) 250 (D) 1000

ME 13.

2012 A solid steel cube constrained on all six face is heated so that the temperature rises uniformly by ∆T. If the thermal coefficient of the material is α young’s modulus is E nd the Poisson’s r tio is , the thermal stress developed in the cube due to heating is

1mm

2m

(A) –

1m

(B) –

∆

(C) –

∆

(D) –

∆ ∆

2R

9.

The axial and circumferential stress ( , ) experienced by the cylinder wall at mid-depth (1 m as shown) are (A) (10, 10) MPa (C) (10, 5) MPa (B) (5, 10) MPa (D) (5, 5) MPa

10.

If the Young’s modulus nd Poisson’s ratio of the container material are 100GPa and 0.3, respectively, the axial strain in the cylinder wall at mid-depth is (A) 2 × (C) (D) 1.2 × (B) 6 ×

14.

A thin walled spherical shell is subjected to an internal pressure. If the radius of the shell is increased by 1% and the thickness is reduced by 1%, with the internal pressure remaining the same, the percentage change in the circumferential (hoop) stress is (A) 0 (C) 1.08 (B) 1 (D) 2.02

ME 15.

2013 A rod of length L having uniform cross – sectional area A is subjected to a tensile force P as shown in the figure below. If the Young’s modulus of the m teri l varies linearly from E to E along the

th

th

th



length of the rod, the normal stress developed at the section – SS is S

E

19.

A 200 mm long, stress free rod at room temperature is held between two immovable rigid walls. The temperature of the rod is uniformly raised by 250°C. If the Young’s modulus nd coefficient of thermal expansion are 200 GPa and 1× /°C, respectively, the magnitude of the longitudinal stress (in MPa) developed in the rod is ____________.

20.

A steel cube, with all faces free to deform, h s Young’s modulus E Poisson’s r tio ν nd coefficient of therm l exp nsion α. The pressure (hydrostatic stress) developed within the cube, when it is subjected to a uniform increase in temper ture ΔT is given by α ΔT E (A) 0 C – v α ΔT E α ΔT E B D v v

21.

If the Poisson's ratio of an elastic material is 0.4, the ratio of modulus of rigidity to Young's modulus is ____________.

22.

The number of independent elastic constants required to define the stressstrain relationship for an isotropic elastic solid is ____________.

23.

A thin gas cylinder with an internal radius of 100 mm is subject to an internal pressure of 10 MPa. The maximum permissible working stress is restricted to 100 MPa. The minimum cylinder wall thickness (in mm) for safe design must be ____________.

E

P

P L/2

S L

(A)

(C)

(B)

(D)

16.

A longh thin walled cylindrical shell, closed at both the ends, is subjected to an internal pressure. The ratio of the hoop stress (circumferential stress) to longitudinal stress developed in the shell is (A) 0.5 (C) 2.0 (B) 1.0 (D) 4.0

ME 17.

2014 A circul r rod of length ‘L’ nd re of cross-section ‘A’ h s modulus of el sticity ‘E’ nd coefficient of therm l exp nsion ‘α’. One end of the rod is fixed and other end is free. If the temperature of the rod is incre sed by ΔT then (A) stress developed in the rod is E α ΔT and strain developed in the rod is α ΔT (B) both stress and strain developed in the rod are zero (C) stress developed in the rod is zero and strain developed in the rod is α ΔT (D) stress developed in the rod is E α ΔT and strain developed in the rod is zero

18.

SOM

A metallic rod of 500 mm length and 50 mm diameter, when subjected to a tensile force of 100 KN at the ends, experiences an increase in its length by 0.5 mm and a reduction in its diameter by 0.015 mm. The Poisson’s r tio of the rod material is____________. th

th

th



SOM


2.

[Ans. A] If a body is allowed to expand or contract freely with rise or fall in temperature then no stress are introduced in body. i.e. inst ed of nd (but strain is not zero)

Since rod is fixed at both ends, so thermal strain will be zero but there will be thermal stresses. 7.

[Ans. A] Here, Twisting moment, T = 600 × 500 × = 300 Nm Bending momentum, M = 600×300× =180 Nm T = shear stress, Normal stress.

[Ans. A] 63 kN

21 kN

35 kN 49 kN R

Q

P

S

T

kN i. e.

d T d

kN i. e.

kN

kN

.

Stress = =

8.

N m

4.

[Ans. A] PL l EA

. MP

[Ans. B] Maximum principle stress

=4× N m = 40MPa 3.

MP

M d

(

)

√(

√( .

mm

.

√(

[Ans. C]

)

) .

)

.

MP 5.

t n

[Ans. B] WE know that Δ E Δ

t n

[

t n

[ .

. Δ 6.

9.

[Ans. C] Since rod is free to expand, therefore ∆L elong tion Lα∆t str in

. .

]

.

[Ans. B] Pressure at mid-depth ρgh = 10³ 10 1=104 N/m2

mm

∆

]

a=

=

= 5 106 N/m2 =5MPa

∝ ∆t

c=

Thermal stress = E α ∆t th

=

th

=10MPa

th



10.

.

E

Axi l str in t mid depth .

12.

PD tss Change in hoop stress= 2.0202%

[Ans. A] Axi l str in is given by

11.

[Ans. D] For liner state of stress only two modulus of elasticity is required. Independent elastic const = E & from which others can be derived [Ans. B] Circumferential hoop stress Where P = pressure = 5 MPa d = diameter = 1000 mm t = thickness = 10mm

15.

[Ans. A] Do the force balance. Force at section is P. P re ction A Are of cross section P A

16.

[Ans. C] pd t t pd

17.

[Ans. C] Strain = change in length ∝ ΔT Stress = 0 (free to expand, no restraint)

18.

[Ans. *] Range 0.29 to 0.31 l ter l str in longitudi l str in

MP 13.

[Ans. A]

( after

L

L

P

L

K(

Δ

)

L

K(

∝ ΔT

19.

L K[

[Ans. D] D t P . D . t

.

.

.

.

[Ans. *] Range 499 to 501 E ∝ ΔT

)

MP 20.

[Ans. A] Free expansion. No restriction No stress in any direction

21.

[Ans. *]Range 0.35 to 0.36 E B E G G E . G . E

22.

[Ans. *]Range 1.9 to 2.1 Total number of elastic constant are 4 i.e., (E, G, K and )

E K Thermal stress are compressive so ∝ ΔT E P 14.

.

ΔL

∝ ΔT ] L exp nd by t ylor series P K ∝ ΔT ∝ ΔT P K ∝ ΔT E ∝ ΔT ∝ ΔTE P P

)

Δl l

L

Let

SOM

th

th

th



SOM

But total number of independent elastic constants are 2 i.e., [E and ] Other are inter related by the formula KG E G k K G 23.

[Ans. *]Range 1.7 to 1.8 Yield stress (k) in pure shear (torsion) according to von-Mises criteria k

k

where √ un xi l torsion here the yield strength √

. MP developed

F ctor of s fety n

MP .

.

th

th

th



SOM

Shear Force and Bending Moment ME 1.

2005 A cantilever beam carriers the anti symmetric load shown, where w is the peak intensity of the distributed load .Qualitatively ,the correct bending moment diagram for this beam is

ME 3.

2007 In a simply – supported beam loaded as shown below, the maximum bending moment in Nm is 1000 mm 500 mm

100 N

B A

100 mm L

L

(A) 25 (B) 30

(A) ME 4. (B)

(C) 35 (D) 60

2011 A simple supported beam PQ is loaded by a moment of 1 kN-m at the mid-span of the beam as shown is the figure. The reaction forces RP and RQ at supports P and Q respectively are 1kN-m P

(C)

Q

1m

(A) (B) (C) (D)

(D)

2.

A Beam is made up of two identical bars AB and BC by hinging them together at B .The end A is built in (cantilevered) and the end C is simply supported .with the load P acting as shown, the bending moment at A is

ME 5.

P A

B

L

(A) zero (B) PL/2

C L/2

1 kN downward, 1 kN upward 0.5 kN upward, 0.5 kN downward 0.5 kN downward, 0.5 kN upward 1 kN upward, I kN downward

2013 A simply supported beam of length L is subjected to a varying distributed load sin x⁄L Nm , where the distance x is measured from the left support. The magnitude of the vertical reaction force in N at the left support is (A) Zero (C) L (B) L (D) L

L

(C) 3PL/2 (D) Indeterminate th

th

th



SOM


[Ans. C] Here make hit and trial method. Consider first option (A) of moment diagram.

B

2.

[Ans. B] MB = 0 x L – P x L/2 = 0 = P/2 MA = x 2L – P x 3L/2 = PL/2

3.

[Ans. B] 100 N

A L

M = 10 Nm

L

A

B

0.5m

V= At right end B, slope must be zero as there is no shear force at B so option A wrong. Now, consider option (B) Due to symmetric load intensity at left side too the shear force equal to zero so slope at left side must be zero. So option (B) wrong.

1m

Taking moment about A, 100 × 0.5 + 10 = N

30 Nm

1

20 Nm

A

2

Now, consider option (C) Here at both the ends, slope is zero means shear force at both the end is zero and also when we move from right to left, rate of increase of shear force decrease due to triangular shape of load intensity and at middle slope should be maximum and there after decreases so in option (C) all these criterion fulfills. Here (C) is the correct option.

BM diagram

B

Maximum bending moment = = . V = 30 Nm 4.

[Ans. A] 1 kN-m

P

Q

Assume direction as shown is in figure o .. i T king momentum bout M t

3

As far as (D) option is concerned its middle part slope and right most part slope is strictly not agreeing with the load shown.

Since direction of is negative so our assumed direction is wrong so downward.

th

th

th



5.

SOM

[Ans. B]

As loading is symmetric, reactions are going to be same. 2R = ∫ wdx

∫ sin (

) dx

2R = R=

th

th

th



SOM

Stresses in Beams ME 1.

2005 A cantilever beam has the square cross section 10 mm x 10mm. It carries a transverse load of 10N. Considering only the bottom fibers of the beam ,the correct representation of the longitudinal variation of the bending stress is

ME 4.

2008 For the component loaded with a force F as shown in the figure, the axial stress at the corner point P is

P

10N L

10mm 1m

1m

F L-b

10mm

(A) L

60 MPa 2b

(B)

2b

(A)

60 MPa

400 MPa

The transverse shear stress acting in a beam of rectangular cross-section, subjected to a transverse shear load, is (A) variable with maximum at the bottom of the beam (B) variable with maximum at the top of the beam (C) uniform (D) variable with maximum at the neutral axis

6.

An axial residual compressive stress due to a manufacturing process is present on the outer surface of a rotating shaft subjected to bending. Under a given bending load. The fatigue life of the shaft in the presence of the residual compressive stress is (A) Decreased (B) Increased or decreased, depending on the external bending load (C) Neither decreased nor increased (D) Increased

400 MPa

3.

2006 Statement for Linked Answer Questions 2 & 3: A simply supported beam of span length 6 m and 75 mm diameter carriers a uniformly distributed load of 1.5kN/m What is the maximum value of bending moment? (A) 9 kNm (C) 81 kNm (B) 13.5 kNm (D) 125 kNm

(D)

5.

(D)

2.

(C)

2b

(B)

(C)

ME

2b

What is the maximum value of bending stress? (A) 162.98 MPa (C) 625.95 MPa (B) 325.95 MPa (D) 651.90 MPa

th

th

th



ME

2010 Statement for Linked Answer Questions 7 & 8: A massless beam has a loading pattern as shown in the figure. The beam is of rectangular cross – section with a width of 30 mm and height of 100mm 3000 N m B

A

7.

8.

The maximum magnitude of bending stress (in MPa) is given by (A) 60.0 (C) 200.0 (B) 67.5 (D) 225.0

ME 9.

2014 Consider a simply supported beam of length, 50h, with a rectangular crosssection of depth h, and width 2h. The beam carries a vertical point load P, at its mid-point. The ratio of the maximum shear stress to the maximum bending stress in the beam is (A) 0.02 (C) 0.05 (B) 0.10 (D) 0.01

C

2000 mm

SOM

2000 mm

The maximum bending moment occurs at (A) Location B (B) 2675 mm to the right of A (C) 2500 mm to the right of A (D) 3225 mm to the right of A


[Ans. A] The bending moment varies from zero to 10 N-m along the length of the beam from the centre of the beam.

3.

[Ans. A] Bending stress M ximum lo d dist nce or Bending moment ection modulus . Nm . .

4.

[Ans. D] M I y F L b b b F L b b due to bending F b due to xi l force Tot l xi l stress F KL b F b b F L b Fb b b FL Fb Fb F L b b b

10N B

A M

C

10 N M

10 M

M I

y MP

Similarly 2.

MP

[Ans. *] Bending moment (B.M)=w x Where, w=weight /unit length And L= length of rod B. M

. . (x .

Here, L=6m, B. M

)

.

kN

m

th

th

th



5.

SOM

[Ans. D]

N

. . MP

+

Maximum at the central neutral axis

9.

m

[Ans. D] P

6.

[Ans. D] Shaft subjective to compressive load

P

P h

h

For bending, stress variation

h

Max shear stress will be at mid of depth h:-h h

NA

Strong in compression 7.

b

[Ans. C]

F

3000 N/m B

A

C

M x. she r stress A

N

y S. F. equation at any section x from end A. x {for x > 2m} x . m 8.

I

tx

.x

z

h

h

h h h

h

b

h h

P h Max. Bending moment will be at middle of Beam (m): P Ph h m h (y )

x

.

. N. m bending stress M

h

FA y Ib

M x. she r stress

[Ans. B] Maximum bending moment BM

P

he r force

.

M x. Bending stress Ph h h P h M x. she r stress M x. Bending stress

. bd

th

th

th

My I

P h

h P

.



SOM

Deflection of Beams ME 1.

2005 Two identical cantilever beams are supported as shown, with their free ends in contact through a rigid roller. After the load P is applied, the free ends will have

ME 3.

P

(A) equal deflections but not slopes (B) equal slopes but not deflections (C) equal slopes as well as deflections (D) neither equal slopes nor deflections ME 2.

2009 A frame of two arms of equal length L is shown in the below figure. The flexural rigidity of each arm of the frame is EI. The vertical deflection at the point of application of load P is

L

equal

P

equal equal

L

equal

2007 A uniformly loaded propped cantilever beam and its free body diagram are shown below. The reactions are q

ME

(A)

(C)

(B)

(D)

2011 Linked Data Question 4 and 5. A triangular-shaped cantilever beam of uniform-thickness is shown in the figure. The young’s modulus if the m teri l of the beam is E. A concentrated load P is applied at the free end of the beam.

b

l q

P

t

M

1

(A)

M

(B)

M

(C)

M

(D)

M

4.

The maximum deflection of the beam is

th

(A)

(C)

(B)

(D)

th

th



5.

ME 6.

ME 7.

8.

The area moment of inertia about the neutral axis of a cross-section at a distance x measure from the free end is (A)

(C)

(B)

(D)

9.

A force P is applied at a distance x from the end of the beam as shown in the figure. What would be the value of x so th t the displ cement t ‘A’ is equ l to zero? L A

2012 A cantilever beam of length L is subjected to a moment M at the free end. The moment of inertia of the beam cross section about the neutral axis is I and the young’s modulus is E. the m gnitude of the maximum deflection is (A)

(C)

(B)

(D)

SOM

P

P

L

(A) 0.5L (B) 0.25L 10.

2014 A cantilever beam of length, L, with uniform cross-section and flexural rigidity, EI, is loaded uniformly by a vertical load, w per unit length. The maximum vertical deflection of the beam is given by wL wL A C EI EI wL wL D B EI EI

(C) 0.33L (D) 0.66L

The flexural rigidity (EI) of a cantilever beam is assumed to be constant over the length of the beam shown in figure. If a load P and bending moment PL/2 are applied at the free end of the beam then the value of the slope at the free end is P PL

L

PL EI PL B EI A

C D

PL EI PL EI

A frame is subjected to a load P as shown in the figure. The frame has a constant flexural rigidity EI. The effect of axial load is neglected. The deflection at point A due to the applied load P is

L A L P

A B

PL EI PL EI

PL EI PL D EI C

th

th

th



SOM


[Ans. A]

Alternately A

2.

[Ans. A] L

2

q

1

P

B

For BC M P x Strain energy, M dx P L ∫ EI EI For AB, M P L Strain energy, M dy P L ∫ EI EI Total strain energy, P L P L P L EI EI EI

+ = qL .. This is equal to UDL cantilever Plus cantilever having lo d t So Deflection at 2 due to UDL qL EI L EI For zero deflection at 2 point two deflections should equal qL L EI EI qL

P From (i) qL

M 3.

[Ans. D] d y EI dn or M EI

deflection

qL

Moment M =

L

qL

Or Deflection

qL qL

4.

M x

P L

x

∫

M dx M El P. x dx E

LP L . Ebt L

x

PL EI

P L EI

[Ans. D] Strain Energy, ∫

P L

C

L

P

5.

Px

∫

LP . xdx Ebt

LP ⁄ Ebt L PL p Ebt Ebt

[Ans. B] At any distance x

L

h= x

Integrating two time and putting x=L, we get y=

th

th

th



SOM

P h

b

x

L

l L

Area moment of inertia, = x. 6.

B P

[Ans. A]

M = PL

M

M

M L BMD Deflection

EI

ML EI EI Chapter Name: Stresses in beams M

L

d dx

PL

dy dx

PL x EI

dy dx

PL x tx EI

C

slope

(Moment of area of BMD)

7.

PL EI

L

9.

w

tx

8.

d dx

dy dx

C

dy PL | dx EI dy PL | L dx EI PL EI

L

L

PL EI

P(L x)

L

wL EI

EI

[Ans. C] After transferring force P to A We get

[Ans. A]

Δ

A

t nd rd deriv tion

A

Displacement at A only due to point load P

[Ans. D] Net deflection PL EI A moment will act at point B due to load P at A

is given by

(Downward)

Displacement at A only due to bending moment (P(L-x) is given by : _________________ . P L

. L

A Are moment method

moment of B. M. D. re t A EI P L x L Δ upw rd EI Resultant displacement at A is zero whenΔ B l nces Δ Δ

th

th

th



Δ Δ PL P L L EI EI L L x 10.

L

x

L L

L .

SOM

x L

[Ans. B] P

PL b

PL EI ML PL L lope t free end in b is ( ) EI EI PL EI PL PL slope slope slope EI EI PL EI lope t free end in

is

th

th

th



SOM

Torsion ME 1.

2.

2005 A weighing machine consists of a 2 kg pan resting on spring. In this condition, with the pan resting on the spring, the length of the spring is 200mm. When a mass of 20 kg is placed on the pan, the length of the spring becomes 100 mm. For the spring, the un-deformed length and the spring constant k(stiffness) are (A) l₀ mm k N m (B) l₀ mm k N m (C) l₀ mm k N m (D) l₀ mm k N m

Coupling

A

(A) T (B) T

L B

The maximum shear stress in the shaft is T T A C d d T T D B d d

T T

ME 6.

C

(C) T (D) T

3L/4

L/4

C B L

2009 A solid shaft of diameter, d and length L is fixed at both the ends. A torque, T0 is applied at a distance, L/4 from the left end as shown in the figure given below. T0

2d

A

ME 4.

ME 5.

The two shafts AB and BC , of equal length and diameter d and 2d, are made of the same material they are joined at B through a shaft coupling, while the ends A and C are built in (cantilevered). A twisting moment T is applied to the coupling .If T and T represent twisting moments at the ends A and C respectively, then d

ME 3.

diameter is 20mm, free length is 40mm and the number of active coils is 10. If the mean coil diameter is reduced to 10mm, the stiffness of the spring is approximately (A) decreased by 8 times (B) decreased by 2 times (C) increased by 2 times (D) increased by 8 times

T T

2006 For a circular shaft of diameter d subjected to torque T, the maximum value of the shear stress is: T T A C d d T T B D d d

2011 A torque T is applied at the free end of a stepped rod of circular cross-section as shown in the figure. The shear modulus of the material of the rod is G. the expression of d to produce an angular twist t the free end is L/2

L

T 2d

2008 A compression spring is made of music wire of 2mm diameter having a shear strength and shear modulus of 800 MPa and 80 GPa respectively. The mean coil th

d

(A) (

)

(C) (

(B) (

)

(D) (

th

th

) )



ME 7.

2012 A solid circular shaft needs to be designed to transmit a torque of 50N.m. If the allowable shear stress of the material is 140MPa, assuming a factor of safety of 2, the minimum allowable design diameter in mm is (A) 8 (C) 24 (B) 16 (D) 32

ME 8.

SOM

2014 Two solid circular shafts of radii and are subjected to same torque. The maximum shear stresses developed in the two shafts are nd . If then is


2.

[Ans. B] Let initi l length is without lo d nd stiffness is k s F KΔm g k . i g k . ii Solving equ i nd ii , we get mm nd k N m

[Ans. B] T0 1

T1

T T l

L

G

T

d

G

T∝ TL

L

T

T G L L

T2

T

.. i

r

TL T (

L

)

T T ii From equation (i) and (ii) T T T T T T T T

T

[Ans. C] Let T = torque; d = diameter of shaft and m ximum v lue of she r stress

nd T

m x 4.

T0

T ( )

T

T

3L/4 L2

L

d

2

d

L/4 L1

≈

[Ans. C] Angular deflection in bot shafts are same T

3.

5.

T

T

T T she r stress ∝ T Therefore maximum shear stress in the shaft will be due to torque T r l

[Ans. D] Gd k D N k D ( ) k D k ( ) k

( d

) d T d

th

th

th



6.

[Ans. B] 2

T J {

1

G r L TL GJ T L d TL d

T d TL d TL G

d 7.

SOM

d

[

TL ] G

[ Ans. B]

d

T d T

d d . So closest = 16 mm 8.

[Ans. *] Range 7.9 to 8.1 Given Maximum shear stress developed T in sh ft is given by d T T o d d T d d d T d (

)

th

th

th



SOM

Mohr’s Circle ME 1.

2005 The Mohr’s circle of pl ne stress for point in a body is shown .The design is to be done on the basis of the maximum shear stress theory for yielding. Then, yielding will just begin if the designer chooses a ductile material whose yield strength is MP

5.

If the principal stresses in a plane stress problem, are the magnitude of the maximum shear stress (in MPa) will be (A) 60 (C) 30 (B) 50 (D) 20

ME 6.

2010 The state of plane given by x = nd xy = 100 MPa. stress (in MPa) is (A) 111.8 (B) 150.1

MP

(A) 45 MPa (B) 50 MPa

(C) 90 MPa (D) 100MPa

ME 2.

2008 A two dimensional fluid element rotates like a rigid body. At a point within the element, the pressure is 1 unit. Radius of the Mohr’s circle ch r cterizing the st te of stress at the point, is (A) 0.5 unit (C) 1 unit (B) 0 unit (D) 2 unit

3.

A solid circular shaft of diameter 100 mm is subjected to an axial stress of 50 Mpa. It is further subjected to a torque of 10 kNm. The maximum principal stress experienced on the shaft is closest to (A) 41 MPa (C) 164 MPa (B) 82 MPa (D) 204 MPa

ME 4.

2009 A solid circular shaft of diameter d is subjected to a combined bending moment M and torque, T. The material property to be used for designing the shaft using the relation (A) (B) (C) (D)

√M

stress at a point is MP y = 100 MPa The maximum shear (C) 180.3 (D) 223.6

ME 7.

2012 The state of stress at a point under plane stress condition is MP MP nd MP . The r dius of the Mohr’s circle representing the given state of stress in MPa is (A) 40 (C) 60 (B) 50 (D) 100

ME 8.

2014 The state of stress at a point is given by MP MP nd MP . The maximum tensile stress (in MPa ) at the point is ________

T is

ultimate tensile strength (Su) tensile yield strength (Sy) torsional yield strength (Ssy) endurance strength (Se) th

th

th



SOM


[Ans. C]

4.

[Ans. C] Material subjected to torsion fails by shear stresses (Ssy)

5.

[Ans. C] Maximum shear stress,

MP MP 2.

[Ans. B] Since the pressure in fluid is of hydrodynamic type p p p normal stress in all directions is same and shear stress on any plane is zero. Hence r dius of Mohr’s circle is zero.

MP 6.

[Ans. C] MP MP

MP

√(

)

√(

) . MP .

p

7.

p

[Ans. B] √(

3.

[Ans. B]

√(

) )

√ 8.

[Ans. *] Range 8.4 to 8.5

. Maximum principal stress

√(

√( ) √( √

√( )

MP

√

√ . .

. .

th

th

)

)

MP

th



SOM

Strain Energy Methods ME 1.

2007 A stepped steel shaft shown below is subjected to 10 Nm torque. If the modulus of rigidity is 80 GPa, the strain energy in the shaft in N mm is T = 10 Nm

50 mm

25 mm

100 mm

100 mm

(A) 4.12 (B) 3.46 ME 2.

ME 3.

(C) 1.73 (D) 0.86

2013 Two threaded bolts A and B of same material and length are subjected to identical tensile load. If the elastic strain energy stored in bolt A is 4 times that of bolt B and the mean diameter of bolt A is 12 mm, the mean diameter of bolt B in mm is (A) 16 (C) 36 (B) 24 (D) 48

2008 The strain energy stored in the beam with flexural rigidity EI and loaded as shown in the figure is P

P 2L

L

L

(A)

(C)

(B)

(D)


[Ans. C] ∫

Strain energy = T L ( Gj J

J

)

[(

. 2.

P ∫ x dx EI

)] [ N

]

mm

3.

PL

L EI

P L EI

P EI

L

P L EI

[Ans. B] (

[Ans. C] PL EI

Px dx EI

)

(

)

P L E re s me

PL EI

d

d

mm

Total strain energy stored is given by

th

th

th



SOM

Columns & Struts ME 1.

ME 2.

2006 A pin-ended column of length L, modulus of elasticity E and second moment of the cross-sectional area I is loaded centrically by a compressive load P. The critical bucking load( ) is given by (A) P =

(C) P =

(B) P =

(D) P =

2008 The rod PQ of length L with flexural rigidity EI is hinged at both ends. For what minimum force F is it expected to buckle? p

Q F

(A) (B)

(C) √

√

(D)

ME 3.

2011 A column has a rectangular cross- section of 10 mm × 20 mm and a length of 1m. the slenderness ratio of the column is close to (A) 200 (C) 477 (B) 346 (D) 1000

ME 4.

2012 For a long slender column of uniform cross section, the ratio of critical buckling load for the case with both ends clamped to the case with both ends hinged is (A) 1 (C) 4 (B) 2 (D) 8

th

th

th



SOM


2.

[Ans. D] The critical buckling load P for a column having length L, modulus of elasticity E and second moment of cross sectional area I is loaded centrically. Condition: Both ends are having pin joint. i.e., hinged So n = 1 EI EI P n L L

lenderness r tio 4.

.

[Ans. C] Euler’s buckling lo d n El PE L n = 4 for both end clamped (fixed) n= 1 for both end hinged tio

[Ans. B] Since both ends hinged, therefore Le=L Buckling load, W = F

F

Also

W=Fcos 45 EI l EI EI √ ⁄cos L L

F 3.

[Ans. B] lenderness r tio Effective length Le st r dius of gyr tion Least radius of gyration will be along the major axis 20 10

r

√

r

√

l A

⁄

√

.

but l

bh

mm

th

th

th


GATE QUESTION BANK

Thermodynamics

Basic Thermodynamics ME –2005 1. A reversible thermodynamic cycle containing only three processes and producing work is to be constructed. The constraints are (i) there must be one isothermal process, (ii) there must be one isentropic process, (iii) the maximum and minimum cycle pressures and the clearance volume are fixed, and (iv) polytropic processes are not allowed. Then the number of possible cycles are (A) 1 (B) 2 (C) 3 (D) 4 2.

The following four figures have been drawn to represent a fictitious thermodynamic cycle, on the P-V and T-S planes. P

V Figure 1 T

T

S Figure 4

According to the first law of thermodynamics, equal areas are enclosed by (A) Figures 1 and 2 (C) Figures 1 and 4 (B) Figures 1 and 3 (D) Figures 2 and 3 ME –2006 Statement for Linked Answer Questions 3 & 4: A football was inflated to a gauge pressure of 1 bar when the ambient temperature was C. When the game started next day, the air temperature at the stadium was C. Assume that the volume of the football remains constant at 2500 cm3. 3. The amount of heat lost by the air in the football and the gauge pressure of air in the football at the stadium respectively is (A) 30.6 J, 1.94 bar (C) 61.1J , 1.94 bar (B) 21.8J , 0.93 bar (D) 43.7J ,0.93 bar 4.

Gauge pressure of air to which the ball must have been originally inflated so that it would equal 1 bar gauge at the stadium is (A) 2.23 bar (C) 1.07 bar (B) 1.94 bar (D) 1.00 bar

S Figure 2 P

V Figure 3

ME –2007 5. Which of the following relationships is valid only for reversible processes undergone by a closed system of simple compressible substance (neglect changes in kinetic and potential energy)? (A) Q U W (C) T S U W (B) T S U p V (D) Q U p V th

th

th



of the following statements is TRUE at the end of above process? (A) The internal energy of the gas decreases from its initial value, but the enthalpy remains constant (B) The internal energy of the gas increases from its initial value, but the enthalpy remains constant (C) Both internal energy and enthalpy of the gas remain constant (D) Both internal energy and enthalpy of the gas increase

ME –2008 6. In a steady state flow process taking place in a device with a single inlet and a single outlet, the work done per unit mass flow v p , where ∫ v is the specific volume and p is the pressure. The expression for w given above (A) is valid only if the process is both reversible and adiabatic (B) is valid only if the process is both reversible and isothermal (C) is valid for any reversible process (D) is incorrect; it must be

rate is given by w =

w =∫ 7.

p v

A rigid, insulated tank is initially evacuated. The tank is connected with a supply line through which air (assumed to be ideal gas with constant specific heats) passes at 1 MPa, 350C . A valve connected with the supply line is opened and the tank is charged with air until the final pressure inside the tank reaches 1 MPa. The final temperature inside the tank Air supply line

valve

Tank

(A) (B) (C) (D)

8.

is greater than is less than is equal to may be greater than, less than, or equal to , depending on the volume of the tank

Thermodynamics

ME –2009 9. A compressor undergoes a reversible, steady flow process. The gas at inlet and outlet of the compressor is designated as state 1 and state 2 respectively. Potential and kinetic energy changes are to be ignored. The following notation are used: v = specific volume and P = pressure of the gas. The specific work required to be supplied to the compressor for this gas compression process is (C) v P P (A) ∫ P v (D) P v v (B) ∫ v p ME –2011 10. Heat and work are (A) Intensive properties (B) Extensive properties (C) Point functions (D) Path functions 11.

A balloon containing an ideal gas is initially kept in an evacuated and insulated room. The balloon ruptures and the gas fills up the entire room. Which one th

The contents of a well – insulated tank are heated by a resistor of 23 in which 10 A current is flowing. Consider the tank along with its contents as a thermodynamic system. The work done by the system and the heat transfer to the system are positive. The rates of heat (Q), work (W) and change in internal energy ∆U uring the process in kW are

th

th



(A) (B) (C) (D)

Q = 0, W = 2. , ∆U Q 2. , W , ∆U Q = 2. , W , ∆U Q ,W 2. , ∆U

2. 2. 2.3 2.3

ME –2012 Statement for linked answer Questions 12 & 13: Air enters an adiabatic nozzle at 300kPa, 500K with a velocity of 10 m/s. It leaves the nozzle at 100kPa with a velocity of 180m/s. The inlet area is 80cm . The specific heat of air is 1008 J/kg. K 12. The exit temperature of the air is (A) 516K (C) 484 K (B) 532 K (D) 468 K 13.

The exit area of the nozzle in cm is (A) 90.1 (C) 4.4 (B) 56.3 (D) 12.9

ME –2013 14. A cylinder contains 5 m3 of an ideal gas at a pressure of 1 bar. This gas is compressed in a reversible isothermal process till its pressure increases to 5 bar. The work in kJ required for this process is (A) 804.7 (C) 981.7 (B) 953.2 (D) 1012.2 15.

Specific enthalpy and velocity of steam at inlet and exit of a steam turbine running under steady state , are as given below : Specific Velocity(m/s) enthalpy (kJ/kg) Inlet 3250 180 steam condition Exit 2360 5 stean condition The rate of heat loss from the turbine per kg of steam flow rate is 5 kW. Neglecting changes in potential energy of steam, the

Thermodynamics

power developed in kW by the steam turbine per kg of steam flow rate, is (A) 901.2 (C) 17072.5 (B) 911.2 (D) 17082.5 16.

The pressure , temperature and velocity of air flowing in a pipe are 5 bar , 500 K and 50 m/s , respectively. The specific heats of air at constant pressure and at constant volume are 1.005 kJ/kgK and 0.718 kJ/kgK, respectively. Neglect potential energy. If the pressure and temperature of the surrounding are 1 bar 300 K, respectively , the available energy in kJ/kg of the air stream is (A) 170 (C) 191 (B) 187 (D) 213

ME –2014 17. The maximum theoretical work obtainable, when a system interacts to equilibrium with a reference environment, is called (A) Entropy (C) Exergy (B) Enthalpy (D) Rothalpy 18.

In a power plant, water (density = 1000 kg/m ) is pumped from 80 kPa to 3 MPa. The pump has an isentropic efficiency of 0.85. Assuming that the temperature of the water remains the same, the specific work (in kJ/kg) supplied to the pump is (A) 0.34 (C) 2.92 (B) 2.48 (D) 3.43

19.

1.5 kg of water is in saturated liquid state at 2 bar V . m kg, u . kJ kg, h kJ kg .Heat is added in a constant pressure process till the temperature of water reaches 400°C (v = 1.5493 m /kg, u = 2967.0 kJ/kg, h = 3277.0 kJ/kg). The heat added (in kJ) in the process is _______

th

th

th



20.

A reversible heat engine receives 2 kJ of heat from a reservoir at 1000 K and a certain amount of heat from a reservoir at 800 K. It rejects 1 kJ of heat to a reservoir at 400 K. The net work output (in kJ) of the cycle is (A) 0.8 (C) 1.4 (B) 1.0 (D) 2.0

21.

A pure substance at 8 MPa and 400 °C is having a specific internal energy of 2864 kJ/kg and a specific volume of 0.03432 m /kg. Its specific enthalpy (in kJ/kg) is _______

22.

Which one of the following pairs of equations describes an irreversible heat engine? (A)

an

(B)

an

(C)

an

(D)

an

Thermodynamics

23.

A source at a temperature of 500 K provides 1000 kJ of heat. The temperature of environment is 27°C. The maximum useful work (in kJ) that can be obtained from the heat source is _______

24.

A certain amount of an ideal gas is initially at a pressure p and temperature T . First, it undergoes a constant pressure process 1-2 such that T = T /4. Then, it undergoes a constant volume process 2-3 such that T = T /2. The ratio of the final volume to the initial volume of the ideal gas is (A) 0.25 (C) 1.0 (B) 0.75 (D) 1.5

25.

An amount of 100 kW of heat is transferred through a wall in steady state. One side of the wall is maintained at 127°C and the other side at 27°C. The entropy generated (in W/K) due to the heat transfer through the wall is _______

th

th

th



Thermodynamics


2 2 . KPa Gauge pressure on the day of play =194.33 – 101.325 = 93kPa = 0.93bar

[Ans. B] One cycle having constant volume process (i) and one cycle having constant pressure process (ii) can be formed. P

2

3 2

P

1

2

4.

[Ans. C] We know that P V mRT PV mRT V V P T P T P 2.08 bar (absolute) Gauge pressure =2.08 1. =1.07 bar

5.

[Ans. C] For reversible processes, by I law Q U W

V

(i) 3

1 V (ii ) Thermodynamic

Note – cycle producing work is alway’s clockwise cycle. 2.

3.

[Ans. A] The area enclosed by cycle on P-V diagram give net work transfer where as the area enclosed by cycle on T-S diagram gives net heat transfer According to first law of thermodynamics for cycle. Net heat transfer = Net work transfer [Ans. D] P . 2 2 . 2 kPa T 2 K, V 2 . PV m RT 2 . 2 2 2 2 . kg As volume is constant, dW =0 Q U m T . . . J P

. 2

Also by definition of entropy, S Hence, T s 6.

U

W

[Ans. C] For steady flow, W=

v p ∫ The expression is valid only for reversible process. 7.

[Ans. A] The fixed temperature inside the tank is given by γT1 = 1.4 × (350 +273) = 872.2 K = 599.2 C

8.

[Ans. C] Room is insulated so dQ = 0 Room is evacuated so the expansion is free expression. for free expansion dW = 0 According to first law of thermodynamics dQ = dU +dW

pressure on the ay o play T P T

th

th

th



U According to joules law for ideal gas U = f(T) So T T We know H = U + Pv H = U + mRT H = f(T) dH = 0 So both internal energy and enthalpy are constant. 9.

11.

12.

h

15.

)

R

P v ln

P P ln

ln . kJ

[Ans. A] Applying steady flow energy equation, Q

h

gz 2

w 16.

)

V

W

[Ans. A] Heat added to the contents of the system. Q=I R 2 2 W 2. kW Wall is insulated Q = 0 Work done on the system, W = 2.3 kW According first law dQ = dU + dW dU = +2.3 kW

w

[Ans. B] Available energy availability h

h

h

gz

2

is

( T

h

w

2 kW⁄kg

.

A= h

V 2

V 2

2

difference )

T

T S

in

S

. 2

kJ kg

.2 kJ kg

T P R In ] T P = 15.44 kJ/kg A = 201 +1.25 – 15.44 = 187 kJ/kg T S

[Ans. C]

kPa K m⁄s

2

[Ans. A] W

[Ans. D] Heat and work are path functions.

P T V

V 2

V

K

[Ans. D] V

∫ v P

P V T

V (

T V (

T

T 13.

V 2

h T

T

14.

h ∫ v P Q v P P. K. W Since K.E. and P.E are constant

10.

V 2

h

R 2. cm

[Ans. B] h u Pv h u P v v P But dQ = du + Pdv u Q P v h Q p v P v v P h Q v P

W

Thermodynamics

kPa m⁄s

S

17.

[Ans. C]

18.

[Ans. D] W

T [

In

∫ V p

P

P

J⁄kgK

th

th

th



2. 2 kJ kg W

W

19.

22.

[Ans. A] For an irreversible heat engine Q , Also, entropy change for irreversible heat engine Q s T s , ecause its point unction Q Q T T

23.

[Ans. *]Range 399 to 401 For maximum work, engine should be reversible, So, Efficiency of reversible engine

2. 2 .

.

kJ kg

[Ans. *] Range 4155 to 4160

2

u P v h v p s pressure , Q h h

Q

Q 20.

m h

h

Q h 2 . 2 2 kJ kg . 2 2

=n T T

kJ

K 2

2

K

[Ans. C]

.

K

K

Q =2 kJ

W Q

Q =?

W

H

Q

Q =1 kJ K

We know that for reversible heat engine change in entropy is always zero that is S Q Q Q ( ) T T T 2 Q Q . kJ W = Q Q

Q = 2

Q

W 24.

21.

Thermodynamics

.

= . kJ

kJ Q

.

kJ

[Ans. B] or constant pressure proces V V T ( ) T T T V V

2

or constant volume process 2 V V V V V V 25.

.

[Ans. *] Range 80 to 85 2

[Ans. *]Range 3135 to 3140 h u pv 2 . 2 . kJ kg

Q

Q

K

K T

T th

th

kw

th



Thermodynamics

Entropy change of face 1 of wall: Q S .2 T ntropy change o ace 2 o wall Q S . T Entropy change of wall will be zero , due to steady state S S S S S .2 . . S ,means process of heat flow through finite temperature difference is irreversible irreversibility is responsible for entropy generation So, Ṡ . W K

th

th

th



Thermodynamics

Irreversibility & Availability r.

The work done by a closed system in an adiabatic process is a point function. s. A liquid expands upon freezing when the slope of its fusion curve on Pressure-Temperature diagram is negative. (A) R and S (C) Q, R and S (B) P and Q (D) P,Q and R

ME –2005 1. Nitrogen at an initial state of 10 bar, 1m and 300 K is expanded isothermally to a final volume of 2m . The p - v - T relation is p v = RT, where a . The final pressure (A) will be slightly less than 5 bar (B) will be slightly more than 5 bar (C) will be exactly 5 bar (D) cannot be ascertained in the absence of the value of a. ME –2007 2. Which combination of the following statements is correct? p. A gas cools upon expansion only when its Joule-Thomson co-efficient is positive in the temperature range of expansion. q. For a system undergoing a process, its entropy remains constant only when the process is reversible.

ME –2010 3. Consider the following two processes; a. A heat source at 1200 K loses 2500 kJ of heat to a sink at 800 K b. A heat source at 800 K loses 2000 kJ of heat to a sink at 500 K Which of the following statements is true? (A) Process I is more irreversible than Process II (B) Process II is more irreversible than Process I (C) Irreversibility associated in both the processes are equal (D) Both the processes are reversible


[Ans. B] (p

3.

[Ans. B] for irreversible process

)V=RT

For Isothermal Process a a (P )V (P )V V V V a a P P V VV V a a a 2 2 2 2.

i. e.

0

Process I Q 2 2 ∮ . 2 T 2 Process II Q 2 2 ∮ . . T process II is more irreversible than process I

[Ans. A]

th

th

th



Thermodynamics

Properties of Pure Substances ME –2006 1. Given below is an extract from steam tables. Temperature in (bar)

Specific volume(

Enthalpy (kJ/kg)

Saturated liquid

Saturated vapour

Saturated liquid

Saturated vapour

45

0.09593

0.001010

15.26

188.45

2394.8

342.24

150

0.001658

0.010337

1610.5

2610.5

Specific enthalpy of water in kJ/kg at 150 bar and 45 C is (A) 203.60 (C) 196.38 (B) 200.53 (D) 188.45

ME –2014 3. A spherical balloon with a diameter of 10 m, shown in the figure below is used for advertisements. The balloon is filled with helium ( R = 2.08 kJ/kg.K) at ambient conditions of 15°C and 100 kPa. Assuming no disturbances due to wind, the maximum allowable weight (in newton) of balloon material and rope required to avoid the fall of the balloon (R = 0.289 kJ/kg.K) is _______ G T 2

ME –2007 2. Water has a critical specific volume of 0.003155 m3/kg. A closed and rigid steel tank of volume 0.025 m3 contains a mixture of water and steam at 0.1 MPa. The mass of the mixture is 10 kg. The tank is now slowly heated. The liquid level inside the tank (A) will rise (B) will fall (C) will remain constant (D) may rise or fall depending on the amount of heat transferred

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Thermodynamics

Answer Keys and Explanations 1.

2.

[Ans. A] Specific volume in compressed liquid region does not change with pressure as water is incompressible. U h P V = 188.45 – 9.593 x 0.0010 = 188.44 kJ/kg h U P V = 188.44 + 150 x 100 x 0.00101 = 203.6 kJ/kg [Ans. A] V . m ⁄kg V . 2 m m kg Rigid means volume Specific Vol. (ν)=

3.

[Ans. *] Range 5300 to 5330

w w

w

upwar thrust w

w w is

w Vg

Vg

=

constant.

w

.

(

P R

T

) Vg

PVg [ T R

= . 2 m ⁄kg We are given initially a mixture of water and steam at 0.117Pa. After that it is a constant vol.Process. Since it is a constant vol heating process. Point (2) will touch the saturated liquid line and hence the liquid level will rise.

P

r

(

T

R

T

) Vg

]

R g

P

[ R

]

R .

2

.2

]

2.

(2)

[

. 2N N

0.1MPa (1)

V

th

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Thermodynamics

Work, Heat & Entropy ME –2006 1. Match items from groups 1, 2, 3, 4 and 5. Group 1

Group 2 Group 3 When added Differential to the system, is E-Heat G Positive I Exact F Work H Negative J Inexact

3.

The above cycle is represented on T-S plane by (A) T

Group 4 Group 5 Function Phenomenon

K Path L Point

3

M Transient N Boundary

2.

2

1

(A) F – G – J – K – M E–G–I–K–N (B) E – G – I – K – M F–H–I–K–N (C) F – H – J – L – N E–H–I–L–M (D) E – G – J – K – N F–H–J–K–M

S

(B)

T

2

3

1 S

(C) T

A 100 W electric bulb was switched on in a 2.5 m × 3 m × 3 m size thermally insulated room having a temperature of 2 . The room temperature at the end of 24 hours will be (A) 2 (C) (B) (D)

3

1 2

(D)

S T

ME –2007 Common Data Questions: 3 & 4 A thermodynamic cycle with an ideal gas as working fluid is shown below.

3

1

2

P 100 kPa

S

3

4.

If the specific heats of the working fluid are constant and the value of specific heat ratio γ is . , the thermal e iciency % o the cycle is (A) 21 (C) 42.6 (B) 40.9 (D) 59.7

5.

A heat transformer is a device that transfers a part of the heat, supplied to it at an intermediate temperature, to a high temperature reservoir while rejecting the remaining part to a low temperature heat sink. In such a heat transformer, 100 kJ of

PV = Constant 2

100 kPa m

1 V

V

th

th

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heat is supplied at 350 K. The maximum amount of heat in kJ that can be transferred to 400 K, when the rest is rejected to a heat sink at 300 K is (A) 12.50 (C) 33.33 (B) 14.29 (D) 57.14 ME –2008 Common Data for Questions 6, 7 and 8: In the figure shown, the system is a pure substance kept in a piston- cylinder arrangement. The system is initially a twophase mixture containing 1 kg of liquid and 0.03 kg of vapour at a pressure of 100 kPa. Initially, the piston rests on a set of stops, as shown in the figure. A pressure of 200kPa is required to exactly balance the weight of the piston and the outside atmospheric pressure. Heat transfer takes place into the system until its volume increases by 50%. Heat transfer to the system occurs in such a manner that the piston, when allowed to move, does so in a very slow (quasi – static / quasi – equilibrium) process. The thermal reservoir from which heat is transferred to the system has a temperature of . Average temperature of the system boundary can be taken as . The heat transfer to the system is 1kJ, during which its entropy increase by 10 J/K.

6.

At the end of the process, which one of the following situations will be true? (A) superheated vapour will be left in the system (B) no vapour will be left in the system (C) a liquid + vapour mixture will be left in the system (D) the mixture will exist at a dry saturate vapour state

7.

The work done by the system during the process is (A) 0.1 kJ (C) 0.3 kJ (B) 0.2 kJ (D) 0.4 kJ

8.

The net entropy generation (considering the system and the thermal reservoir together) during the process is closest to (A) 7.5 J/K (C) 8.5 J/K (B) 7.7 J/K (D) 10 J/K

9.

A gas expands in a frictionless piston cylinder arrangement. The expansion process is very slow, and is resisted by an ambient pressure of 100kPa. During the expansion process, the pressure of the system (gas) remains constant at 300kPa. The change in volume of the gas is 0.01m3. The maximum amount of work that could be utilized from the above process is (A) 0 kJ (C) 2 kJ (B) 1 kJ (D) 3 kJ

10.

2 moles of oxygen are mixed adiabatically with another 2 moles of oxygen in mixing chamber, so that the final total pressure and temperature of the mixture become same as those of the individual constituents at their initial states. The universal gas constant is given as R. The change in entropy due to mixing, per mole of oxygen, is given by (A) –R ln2 (C) R ln2 (B) 0 (D) R ln4

Atmospheric pressure Piston

g

Stop

System

Specific volume of liquid (vf) and vapour (vg) phases, as well as values of saturation temperatures, are given below. Pressure (kPa)

Vf(m3/kg)

Vg (m3/kg)

100

Saturation Temperature, Tsat ) 100

0.001

0.1

200

200

0.0015

0.002

Thermodynamics

th

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conditions maintained at P and Q. The condition at the P is 150 kPa and 350 K. The temperature at station Q is 300 K. The following are the properties and relations pertaining to air: Specific heat at constant pressure, . kJ kgK Specific heat at constant volume, . kJ kgK Characteristic gas constant,R .2 kJ/kgK. Enthalpy, h T. Internal energy, u T.

ME –2009 11. A frictionless piston-cylinder device contains a gas initially at 0.8MPa and 0.015 m3. It expands quasi-statically at constant temperature to a final volume of 0.030 m3. The work output (in kJ) during this process will be (A) 8.32 (C) 554.67 (B) 12.00 (D) 8320.00 12.

If a closed system is undergoing an irreversible process, the entropy of the system (A) Must increase (B) Always remains constant (C) Must decrease (D) Can increase, decrease or remain constant

ME –2010 13. A mono – atomic i eal gas γ . , molecular weight = 40) is compressed adiabatically from 0.1 MPa, 300 K to 0.2 MPa. The universal gas constant is 8.314 kJ k mol K . The work of compression of the gas (in kJ kg ) is (A) 29.7 (C) 13.3 (B) 19.9 (D) 0 14.

One kilogram of water at room temperature is brought into contact with a high temperature thermal reservoir. The entropy change of the universe is (A) equal to entropy change of the reservoir (B) equal to entropy change of water (C) equal to zero (D) always positive

ME –2011 Common Data Question 15 and 16 In an experimental set-up, airflows between two station P and Q adiabatically. The direction of flows depends on the pressure and temperature

Thermodynamics

15.

If the air has to flow from station P to station Q, the maximum possible value of pressure in kPa at station Q is close to (A) 50 (C) 128 (B) 87 (D) 150

16.

If the pressure at station Q is 50 kPa, the change in entropy (S S ) in kJ/kgK is (A) . (C) 0.160 (B) (D) 0.355

ME –2012 17. An ideal gas of mass m and temperature T undergoes a reversible isothermal process from an initial pressure P to final pressure P . The heat loss using the process is Q. the entropy change ∆S o the gas is (A) mRln( )

(C) mRln( )

(B) mRln( )

(D) Zero

ME –2014 18. A closed system contains 10 kg of saturated liquid ammonia at 10°C. Heat addition required to convert the entire liquid into saturated vapour at a constant pressure is 16.2 MJ. If the entropy of the saturated liquid is 0.88 kJ/kg.K, the entropy (in kJ/kg.K) of saturated vapour is_______

th

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Thermodynamics


2.

3.

4.

[Ans. D] When added to the system Differential Function Phenomenon

Heat

P P

Work

. ( Positive Inexact Path Boundary

Negative Inexact Path Transient

5.

Where

%

[Ans. D] Consider maximum heat transfer to 400K is Q kJ 400

Q Q=100

350

Q 300

Q Q Q Q Q Apply Clausicus equality Q Q Q T T T Q [ Q 6.

.

] kJ

[Ans. A] Given data: Mass of liquid mw = 1kg Mass of vapour mv = 0.03kg Pressure of two-phase mixture = 100 kPa Dryness fraction of the steam x1 =

x1 =

. .

= 0.029

Dryness fraction of the steam at 200 kPa (x2) x v x v . 2 . x . 2 x . that means, superheated vapour will be left in the system.

[Ans. A] The given cycle is Lenoir cycle for which thermal efficiency is given by γ(

)

2 .

[Ans. D] Heat generated by bulb = 100 × 24 × 60 × 60 Joule. Density of the air =1.2 kg/m3 Heat issipate V × (T 20) Volume of the room 2. Cv of air = .717 kJ/kgK ⟹ 100 × 24 × 60 × 60 = 27 × 0.717 × 1000 (T 20) T2 = [Ans. C] Since 3 1 is adiabatic process, it will be represented by a straight line on T s plane. 1 2 is isobaric process, so with decrease of volume, temperature will also decrease. We can observe in the pv diagram that temperature is not constant during any stage hence options (b) & (d) are rejected as temperature is constant during the stage 3 – 1 in both the options which is not possible option (a) is rejected because clockwise process in P v diagram cannot have anticlockwise T-s diagram. Hence the correct option is (c)

.

)

γ = specific heat ratio = 1.4

th

th

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7.

8.

[Ans. D] W. D. by the system W = W. D. in constant volume process + W.D. in constant pressure process W P v P v v . Pv since v . v Where v m v v v xv . . 2 . . . m kg v . . m kg w . 2 . kJ [Ans. C] Net entropy (∆S)univ = (∆S)system + (∆S)surrounding (∆S)univ= 10

9.

10.

R P T P

∫v p

. .

.

. .

14.

[Ans. D] The entropy change of the universe is always positive.

15.

[Ans. B] Since it talks about maximum possible value of pressure hence we take it case of reversible case. T s h v p T P s In ( ) R In ( ) T P Since it is adiabatic reversible process for the maximum possible value of pressure at station Q. T P In ( ) R In ( ) T P 1.005 In (

)

P 16.

13.

. 2

= 2 . kJ⁄kg Negative sign shows compression work.

= 8.317 8.32 kJ Work output . 2 kJ 12.

2

Work done/kg =

[Ans. A] In isothermal process V W P V ln V 103 0.015 ln (

kPa k kPa

K

[Ans. B]

= 0.8

kJ/kg K

mR T T γ ⁄ T P ( ) T P T

Since, ∆Q = 0, therefore S = ∫ 11.

.2

Work done

8.5 J/K

[Ans. C] Maximum amount of work = = 0.01 (300 100) = 0.01 200 = 2 kJ

Thermodynamics

[Ans. D] If a closed system is undergoing an irreversible process, the entropy of the system can increase, decrease or remain constant.

.

)

.2

S ) .

)

kPa

[Ans. C] By using Tds we get: (S

In (

ln (

h

V p, after simplify

T ln ( ) T

P R ln ( ) P

)

ln (

.2

)

=0.160

[Ans. A] Gas constant R ̅ R . R M th

th

th



17.

Thermodynamics

[Ans. B] Q=U+W U=0 Since isothermal process by ideal gas Q=W W = mRT ln ( ) Q

T ∆S T ∆S ∆S

18.

mRT ln ( ) mRln ( )

[Ans. *]Range 6.4 to 6.7 Q T S m .2 2 S S

S

.

. kJ kgK . kJ kgK

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Thermodynamics

Psychrometrics moisture in kg per kg of dry air will be approximately (A) 0.002 (C) 0.25 (B) 0.027 (D) 0.75

ME –2005 1. Various psychrometric processes are shown in the figure below.

5

W(kg/kg)

4 0

3

1 2

t

Process in Figure P. 0 1

Name of the process 1.

Chemical dehumidification Q. 0 2 2. Sensible heating R. 0 3 3. Coooling and dehumidification S. 0 4 4. Humidification with steam injection T. 0 5 5. Humidification with water injection The matching pairs are (A) P 1, Q 2, R 3, S 4, T 5 (B) P 2, Q 1, R 3, S 5, T 4 (C) P 2, Q 1, R 3, S 4, T 5 (D) P 3, Q 4, R 5, S 1, T 2 2.

Water at 42 is sprayed into a stream of air at atmospheric pressure, dry bulb temperature of 40 C and a wet bulb temperature of 20 C. The air leaving the spray humidifier is not saturated. Which of the following statements is true? (A) Air gets cooled and humidified (B) Air gets heated and humidified (C) Air gets heated and dehumidified (D) Air gets cooled and dehumidified

3.

For a typical sample of ambient air (at 35 , 75% relative humidity and standard atmospheric pressure), the amount of

ME –2006 4. The statement concern psychrometric chart. 1. Constant relative humidity lines are uphill straight lines to the right 2. Constant wet bulb temperature lines are downhill straight lines to the right. 3. Constant specific volume lines are downhill straight lines to the right. 4. Constant enthalpy lines are coincident with constant wet bulb temperature lines. Which of the statements are correct? (A) 2 and 3 (C) 1 and 3 (B) 1 and 2 (D) 2 and 4 5.

Dew point temperature is the temperature at which condensation begins when the air is cooled at constant (A) volume (C) pressure (B) entropy (D) enthalpy

6.

A thin layer of water in a field is formed after a farmer has watered it. The ambient air conditions are: temp 2 C and relative humidity 5%. An extract of steam table is given below Temperature Saturation pressure (kPa) 0.10 0.26 0.40 . 0.61 0.87 1.23 1.71 2 2.34 Neglecting the heat transfer between the water and the ground, the water th

th

th



temperature in the field after phase equilibrium is reached equals (A) . (C) . (B) . (D) . ME –2007 7. Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters a cooling and dehumidifying coil with an enthalpy of 85 kJ/kg of dry air and a humidity ratio of 19 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio of 8 grams/kg of dry air. If the condensate water leaves the coil with an enthalpy of 67 kJ/kg, the required cooling capacity of the coil in kw is (A) 75.0 (C) 128.2 (B) 123.8 (D) 159.0 8.

A building has to be maintained at 2 C (dry bulb) and . C (wet bulb). The dew point temperature under these conditions is . C. the outside temperature is 2 C (dry bulb) and an internal and external heat transfer coefficients are 8 W/ m K and 23 W/m K respectively. If the building wall has the thermal conductivity of 1.2 W/mK. The minimum thickness(in m) of wall required to prevent condensation is (A) 0.471 (C) 0.321 (B) 0.407 (D) 0.125

ME –2008 9. Air (at atmospheric pressure) at a dry bulb temperature of 40C and wet bulb temperature of 20C is humidified in an air washer operating with continuous water recirculation. The wet bulb depression (i.e. the difference between the dry and wet bulb temperature) at the exit is 25% of that at the inlet. The dry bulb temperature at the exit of the air washer is closest to (A) 10 oC (C) 25 oC (B) 20 oC (D) 30 oC

10.

Thermodynamics

Moist air at a pressure of 100 kPa is compressed to 500 kPa and then cooled to 35C in an after cooler. The air at the entry to the after cooler is unsaturated and becomes just saturated at the exit of the after cooler. The saturation pressure of water at 35C is 5.628 kPa. The partial pressure of water vapour (in kPa) in the moist air entering the compressor is closest to (A) 0.57 (C) 2.26 (B) 1.13 (D) 4.52

ME –2010 11. A moist air sample has dry bulb temperature o an speci ic humidity of 11.5 g water vapour per kg dry air. Assume molecular weight of air as 28.93. If the saturation vapour pressure o water at is .2 kPa an the total pressure is 90 kPa, then the relative humidity (in %) of air sample is (A) 50.5 (C) 56.5 (B) 38.5 (D) 68.5 ME –2011 12. If a mass of moist air in an airtight vessel is heated to a higher temperature, then (A) specific humidity of the air increases (B) specific humidity of the air decreases (C) relative humidity of the air increases (D) relative humidity of the air decreases ME –2012 13. A room contains 35kg of dry air & 0.5kg water vapor. The total pressure and temperature of air in the room are 100kPa and 2 respectively. Given that the saturation pressure for water at 2 . is 3.17kPa, the relative humidity of the air in the room is (A) 67% (C) 83% (B) 55% (D) 71%

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ME –2013 14. The pressure , dry bulb temperature and relative humidity of air in a room are 1 bar , C and 70% , respectively. If the saturated steam pressure at C is 4.25kPa, the specific humidity of the room air in kg water vapour /kg dry air is (A) 0.0083 (C) 0.0191 (B) 0.0101 (D) 0.0232

Thermodynamics

relative humidity of 65%, the absolute humidity (in gram) of water vapour per kg of dry air is _______ 16.

ME –2014 15. A sample of moist air at a total pressure of 85 kPa has a dry bulb temperature of 30°C (saturation vapour pressure of water = 4.24 kPa). If the air sample has a

Moist air at 35°C and 100% relative humidity is entering a psychrometric device and leaving at 25°C and 100% relative humidity. The name of the device is (A) Humidifier (B) Dehumidifier (C) Sensible heater (D) Sensible cooler


[Ans. B]

Now

2.

[Ans. B] Here, t = , t =2 Water sprayed at temperature = 2 Since, twater spray >tDBT so heating and humidification

From table P 2. kPa P . 2. P . kPa Now we have to find the temperature at which P becomes saturated pressure by interpolation method

3.

[Ans. B]

T

4.

[Ans. A]

5.

.

.2 2

.

.

.

.

. .

[Ans. C] Constant pressure line

7.

[Ans. B] m .h Q Q m h

T dpt

2 . Dew point temp

m .h m . h m . kW

kJ kg

Air is cooled at constant pressure to make unsaturated air to saturated one.

h h .

kJ⁄kg

h

h

6.

Given

.

× ×

[Ans. C] P Partial pressure of water vapour at temperature 2

h

th

th

. kJ at kg th

P



8.

[Ans. B] To avoid condensation in the building, the inside wall temp should be greater than or equal to DPT .

T

11.

Thermodynamics

[Ans. B] . 22

P P

P

.

2

P

. 22

P

.01847= h

h T

2

k

2

18.1849 P P

.2

H.T from inside air to inside wall = H.T from inside wall surface to outside air 2 . . 2 .

. .

.2

9.

10.

2 .

m

.

%

[Ans. D] R H Decreases

T

13.

˚

S

.

.

.

.

2

P . 2 . 22

P . 22

2

P

. 22

P

14.

[Ans. C]

p p

. 22

p p

. 22

. 22 . . . th

.

kPa or

%

p pp

. 22

th

P

P

. 22

1.13kPa

P

. 22

–P

. 22 . 22 P 2.2 . 22 P 2.2 RH . P . at 2 P . kPa

3 1

[Ans. D ] Specific humidity =

2 5.628kPa

.

or

Heating

T

p

.

%

[Ans. B] Assuming that compression is isentropic in air compressors, the process can be described on the T-s diagram. The process in the intercooler is constant pressure.

But

kPa.

.2

. 12.

.

. .

[Ans. C] Wet Bulb Depression at the exit = 0.25 Wet Bulb Depression at the inlet (DBT - WBT)exit = 0.25 ( DBT - WBT)inlet Texit = 0.25(40-20) + 20 = 25

p p p

.

P P

.

. .

P p

P

.2 .2

.

th



15.

[Ans. *]Range 19 to 22 Total pressure = P =85 kPa Saturation vapour pressure of water at dry bulb temperature of P .2 kPa Relative humidity = RH = 65%= 0.65 P R. H. P P . .2 2. kPa P solute humi ity w . 22 P P 2. w . 22 2. . 2 kg o water vapour per kg o ry air w . 2 2 . gram o water vapour per kg o ry air

16.

[Ans. B] It is a dehumidifier RH

Thermodynamics

%

2

T Psychrometric chart

th

th

th



Thermodynamics

Power Engineering (D)

ME –2005 1. In the velocity diagram shown below, u=blade velocity, C=absolute fluid velocity and w=relative velocity of fluid and the subscripts 1 and 2 refer to inlet and outlet. This diagram is for u

p

P

P

u

v

(A) (B) (C) (D) 2.

an impulse turbine a reaction turbine a centrifugal compressor an axial flow compressor

A p-v diagram has been obtained from a test on a reciprocating compressor. Which of the following represents that diagram? (A) p P

P

ME –2006 3. Determine the correctness or otherwise of the following Assertion [a] and the Reason [r]. Assertion [a]: Condenser is an essential equipment in a steam power plant. Reason [r]: For the same mass flow rate and the same pressure rise, a water pump requires substantially less power than a steam compressor. (A) Both [a] and [r] are true and [r] is the correct reason for [a]. (B) Both [a] and [r] are true but [r] is NOT the correct reason for [a]. (C) [a] is true but [r] is false (D) [a] is false but [r] is true.

v

4.

(B)

Determine the correctness or otherwise of the following Assertion [a] and the Reason [r]. Assertion [a]: In a power plant working on a Rankine cycle, the regenerative feed water heating improves the efficiency of the steam turbine. Reason [r]: The regenerative feed water heating raises the average temperature of heat addition in the Rankine cycle. (A) Both [a] and [r] are true and [r] is the correct reason for [a]. (B) Both [a] and [r] are true but [r] is NOT the correct reason for [a]. (C) Both [a] and [r] are false (D) [a] is false but [r] is true.

p P

P v v (C)

p

P

P v

v

th

th

th



ME –2007 5. Which combination of the following statements is correct? The incorporation of reheater in a steam power plant: P: always increases the thermal efficiency of the plant. Q: always increases the dryness fraction of steam at condenser inlet. R: always increases the mean temperature of heat addition. S: always increases the specific work output. (A) P and S (C) P, R and S (B) Q and S (D) P,Q,R and S ME –2008 6. A thermal power plant operates on a regenerative cycle with a single open feed water heater, as shown in the figure. For the state points shown, the specific enthalpies are: h1 = 2800 kJ/kg and h2 = 200 kJ/kg. The bleed to the feed water heater is 20% of the boiler steam generation rate. The specific enthalpy at state 3 is Boiler feed pump

Boiler

Turbine Condenser

1 3

2 Open feedwater heater

(A) 720 kJ/kg (B) 2280 kJ/kg

Condensate extraction pump

(C) 1500 kJ/kg (D) 3000 kJ/kg

ME –2009 Common Data Questions: 7 & 8 The inlet and the outlet conditions of steam for an adiabatic steam turbine are as indicated in the notations are as usually followed

Thermodynamics

h = 3200kJ/kg V = 160m/s = 10m P = 3MPa

h = 2600kJ/kg V = 100m/s = 6m P = 70kPa

7.

If mass flow rate of steam through the turbine is 20kg/s, the power output of the turbine (in MW ) is (A) 12.157 (C) 168.001 (B) 12.941 (D) 168.785

8.

Assume the above turbine to be part of a simple Rankine cycle. The density of water at the inlet to the pump is 1000kg/m3. Ignoring kinetic and potential energy effects, the specific work (in kJ/kg) supplied to the pump is (A) 0.293 (C) 2.930 (B) 0.351 (D) 3.510

ME –2010 Common Data Questions: 9 & 10 In a steam power plant operating on the Rankine cycle, steam enters the turbine at 4MPa, 350 and exits at a pressure of 15 kPa. Then it enters the condenser and exits as saturated water. Next, a pump feeds back the water to the boiler. The adiabatic efficiency of the turbine is 90%. The thermodynamic states of water and steam are given in the table. State Steam: 4MPa, Water: 15 kPa 350 h 225.94 3092.5 h (kJ kg ) h 2599.1 s s(kJ kg K ) 6.5821 0.7549 s 8.0085 v 0.06645 v (m kg ) 0.001014 v 10.02 th

th

th



9.

10.

Thermodynamics

h is specific enthalpy, s is specific entropy and v the specific volume; subscripts f and g denote saturated liquid state and saturated vapour state. The net work output (kJ kg ) of the cycle is (A) 498 (C) 860 (B) 775 (D) 957

13.

The values of enthalpy of steam at the inlet and outlet of a steam turbine in a Rankine cycle are 2800 kJ/kg and 1800 kJ/kg respectively. Neglecting pump work, the specific steam consumption in kg/kW-hour is (A) 3.60 (C) 0.06 (B) 0.36 (D) 0.01

Heat supplied (kJ kg ) to the cycle is (A) 2372 (C) 2863 (B) 2576 (D) 3092

14.

A pump handing a liquid raises its pressure form 1 bar to 30 bar. Take the density of the liquid as 990 kg/ . The isentropic specific work done by the pump in kJ/kg is (A) 0.10 (C) 2.50 (B) 0.30 (D) 2.93

15.

An ideal Brayton cycle, operating between the pressure limits of 1 bar and 6 bar, has minimum and maximum temperatures of 300 K and 1500 K. The ratio of specific heats of the working fluid is 1.4. The approximate final temperatures in Kelvin at the end of the compression and expansion processes are respectively (A) 500 and 900 (C) 500 and 500 (B) 900 and 500 (D) 900 and 900

ME –2011 Statement for linked answer questions: 11 & 12 The temperature and pressure of air in a large reservoir are 400k and 3 bar respectively. A converging-diverging nozzle of exit area 0.005m2 is fitted to the wall of reservoir shown in figure. The static pressure of air at the exit section for isentropic flow through the nozzle is 50kPa. The characteristic gas constant and the ratio of specific heats of air are 0.287 kJ/kgK and 1.4 respectively.

Flow from the reservoir reservoir

Nozzle exit

11.

The density of air in kg/ at the nozzle exit is (A) 0.560 (C) 0.727 (B) 0.600 (D) 0.800

12.

The mass flow rate of air through the nozzle in kg/s is (A) 1.30 (C) 1.85 (B) 1.77 (D) 2.06

ME –2012 16. Steam enters an adiabatic turbine operating at steady state with an enthalpy of 3251.0kJ/kg and leaves as a saturated mixture at 15 kPa with quality (dryness fraction) 0.9. The enthalpies of the saturated liquid and vapour at 15 kPa are hf = 225.94kJ/kg and hg = 2598.3 kJ/kg respectively. The mass flow rate of steam is 10kg/s. Kinetic and potential energy changes are negligible. The power output of the turbine in MW is (A) 6.5 (C) 9.1 (B) 8.9 (D) 27.0

th

th

th



Thermodynamics

ME –2013 Statement for linked answer questions 17 & 18: In a simple Brayton cycle, the pressure ratio is 8 and temperature at the entrance of compressor and turbine are 300 K and 1400 K, respectively. Both compressor and gas turbine isentropic efficiencies equal to 0.8 for the gas , assume a constant value of (specific heat at constant pressure) equal to 1 kJ/kg K and ratio of specific heat as 1.4. Neglect changes in kinetic and potential energies. 17. The power required by the compressor in kW/kg of gas flow rate is (A) 194.7 (C) 304.3 (B) 243.4 (D) 378.5

20.

21.

In an ideal Brayton cycle, atmospheric air (ratio of specific heats, = 1.4, specific heat at constant pressure = 1.005 kJ/kg.K) at 1 bar and 300 K is compressed to 8 bar. The maximum temperature in the cycle is limited to 1280 K. If the heat is supplied at the rate of 80 MW, the mass flow rate (in kg/s) of air required in the cycle is _______

18.

22.

For a gas turbine power plant, identify the correct pair of statements. P. Smaller in size compared to steam power plant for same power output Q. Starts quickly compared to steam power plant R. Works on the principle of Rankine cycle S. Good compatibility with solid fuel (A) P, Q (C) Q, R (B) R, S (D) P, S

23.

Steam with specific enthalpy (h) 3214 kJ/kg enters an adiabatic turbine operating at steady state with a flow rate 10 kg/s. As it expands, at a point where h is 2920 kJ/kg, 1.5 kg/s is extracted for heating purposes. The remaining 8.5 kg/s further expands to the turbine exit, where h = 2374 kJ/kg. Neglecting changes in kinetic and potential energies, the net power output (in kW) of the turbine is _______

24.

In a compression ignition engine, the inlet air pressure is 1 bar and the pressure at the end of isentropic compression is 32.42 bar. The expansion ratio is 8. Assuming ratio of speci ic heats γ as . , the air stan ar efficiency (in percent) is _______

The thermal efficiency of the cycle in percentage (%) is (A) 24.8 (C) 44.8 (B) 38.6 (D) 53.1

ME –2014 19. An ideal reheat Rankine cycle operates between the pressure limits of 10 kPa and 8 MPa, with reheat being done at 4 MPa. The temperature of steam at the inlets of both turbines is 500°C and the enthalpy of steam is 3185 kJ/kg at the exit of the high pressure turbine and 2247 kJ/kg at the exit of low pressure turbine. The enthalpy of water at the exit from the pump is 191 kJ/kg. Use the following table for relevant data. Superheated Steam temperature 500 500 (°C) Pressure (Mpa) 4 8 0.08644 0.04177 ν( /kg) h(kJ/kg) 3446 3399 s(kJ/kg.K) 7.0922 6.7266 Disregarding the pump work, the cycle efficiency (in percentage) is _______

th

The thermal efficiency of an air-standard Brayton cycle in terms of pressure ratio and γ c c is given by

th

r

r

r

r

th

⁄



Thermodynamics


2.

3.

[Ans. B] As u2 and w2>>w1 Velocity of flow is constant throughout the stage and the diagram is symmetrical hence it is diagram of reaction turbine. [Ans. D] At the time of intake, intake valve operates only when the pressure inside the cylinder is lower than the atmospheric pressure. And appreciable large pressure difference at the time of suction starting, causing fluttering at initial. At the time of compression, process followed is polytropic and continues till pressure exceeds the delivery line pressure for opening the delivery valve.

[Ans. A]

5.

[Ans. B] Reheater always increases the dryness fraction of steam at condenser inlet. Always increases the specific work output.

7.

z g

W

= 2600 × 103+ 6 × 9.81+

W

W = 600 × 103+ 39.24 + = 607839.24 × 20 J/s = 12.156 MW. 8.

[Ans. C] Pump work hinlet+q = houtlet+Wpump here, q=0 Wpump = hinlet houtlet W = ∆hpump Reversible adiabatic dq = dh Vdp dh=Vdp WP=∫ v p

[Ans. B] Condenser is on essential equipment in a steam turbine because it is not economical to feed steam directly to the boiler. For same mass flow rate and the same pressure rise, a water pump require very less power because the specific volume of liquid is very less as compare to specific volume of vapour.

4.

6.

c c h z g 2 2 3 (or) 3200 × 10 + 10 × 9.81 + h

h

V p

p kPa

= 2.930 kPa Alternately 1

T Boiler

Turbine

Condenser

2

S

V=

kg

Pump work = p

[Ans. A] h3 =0.2h1 +h2 0.8 = 560 + 160 = 720 kJ/kg

=

p V

kJ kg

= 2.930kJ/kg

[Ans. A] c Q h z g 2 For adiabatic process , Q = 0 h

z g

c 2

W

th

th

th



9.

[Ans. C]

13.

[Ans. A] Specific steam consumption in

1

T

Thermodynamics

kg/kw – hr

Boiler

.

2 4 3

14.

2

15kPa

[Ans. D] Work done

S

Turbine work. h h 2. 2 2. .2kJ kg Actual work=theoretical work x adiabatic = . .2 .2 kJ/kg Pump work =Vdp . . kJ/kg Net work Tur ine work pump work .2 . . kJ⁄kg. 10.

11.

P

P P γ

[Ans. C] Heat supplie Q h h h h pump work 2. 22 . . 2 2. kJ kg [Ans. C] P kPa, T Pressure at exit, P

[Ans. A] ar ar .

)

T

Bar

T ( )

P ( ) P

T

.

( )

T

2

.

16.

√2 h

( )

K

[Ans. B] V

m *h

h

T

√2 V V . m s Mass flow rate, m V . 2 2. kg s

. .

T

. 2 kg m

. V

.

.

th

z g

2

m *h 2

. .

. K

T T

Velocity at exit, V √2

K

Bar

T T

[Ans. D] V

K

S

T 2 . K Density at exit, P RT .2 12.

T T

2

.

(

kJ kg

kJ kg

T

K kPa

Temperature at exit, T T

P

P

2. 15.

ν P

V

+

z g

2

Q +

W

V

h

2

h

h th

kJ kg x hg

h th

THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30 Cross, 10 Main, Jayanagar 4 Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

Page 215


22 . 2 W

17.

. 2 kJ kg h

.

m h . MW

.

22 .

Thermodynamics

T

T T

T T

. 2

2

T .

.

. 2 . %

[Ans. C]

19.

[Ans. *]Range 40 to 42 MPa T

2

T

MPa

2

kPa

S

S

T T T T

h h h

h at pump exit kJ kg h at MPa , kJ kg enthalpy at exit o high pressure tur ine kJ kg h enthalpy at exit of low pressure turbine 22 kJ kg h h at MPa , kJ kg work one heat supplie h h h h h h h h 22

P ( ) P T h h

h h

. T T

T T

k . T

. T P P ṁ

18.

ṁ h

.2 2 K h

h

h

T

T .2 2 . kJ kg

[Ans. A] T P ( ) T P T

. .

.

2.

. .

h h T T

T T

.

h h

2 K

.

[Ans. D] Direct thermal efficiency formula

21.

[Ans. *]Range 105 to 112 In Compression, T

.

h h

h h

. .

mc T m .

W h

th

T

( )

T

.

K

Q 2

.

kg . s kg s

m

q h

%

20.

. T 2. 2 .2 K W

T

.

.

th

th


Page 216


22.

[Ans. A] Gas turbine plant is smaller in size as compared to steam power plant that is why gas turbine is also used in aircraft. It starts very quickly because the highly flammable fuel is used in combustor chamber, after compression process in chamber fuel is supplied to burner.

23.

[Ans.*]Range 7580 to 7582 Net power output =m h h m h h 2 2 2 . 2 2 Net Power output kW

24.

Thermodynamics

2

[Ans. *]Range 59 to 61 2

r r r

V V ,r ,r V V r r V P , ( ) V P

V V (

2. 2

.

)

.

r r

.

. r γr

r .

. .

.

. Hence air st e iciency

.

. .

%

th

th

th


Page 217

GATE QUESTION BANK

Thermodynamics

Refrigeration ME –2005 Statement for Linked Answer Questions 1 and 2: The following table of properties was printed out for saturated liquid and saturated vapour of ammonia. The titles for only the first two columns are available. All that we know is that the other columns (columns 3 to 8) contain data on specific properties, namely, internal energy(kJ/kg), enthalpy (kJ/kg) and entropy(kJ/kg.k). P (kPa) 2 0 20 40

190.2 429.6 587.5 1554.9

88.76 0.3657 179.69 0.7114 272.89 1.0408 368.74 1.3574

89.05 180.36 274.30 371.43

5.6155 5.3309 5.0860 4.8662

1299.5 1318.0 1332.2 1341.0

1418.0 1442.2 1460.2 1470.2

1.

The specific enthalpy data are in columns (A) 3 and 7 (C) 5 and 7 (B) 3 and 8 (D) 5 and 8

2.

When saturated liquid at C is throttled to 2 the quality at exit will be (A) 0.189 (C) 0.231 (B) 0.212 (D) 0.788

3.

the evaporator. The coordinate system used in the figure is 2 3

4

1

(A) (B)

(C) (D)

ME –2008 5. A cyclic device operates between three thermal reservoirs, as shown in the figure. Heat is transferred to/from the cyclic device. It is assumed that heat transfer between each thermal reservoir and the cyclic device takes place across negligible temperature difference. Interactions between the cyclic device and the respective thermal reservoirs that are shown in the figure are all in the form of heat transfer. 300K

500K

1000K

50kJ 100kJ

A vapour absorption refrigeration system is a heat pump with three thermal reservoirs as shown in the figure. A refrigeration effect of 100W is required at 250 K when the heat source available is at 400 K. Heat rejection occurs at 300K. The minimum value of heat required (in W) is

Cyclic device

60kJ

The cyclic device can be (A) a reversible heat engine (B) a reversible heat pump or a reversible refrigerator (C) an irreversible heat engine (D) an irreversible heat pump or an irreversible refrigerator

400K 300K

ME –2009 6.

An irreversible heat engine extracts heat from a high temperature source at a rate of 100kW and rejects heat to a sink at a rate of 50kW. The entire work output of the heat engine is used to drive a reversible heat pump operating between a set of independent isothermal heat reservoirs at C and . The rate

250K

(A) 167 (B) 100 4.

(C) 80 (D) 20

The vapour compression refrigeration cycle is represented as shown in the figure below, with state 1 being the exit of th

th

th



7.

(in kW) at which the heat pump delivers heat to its high temperature sink is (A) 50 (C) 300 (B) 250 (D) 360

9.

The power required for the compressor in kW is (A) 5.94 (C) 7.9 (B) 1.83 (D) 39.5

In an ideal vapour compression refrigeration cycle, the specific enthalpy of refrigerant (in kJ/kg) at the following states is given as: Inlet of condenser: 283 Exit of condenser: 116 Exit of evaporator: 232 The COP of this cycle is (A) 2.27 (C) 3.27 (B) 2.75 (D) 3.75

ME –2014 10. Which one of the following is a CFC refrigerant? (A) R744 (C) R502 (B) R290 (D) R718 11.

ME –2012 Common Data for Questions 8 and 9: A refrigerator operates between 120kPa and 800kPa in an ideal vapor compression cycle with R - 134 a as the refrigerant. The refrigerant enters the compressor as saturated vapor and leaves the condenser as saturated liquid. The mass flow rate of the refrigerant is 0.2 kg/s. properties for R - 134 a are as follows.

Saturated R-134a

P(kPa) T h kJ⁄kg h kJ⁄kg

120 22.32 22.5 237

800 31.31 95.5 267.3

s kJ⁄kg . K s kJ⁄kg . K

0.093 0.95

0.354 0.918

Superheated R-134a P(kPa) 800 T 40 h kJ⁄kg 276.45 s kJ⁄kg . K 0.95

8.

Thermodynamics

A heat pump with refrigerant R22 is used for space heating between temperature limits o 2 an 2 . The heat required is 200 MJ/h. Assume specific heat of vapour at the time of discharge as 0.98 kJ/kg.K. Other relevant properties are given below. The enthalpy (in kJ/kg) of the refrigerant at isentropic compressor discharge is _______ Saturation 20 25 temperature Pressure 0.2448 1.048 P MN m Specific h kJ kg 177.21 230.07 enthalpy h kJ kg 397.53 413.02 Specific entropy

s kJ kg. K s kJ kg. K

0.9139 1.7841

1.1047 1.7183

12.

A reversed Carnot cycle refrigerator maintains a temperature o . The ambient air temperature is 35°C. The heat gained by the refrigerator at a continuous rate is 2.5kJ/s. The power (in watt) required to pump this heat out continuously is _______

13.

Two identical metal blocks L and M (specific heat = 0.4 kJ/kg.K), each having a mass of 5 kg, are initially at 313 K. A reversible refrigerator extracts heat from block L and rejects heat to block M until the temperature of block L reaches 293 K. The final temperature (in K) of block M is _______

The rate at which heat is extracted, in kJ/s from the refrigerated space is (A) 28.3 (C) 34.4 (B) 42.9 (D) 14.6

th

th

th



Thermodynamics


2.

3.

[Ans. D] h = u + pv; h is always greater than u, but just greater than s. h 5 & 8; u 3 and 7; s 4 and 6

6.

[Ans. C] High

50 kW low

2

300K

Q1= 300 kW Alternately Work output irreversible heat engine, W= = 100 – 50 =50 kW Reversible heat pump (348K)

3

Q

Q3

Q2

K

C.O.P

2 W

400 K

2

W =Q1 Q2 =100 50 =50kW

[Ans. C] From question, since refrigeration effect of 100 W is required 2 So, Work o taine 2 2

Pump Q

Engine

[Ans. B] When saturated liquid at 40 is throttled to -20 . The enthalpy remains same. i.e. 37.43 = 89.05 + x(1418 – 89.05) x = 0.212

⟹ Work o taine

QK

100 kW

W = 50 KW

HP Q

WORK Q1 1

(290 K)

250K

Also for irreversible heat engine Q W Q T ∆S T ∆S or ( T T ∆S or (348 -2 ∆S

Now for this amount of work, heat is absorbed from reservoir 3 and rejected to sink 2. W Q ⟹Q W 2 W W 4.

[Ans. D]

5.

[Ans. A] Since heat is taken from the high temperature sources and rejected to low temperature sink, hence the device is a heat engine not heat pump. Since, the temperature differences are negligible, the engine is reversible.

∆ S= Rate of heat rejection, Q T ∆S = 348 ×

th

th

kW

th



7.

[Ans. A]

p

10.

3

[Ans. C] R-744 R

2 2

1

2 work done by the compressor 2 condenser heat rejected at constant pressure throtlling heat addition in evaporator Given: h2 = 283 kJ/kg h3 = 116 kJ/kg = h4 (from ph curve) h1 = 232 kJ/kg We know esire e ect OP work input cooling e ect work one y compressor h h h h 2 2 2.2 2 2 2

9.

[Ans. C]

O 2

2

R 2 R m n P m 2 n | ⌋P m n n + p + q=2 m+2 q H l q H so no chloro loro car on R 2 R m n P n m | ⌋P 2 n m n p q 2m 2 2 q 2 2 q So p & q is present so H l So chlorine and Fluorine is present So best option is C

h

[Ans. A]

M. W 700 + 44

2

4

8.

Thermodynamics

11.

[Ans. *]Range 430 to 440 T 2 2a

2

T 3

S

4

Process 2-3 is isobaric (constant process) T s h v p T s h h s T Applying between 2 and 2a , h ∫ s ∫ T

1 S .2kg⁄sec

h 2 kJ⁄kg m s . kJ⁄kgK h h . kJ⁄kgK s s . kJ⁄kgK h 2 . kJ⁄kgK NRE(kW) m kg⁄sec h h kJ⁄kg = 0.2(273 – 95.5) = 28.3 kW W KW m kg⁄sec h h kJ⁄kg .2 2 . 2 . kW

S

S

∫

(Assume vapour as ideal

gas between 2 and 2a) T S S p ln T S S S at 2 th

th

th

.

kJ kg. K



S

S at 2 .

T h h h 12.

kJ kg. k T . ln T T . ln 2 2 . T . 2 kJ kg . 2

.

.

. k h p T h at 2 . 2 .

. kJ kg

[Ans. *]Range 370 to 375 OP

= 2

2 Power 13.

Thermodynamics

.

2 Watt

2. Power input

[Ans. *]Range 333 to 335 M

to

L

to 2

Heat rejected by block L = heat gained by block M M 2 M T M M 2 T T K

th

th

th



Thermodynamics

Internal Combustion Engines ME –2005 Common Data for Questions 1 & 2: In two air standard cycles - one operating on the Otto and the other on the Brayton cycle - air is isentropically compressed from 300 to 450 K. Heat is added to raise the temperature to 600K in the Otto cycle and to 550K in the Brayton cycle. 1. If and are the efficiencies of the Otto and Brayton cycles, then (A) =0.25, =0.18 (B) = =0.33 (C) =0.5, =0.45 (D) it is not possible to calculate the efficiencies unless the temperature after the expansion is given 2.

If and are work outputs per unit mass, then (A) W >W (B) W
ME –2006 3. Group I shows different heat addition processes in power cycles. Likewise, Group II shows different heat removal processes. Group III lists power cycles. Match items from Groups I, II and III. Group I Group II Group III P. Pressure S. Pressure 1. Rankine constant constant cycle Q. Volume T. Volume 2. Otto constant constant cycle R. U. 3. Carnot Temperature Temperature cycle constant constant 4. Diesel cycle 5. Brayton cycle

(A) (B) (C) (D)

P – S – 5, R – U – 3, P – S – 1, Q – T – 2 P – S – 1, R – U – 3, P – S – 4, P – T – 2 R – T – 3, P – S – 1, P – T – 4, Q – S – 5 P – T – 4, R – S – 3, P – S – 1, P – S – 5

ME –2007 4. The stroke and bore of a four stroke spark ignition engine are 250 mm and 200 mm respectively. The clearance volume is 0.001 m3. If the specific heat ratio   1.4, the air-standard cycle efficiency of the engine is (A) 46.40% (C) 58.20% (B) 56.10% (D) 62.80% ME –2008 5. Which one of the following is NOT a necessary assumption for the airstandard Otto cycle? (A) All processes are both internally as well as externally reversible. (B) Intake and exhaust processes are constant volume heat rejection processes. (C) The combustion process is a constant volume heat addition process. (D) The working fluid is an ideal gas with constant specific heats. ME –2009 6. In an air-standard Otto cycle, the compression ratio is 10. The condition at the beginning of the compression process is 100kPa and 2 C. Heat added at constant volume is 1500 kJ/kg, while 700 kJ/kg of heat is rejected during the other constant volume process in the cycle. Specific gas constant for air = 0.287kJ/kgK. The mean effective pressure (in kPa) of the cycle is (A) 103 (C) 515 (B) 310 (D) 1032

th

th

th



ME –2010 7. A turbo – charged four – stroke direct injection diesel engine has a displacement volume of 0.0259 (25.9 litres). The engine has an output of 950 kW at 2200 rpm. The mean effective pressure(in MPa)is close to (A) 2 (C) 0.2 (B) 1 (D) 0.1 ME –2011 8. The crank radius of a single – cylinder I. C. engine is 60 mm and the diameter of the cylinder is 80 mm. The swept volume of the cylinder in is (A) 48 (C) 302 (B) 96 (D) 603

Thermodynamics

ME –2014 9. In an air-standard Otto cycle, air is supplied at 0.1 MPa and 308 K. The ratio o the speci ic heats γ an the speci ic gas constant (R) of air are 1.4 and 288.8 J/kg.K, respectively. If the compression ratio is 8 and the maximum temperature in the cycle is 2660 K, the heat (in kJ/kg) supplied to the engine is _______ 10.

A diesel engine has a compression ratio of 17 and cut-off takes place at 10% of the stroke. Assuming ratio of specific heats as 1.4 the air-standard efficiency (in percent) is ________


[Ans. B] Efficiency of both the cycle i.e. for otto and brayton is given by

,

5.

[Ans. B] Intake process is the constant volume heat addition process Hence (B) is not the right assumption

6.

[Ans. D] Here , P1 = 100kPa T1 = 27

. Q

3

2

T

4 1

= 10

Const. Volum e

Cp =0.287 kJ/kgK Pm (V1 V2) = work done = h1 h2 = 1500 700 = 800 kJ

Q S

2.

1500kJ 3

[Ans. A] W

.

W

.

4

kJ kg kJ kg

P 700kJ

2

1

3.

[Ans. A]

4.

[Ans. C] Here, = 0.001 m3 = d2 l= (0.2)2 (0.25)=7.85×

V

Compression ratio, r = 1+

Now P1V1 =mRT1 or 100 × 103 × V1 = 1 × 0.287 × 103 × 300 or V1 = 0.861 V2 = 0.0861 Now Pm(0.861-0.0861) = 800 Pm =1032 kPa

= 8.85

= 58.20% th

th

th



Alternately Air standard otto cycle 3

Thermodynamics

TV

TV

V ( ) V

T T T

. P

2

T . K T T 2 Heat supplied, Q V V R T T γ 2 . 2 . = . kJ kg

4

1

V

V

V

Given:

=

.

10.

K T

T

.

[Ans. *]Range 58 to 62 P

.86

2

. Now V V V Work one, ∆ h

. m kg kJ kg

P = 7.

2.

.

V

kPa

Cutoff = V V Stroke V V cut o % o stroke V V . V V V V . V V

[Ans. A] Power =

,

kw

As four stroke n = = 1100rpm P = =2 8.

.

.

106 N/m2 = 2MPa

[Ans. D] Swept volume

o ratio

V V

.2 2

r r

9.

cut

.

ompression ratio

r

V V .

[Ans. *]Range 1400 to 1420

2.

P

[

r .

2

.

]

γ *

2. . 2. .

.

+

V

th

th

th



Theory of Machines

Mechanisms ME –2005 Common Data for Questions 1. 2& 3: An instantaneous configuration of a fourbar mechanism, whose plane is horizontal, is shown in the figure below. At this instant, the angular velocity and angular acceleration of link O A are ω=8 rad/s and = , respectively, and the driving torque( )is zero. The link O A is balanced so that its centre of mass falls at O .

A ω α τ o

Match the items in columns I and II Column I Column II (P) Higher kinematic (1) Grubler’s pair equation (Q) Lower kinematic (2) Line contact pair (R) Quick return ( ) Euler’s mechanism equation (S) Mobility of a (4) Planer linkage (5) Shaper (6) Surface contact (A) P – 2 Q – 6 R – 4 S – 3 (B) P – 6 Q – 2 R – 4 S – 1 (C) P – 6 Q – 2 R – 5 S – 3 (D) P – 2 Q – 6 R – 5 S – 1

7.

The number of inversions for a slider crank mechanism is: (A) 6 (C) 4 (B) 5 (D) 3

8.

For a four-bar linkage in toggle position, the value of mechanical advantage is: (A) 0.0 (C) 1.0 (B) 0.5 (D) ∞

mm

o mm

Which kind of 4 bar mechanism is the O2ABO4? (A) Double crank mechanism (B) Crank rocker mechanism (C) Double rocker mechanism (D) Parallelogram mechanism

2.

At the instant considered, what is the magnitude of the angular velocity of O B? (A) 1 rad/s (C) 8 rad/s (B) 3 rad/s (D) rad/s

4.

6.

mm

1.

3.

ME – 2006 5. In a four-bar linkage, S denotes the shortest link length, L is the longest link length, P and Q are the lengths of other two links. At least one of the three moving links will rotate by if (A) S + L ≤ P + Q (C) S + P ≤ L + Q (B) S + L > P + Q (D) S + P > L + Q

At the same instant, if the component of the force at joint A along AB is 30 N, then the magnitude of the joint reaction at O2 (A) is Zero (B) is 30 N (C) is 78 N (D) Cannot be determined from the given data The number of degrees of freedom of a planar linkage with 8 links and 9 simple revolute joints is (A) 1 (C) 3 (B) 2 (D) 4

ME – 2007 Linked Answer Questions: Q.9 - Q.10 A quick return mechanism is shown below. The crank OS is driven at 2 rev/s in counter – clockwise direction.

th

th

th



freedom of the mechanism, Grubler’s criterion is (A) 0 (C) 2 (B) 1 (D) 3

R

Q

S O 500 mm

If the quick return ratio is 1:2, then the length of the crank in mm is (A) 250 (C) 500 (B) 250√ (D) 500√

10.

The angular speed of PQ in rev/s when the block R attains maximum speed during forward stroke (stroke with slower speed) is (A) ⁄ (C) 2 (B) ⁄ (D) 3

11.

using

ME – 2009 13. Match the approaches given below to perform stated kinematics / dynamics analysis of machine Analysis Approach P. Continuous relative rotation Q. Velocity and acceleration R. Mobility S. Dynamic-static analysis (A) P Q (B) P Q (C) P Q (D) P Q

P

9.

Theory of Machines

14.

The input link O2P of a four bar linkage is rotated at 2 rad/s in counter clockwise direction as shown below. The angular velocity of the coupler PQ in rad/s, at an instant when O4O2P = 1800, is

1.

R R R R

D’ Alembert’s principle 2. Grubler’s criterion 3. Grashoff’s law 4. Kennedy’s theorem S S S S

A simple quick return mechanism is shown in the figure. The forward to return ratio of the quick return mechanism is 2:1. If the radius of the crank O1P is 125mm, then the distance ‘d’(in mm) between the crank centre to lever pivot centre point should be Q

O

O P

PQ = O4Q = √ a O2P = O2O4 = a

d O

P O2

(A) 4 (B) 2√

(A) 144.3 (B) 216.5

O

(C) 240.0 (D) 250.0

(C) 1 (D)

√

ME – 2008 12. A Planar mechanism has 8 links and 10 rotary joints. The number of degrees of

ME – 2010 15. For the configuration shown, the angular velocity of link AB is 10 rad/s counterclockwise. The magnitude of the relative sliding velocity (in ms ) of slider B with respect to rigid link CD is

th

th

th


. GATE QUESTION BANK D B

A

(A) 0 (B) 0.86 16.

AB = 250 BC = 250√ AC = 500

60°

C

ME – 2011 19. For the four-bar linkage shown in the figure, the angular velocity of link AB is 1 rad/s. The length of link CD is 1.5 times the length of link AB. In the configuration shown, the angular velocity of link CD in rad/s is C B

(C) 1.25 (D) 2.50

Which of the following statements is INCORRECT (A) Grashof’s rule states that for a planar crank – rocker four bar mechanism, the sum of the shortest and longest link lengths cannot be less than the sum of the remaining two link lengths. (B) Inversions of a mechanism are created by fixing different links one at a time. (C) Geneva mechanism is an intermittent motion mechanism. (D) Gruebler’s criterion assumes mobility of a planar mechanism to be one.

17.

Mobility of a statically indeterminate structure is (A) ≤ (C) 1 (B) O (D) ≥

18.

There are two points P and Q on a planar rigid body. The relative velocity between he two points (A) should always be along PQ (B) can be oriented along any direction (C) should always be perpendicular to PQ (D) should be along QP when the body undergoes pure translation

Theory of Machines

1 rad/s

A

D

(A) 20.

(C) 1

(B)

(D)

A double – parallelogram mechanism is shown in the figure. Note that PQ is a single link. The mobility of the mechanism is P

(A)

Q

(B) 0

(C) 1

(D) 2

ME – 2012 21. A solid disc roll without slipping on a horizontal floor with angular velocity and angular acceleration . The magnitude of the acceleration of the point of contact on the disc is (A) Zero (B) rα (C) √(rα) + (rω) (D) rω 22.

th

In the mechanism given below, if the angular velocity of the eccentric circular disc is 1 rad/s, the angular velocity (rad/s) of the follower link for the instant shown in the figure is th

th



Note :All dimensions are in mm

fixed link, the input link, the coupler and output link respectively. Which one of the following statements is true about the input and output links? (A) Both links can execute full circular motion (B) Both links cannot execute full circular motion (C) Only the output link cannot execute full circular motion (D) Only the input link cannot execute full circular motion

25

5

45 (A) 0.05 (B) 0.1

(C) 5.0 (D) 10.0

ME – 2013 23. A planar closed kinematic chain is formed with rigid linksPQ =2.0 m, QR = 3.0 m, RS =2.5 m and SP =2.7 m with all revolute joints. The link to be fixed to obtain a double rocker (rocker – rocker ) mechanism is (A) PQ (B) QR (C) RS (D) SP ME – 2014 24. A slider crank mechanism has slider mass of 10 kg, stroke of 0.2 m and rotates with a uniform angular velocity of 10 rad/s. The primary inertia forces of the slider are partially balanced by a revolving mass of 6 kg at the crank, placed at a distance equal to crank radius. Neglect the mass of connecting rod and crank. When the crank angle (with respect to slider axis) is 30°, the unbalanced force (in Newton) normal to the slider axis is _______ 25.

An offset slider-crank mechanism is shown in the figure at an instant. Conventionally, the Quick Return Ratio (QRR) is considered to be greater than one. The value of QRR is _______ mm

27.

A rigid link PQ of length 2 m rotates about the pinned end Q with a constant angular acceleration of 12 rad/ s . When the angular velocity of the link is 4 rad/s, the magnitude of the resultant acceleration (in m/s ) of the end P is _______

28.

Consider a rotating disk cam and a translating roller follower with zero offset. Which one of the following pitch curves, parameterized by t, lying in the interval to π is associated with the maximum translation of the follower during one full rotation of the cam rotating about the center at (x, y)=(0, 0) ? (A) x (t ) = cos t y (t ) = sin t (B) x (t ) = cos t y (t ) = sin t (C) x(t) = + cost y (t) =

sin t

(D) x(t) = + cos t y (t) = sin t 29.

mm mm

26.

Theory of Machines

A slider-crank mechanism with crank radius 60 mm and connecting rod length 240 mm is shown in figure. The crank is rotating with a uniform angular speed of 10 rad/s, counter clockwise. For the given configuration, the speed (in m/s) of the slider is _______

A 4-bar mechanism with all revolute pairs has link lengths l = 20 mm, l = 40 mm, l = 50 mm and l = 60 mm. The suffixes 'f', 'in', 'co' and 'out' denote the th

th

th



30.

Theory of Machines

A uniform slender rod (8m length and 3kg mass) rotates in a vertical plane about a horizontal axis 1m from its end as shown in the figure. The magnitude of the angular acceleration (in rad/S ) of the rod at the position shown is

m

m


[Ans. B] Length of link AB =

3.

[Ans. B] 30 N

) + ( ) IAB = 260mm I + I I + I Since O2O4 is fixed link. Hence it will act as crank – rocker mechanism. If s + l < p +q and link adjacent to shortest link is fixed, then it is a crank – rocker mechanism. √(

2.

B

4

O2

I24

=

1

=

O = 144mm ω I I = = ( ω I I ω =

8

I1 O 44

O

+ 8

)

ω =

ω = 8

8

ma = mrω

5.

[Ans. A] According to Grashoff’s riteria. S + L ≤ P +Q

6.

[Ans. D]

7.

[Ans. C] There are four number of inversions for slider crank mechanism.

8.

[Ans. D] Mechanical advantage Load (output force) = Effort (input force) For a four bar linkage on toggle position, effort= 0 ∴ Mechanical Advantage = ∞

I2

I12

C

O

[Ans. C] Number of degree of freedom, n = (l ) J h ) =( ) ( =

4

2

O R

4.

3 A

A

= R = mrω 30-R = 0 (Since centre of mass is at O r = ) R = 30

[Ans. B] I3

A

rad⁄s

th

th

th



9.

[Ans. A]

Theory of Machines

Angular velocity of coupler Wop =

√

=

π

12.

r o N

M

=

√

/

[Ans. B] y Grubler’s criterion ) F = (L j (8 ) =

13.

[Ans. B]

14.

[Ans. D]

=

P π =

= π

cos =

r cos = OP r

Q

Q Q

r = 250mm 10.

11.

P

[Ans. B] Maximum speed in forward stroke occurs when QR an QS are perpendicular V = 250

2 = 750

√ a

=

–α

i.e PO2O1 = In ∆ PO2O1 tan

= =

O2, O4

for

=

the 15.

=

tan

And sin

a

diagram

Q

Or α = or α = ∴ =

O2

The velocity configuration is

)

P

=

√ a

a

α

(

[Ans. C] At the instant i.e, O4O2P = 8 The linkage is shown in figure O

P

α

.

= ⁄

=

[Ans. D] B

450 O

2a

250

250√

60°

30°

450 P

A

C

500

Angle ABC is , hence, = r. ω = . = . m/s

Velocity of coupler OP is OP = 2a cos = √ th

th

th



16.

[Ans. A] Grashof’s rule states that the sum of the shortest and longest link for a planar Crank – rocker four bar mechanism can not be greater than the sum of the remaining two link lengths

17.

[Ans. A] Mobility or degree of freedom for a statically indeterminate structure is always less than zero. i.e F<0 ∴ only option (A) is negative value ∴ F≤

22.

Theory of Machines

[Ans. B] I Should be in the line joining I and I . Similarly the link 3 is rolling on link 2. So the I – center I will be on the line perpendicular to the link – 2 . So the point C is the intersection of these of two loci. Which is the center of the disc. So ω (I I ) = ω (I I ) ω = ω = . rad/sec Locus of I

B

D

A

18.

C

25

[Ans. C]

Q

23.

O

Planar rigid body

= Relative velocity between P and Q. = Always perpendicular to PQ.

Locus of I

45

I

P

[Ans. C] According to Grashof condition As, we have shortest + longest link < sum of intermediate lengths 2 + 3 < 2.5 +2.7 5 < 5.2 We get double rocker by fixing link opposite to shortest link (PQ) S .

19.

[Ans. D] Velocity of link AB = Velocity of link CD A ω = D ω ω

=

.

=

P

R

Q

Fix RS

20.

[Ans. C]

21.

[Ans. D] Acceleration at point ‘O’ a =a +a +a a and a are linear accelerations With same magnitude and opposite in direction. a =a

.

r

24.

[Ans. *] Range 29 to 31 Crank radius(r) = =

.

= . m,

Ang. velocity (ω) = rad/s Crank angle ( ) = Revolving mass for balance, m = 6 kg, So, unbalanced force normal to slider axis = mω r sin = . sin = N

= rω

th

th

th


. GATE QUESTION BANK 25.

[Ans. *] Range 1.2 to1.3

29.

Theory of Machines

[Ans. *]Range 0.54 to 0.68 A

mm

mm

(position A)

Sin

=

mm

rad/sec

= .

S

= ωr =

= mm/sec = . m/sec = elocity of slider elocity diagram of current configuration

mm mm .

Sin

(Position ) =

=

=

A = .

return + 8 + 8 = . A forward . = t ωQ . QRR = = = = . t ω . 26.

[Ans. A] l+s≤ p+q

+

=8

=( ) +( ) ) = +( = + + = = = . m/sec

.

+

8 So at least one link goes full revolution Also shortest link is fixed. So both input and output execute full revolution

30.

[Ans. *]Range 1.9 to 2.1 I . α = mg. x. x = 3m, I = =

27.

[Ans. *] Range 39 to 41 a

m

= . ∴ . α= .

P ω

mL + m

( )

8 + m m kg m m α=m rad/s

.8

a α

Q a = rα = ( ) = m/s a = rω = ( ) = m/s a = √a + a = √ 28.

+

=

m/s

[Ans. C] In this question we have to find maximum translation of the follower so this can be found choosing the option which has highest value of x and y so we can observe and on the basis of common sense. The most appropriate answer is C.

th

th

th



Theory of Machines

Gear Trains ME – 2005 1. In a cam - follower mechanism, the follower needs to rise through 20 mm during of cam rotation, the first with a constant acceleration and then with a deceleration of the same magnitude. The initial and final speeds of the follower are zero. The cam rotates at a uniform speed of 300 rpm. The maximum speed of the follower is (A) 0.60 m/s (C) 1.68 m/s (B) 1.20 m/s (D) 2.40 m/s ME – 2006 Common Data for Questions 2, 3: A planetary gear train has four gears and one carrier. Angular velocities of the gears are , , and respectively. The carrier rotates with angular velocity .

4.

Match the items in columns I and II. Column I Column II (P) Addendum (1) Cam (Q) Instantaneous (2) Beam Centre of velocity (R) Section modulus (3) Linkage (S) Prime circle (4) Gear (A) P - 4 Q – 2 R – 3 S - 1 (B) P - 4 Q – 3 R - 2 S - 1 (C) P – 3 Q – 2 R – 1 S - 4 (D) P – 3 Q – 4 R – 1 S - 2

ME – 2008 5. In a cam design, the rise motion is given by a simple harmonic motion (SHM) s= ( cos ) where h is total rise, is camshaft angle is the total angle of the rise interval. The jerk is given by (A) ( cos )

Gear 3 20T

Gear 2 45T

(B)

sin ( )

(C)

cos ( )

(D) Carrier 5 Gear 1 15T

2.

3.

Gear 4 40T

What is the relation between the angular velocities of Gear 1 and Gear4? (A) = (B)

=

(C)

=

(D)

=

( )

sin ( )

ME – 2009 6. An epicyclic gear train is shown schematically in the adjacent figure The sun gear 2 on the input shaft is a 20 teeth external gear. The planet gear 3 is a 40 teeth external gear. The ring gear 5 is a 100 teeth internal gear. The ring gear 5 is fixed and the gear 2 is rotating at 60 rpm ccw (ccw = counter-clockwise and cw = clockwise). The arm 4 attached to the output shaft will rotate at 3

For ω1 = 60 rpm clockwise (cw) when looked from the left, what is the angular velocity of the carrier and its direction so that Gear 4 rotates in counterclockwise (ccw) direction at twice the angular velocity of Gear 1 when looked from the left? (A) 130 rpm, cw (C) 256 rpm, cw (B) 223 rpm, ccw (D) 156 rpm, ccw

4

5 2

(A) 10 rpm ccw (B) 10 rpm cw

th

th

(C) 12 rpm cw (D) 12 rpm ccw

th



of the gear R is 2 mm, the center distance in mm between gears P and S is

ME – 2010 7. For the epicyclic gear arrangement shown in the figure, ω = rad/s clockwise (CW) and ω = 8 rad/s counter clockwise (CCW). The angular velocity ω (in rad/s) is

5 3

4

N = Number of teeth for gear i N = N = N = N =8

2

arm

(A) 0 (B) 70 CW 8.

Shaft axis

(C) 140 CCW (D) 140 CW

Tooth interference in an external involute spur gear pair can be reduced by (A) Decreasing center distance between gear pair (B) Decreasing module (C) Decreasing pressure angle (D) Increasing number of gear teeth

Theory of Machines

Q

S

P +

+ R

(A) 40 (B) 80

(C) 120 (D) 160

ME – 2014 11. For the given statements: I. Mating spur gear teeth is an example of higher pair II. A revolute joint is an example of lower pair Indicate the correct answer. (A)Both I and II are false (B) I is true and II is false (C) I is false and II is true (D) Both I and II are true 12.

A pair of spur gears with module 5 mm and a center distance of 450 mm is used for a speed reduction of 5:1. The number of teeth on pinion is _______

13.

Gear 2 rotates at 1200 rpm in counter clockwise direction and engages with Gear 3. Gear 3 and Gear 4 are mounted on the same shaft. Gear 5 engages with Gear 4. The numbers of teeth on Gears 2, 3, 4 and 5 are 20, 40, 15 and 30, respectively. The angular speed of Gear 5 is

ME – 2012 9. The following are the data for two crossed helical gears used for speed reduction: Gear I: pitch circle diameter in the plane of rotation 80mm and helix angle Gear II: pitch circle diameter in the plane of rotation 120mm and helix angle . If the input speed is 1440 rpm. The output speed in rpm is (A) 1200 (C) 875 (B) 900 (D) 720

T

T

ME – 2013 10. A compound gear train with gears P,Q, R and S has number of teeth 20, 40 , 15, and 20, respectively. Gears Q and R are mounted on the same shaft as shown in the figure below. The diameter of the gear Q is twice that of the gear R. If the module

T

T

(A) 300 rpm counterclockwise (B) 300 rpm clockwise (C) 4800 rpm counterclockwise (D) 4800 rpm clockwise

th

th

th



14.

Which one of the following is used to convert a rotational motion into a translational motion? (A) Bevel gears (B) Double helical gears (C) Worm gears (D) Rack and pinion gears

15.

It is desired to avoid interference in a pair of spur gears having a 20° pressure angle.

Theory of Machines

With increase in pinion to gear speed ratio, the minimum number of teeth on the pinion (A)Increases (B) Decreases (C)First increases and then decreases (D)Remains unchanged


[Ans. B] Angular velocity ω = Time taken to move

= =

= = =

Jerk is given by

π 6.

[Ans. A] T = T =

sec

Condition 1. Arm fixed –gear 2 rotates with +1 rotation 2. Arm fixed-gear 2 rotates with +x rotations 3. Adding + y rotations

0.01 = + a * + a = 0.01 2 (60)2 = 72 m/sec2 Vmax = U + at = 0 + 72 = 1.20 m/sec

Given

carrier 5) = (with respect to carrier 5)

3.

7.

[Ans. B]

5.

[Ans. D] S= *

Arm 2 3 4 5

cos

Gear 2 +1

0

+x

y

y+x

x

Gear 3 T T

x

y

T T

x

Gear 4 T T T T

x

T T

y x

=

[Ans. C]

[Ans. D] ω1 = 60 rpm (Clockwise) ω4 = 120 rpm (Counter clockwise) ω = ω ∴ ω5 = 156 i.e counter clockwise i.e, ω = rpm ccω

4.

y

T =

Arm 0

 x = y ……………………… ( ) and y+x= …………………….. ( ) Solving (1) and (2), we get y + 5y = 60  y = 10 rpm ccw

[Ans. A] = (with respect to arm 5 or

As ω3 = ω2 ∴ =

sin

Where = ωt

During this time, follower moves by distance 20mm with initial velocity u=0 now; = Ut + at

2.

=

+ th

th

1. Give + x rotation CW to gear 2. 0 +x N x N N x N N N N x N

2. Give y rotataion CW to arm. y y y y y

th

Total

y x+y N y x N N y x N N y N N x N



( ) x+y= y= 8 Speed of gear 5, ω =

8

=

12.

8

8 =

.

8.

[Ans. D]

9.

[Ans. B] Gear 8 α = Gear α = d = d cos α = 8 d = d cos α = = .8 mm N d = N d .8 = N √ ∴N = rpm

10.

. cos = cos .

√ mm

13.

+

+

N = =

+

rpm anticlock wise

14.

[Ans. D] Rack and pinion gears is used to rotational motion into translational motion

15.

[Ans. A]

d +d +

[Ans. A] N T = N T = N = rpm clock wise N Gear 3 and Gear 4 are mounted on the same shaft Then,N = N = rpm clock wise . N T = N T

)

= 80 11.

)

t= t=

d +d

= (

/N

( t) =

)=

entre distance =

[Ans. * ] Range 29 to 31 ( ) R+r= D R ( ) m= = T T d r ( ) m= = t t Using (2) &(3) in (1) mT mt + = m (T + t) = T T = ( =N t t m ( t + t) =

[Ans. B] d m = = mm T Module of meshing gear pair are equal. d d d d = = T T d d d d = = = = T T d = d = d = d = ( )= Using 3 in 1, d = (

Theory of Machines

[Ans. D] Spur Gear has a line contact. Lower pair has surface contact

th

th

th



Theory of Machines

Flywheel ME – 2005 1. A rotating disc of 1 m diameter has two eccentric masses of 0.5 kg each at radii of 50 mm and 60 mm at angular positions of and , respectively. A balancing mass of 0.1 kg is to be used to balance the rotor. What is the radial position of the balancing mass? (A) 50mm (C) 150mm (B) 120mm (D) 280mm

2600 J. Assuming that the engine runs at an average speed of 200 rpm, the polar mass moment of inertia (in kg.m ) of a flywheel to keep the speed fluctuation within ±0.5% of the average speed is _______ 7.

Consider a flywheel whose mass M is distributed almost equally between a heavy, ring-like rim of radius R and a concentric disk-like feature of radius R/2. Other parts of the flywheel, such as spokes, etc, have negligible mass. The best approximation for α if the moment of inertia of the flywheel about its axis of rotation is expressed as αMR , is _______

8.

An annular disc has a mass m, inner radius R and outer radius 2R. The disc rolls on a flat surface without slipping. If the velocity of the center of mass is v, the kinetic energy of the disc is

ME – 2006 2. If is the coefficient of speed fluctuation of a flywheel then the ratio of ωmax / ωmin will be: (A) (C)

(B)

(D)

ME – 2007 3. The speed of an engine varies from 210 rad/s to 190 rad/s. During a cycle the change in kinetic energy is found to be 400 Nm. The inertia of the flywheel in kgm2 is (A) 0.10 (C) 0.30 (B) 0.20 (D) 0.40 4.

9.

(A)

mv

( )

mv

( )

mv

(D)

mv

Torque and angular speed data over one cycle for a shaft carrying a flywheel are shown in the figures. The moment of inertia (in kg.m ) of the flywheel is _______

A circular solid of uniform thickness 20 mm, radius 200mm and mass 20 kg , is used as a fly wheel. If it rotates at 600 rpm, the kinetic energy of the flywheel, in joules is (A) 395 (C) 1580 (B) 790 (D) 3160

Torque N

m π

ME – 2013 5. A flywheel connected to a punching machine has to supply energy of 400 Nm while running at a mean angular speed of 20 rad/s. if the total fluctuation of speed is not exceed to , the mass moment of inertia of the flywheel in kg – is (A) 25 (C) 100 (B) 50 (D) 125

π/

N

π /

π/ m

rad/s

Angular speed rad/s rad/s

ME – 2014 6. Maximum fluctuation of kinetic energy in an engine has been calculated to be

π

π/

th

th

th

π/

π



Theory of Machines


[Ans. C]

= (

1m

.

= 5.

N

)(

ω

=+ . ω = I( ) ( . )

Let x and y is position of balancing mass along x and y direction. Resolving forces along x-axis . ( cos + cos )ω = . x ∴x= .8 mm . ( sin )ω + . ( sin )ω = . ω y ∴y= mm Position of balancing mass

6.

2.

I=

7.

. kg

=

m

=

+

=

r

8 (

)

[Ans. A] e know ∆E = Iω Cs Here ω = ∆E = CS =

=

=

8.

= .

mR =

4.

∆

=

(

R )

.

=

.

A P

We know that

E = Iω =

mR

[Ans. C]

[Ans. B]

mr

mR

= .

∆E = 400N m ∴I=

m r m ( ) mR =( ) = [ ]= m m = ( )R + R

+I

MR =

3.

m

other half as disc of radius R/ I

ω ω

)= .

[Ans. *] Range 0.55 to 0.57 Half of mass is distributed as ring like rim of radius R m I = m R = ( )R

)

ω

.

=

.

I=I

[Ans. D] ω =

(

[Ans. *] Range 590 to 595 E = Iω π . + . =I ( ) ( )

r = √x + y = √( .8) + ( = . mm

)

m

[Ans. A] E=Iω ω =

I=

π

Kinetic energy = Iω And = rω = R. ω

ω th

th

th



Theory of Machines

ω=

R Moment of Inertia about center I = m (R + R ) = m ( R + R ) = mR moment of inertia at periphery at point P I = mR + m( R) = = Kinetic energy = K. E =

9.

mR + mR mR mR

R

m

[Ans. *]Range 30 to 32 This is a case of only driving torque acting on shaft. We do not have anything else on shaft except fly wheel. So increase in torque would just go as increase in kinetic energy of flywheel. In the part where torque is constant, we apply T = Iα As torque is constant is constant. we can apply ) ω ω = α( I(ω ω ) T= ( ) T(

)=

ω )

I(ω

In π⁄ to π π (π ) = I(

)

)

π = I(

π=I I = 10(3.1415) = 31.415 kg-m2 In π to π, (Just for verification) ( π

π) = I(

π=

I

(

)

)

I= π = 31.415 kg- m

th

th

th



Theory of Machines

Vibrations ME – 2005 1. In a spring – mass system, the mass is 0.1 kg and the stiffness of the spring is 1 kN/m. By introducing a damper, the frequency of oscillation is found to be 90% of the original value. What is the damping coefficient of the damper? (A) 1.2 N.s/m (C) 8.7 N.s/m (B) 3.4 N.s/m (D) 12.0 N.s/m 2.

There are four samples P, Q, R, and S, with natural frequencies 64, 96, 128 and 256 Hz, respectively. These are mounted on test setups for conducting vibration experiments. If a loud pure note of frequency 144 Hz is produced by some instrument, which of the samples will show the most perceptible induced vibration? (A) P (C) R (B) Q (D) S

ME – 2006 Statement for Linked Answer Questions 3 & 4: A vibratory system consists of a mass 12.5 kg, a spring of stiffness 1000 N/m, and a dashpot with damping coefficient of 15 Ns/m. 3. The value of critical damping of the system is: (A) 0.223 Ns/m (C) 71.4 Ns/m (B) 17.88 Ns/m (D) 223.6 Ns/m 4.

The value of logarithmic decrement is (A) 1.35 (C) 0.68 (B) 1.32 (D) 0.66

5.

A machine of 250 kg mass is supported on springs of total stiffness 100 kN/m. Machine has an unbalanced rotating force of 350 N at speed of 3600 rpm. Assuming a damping factor of 0.15, the value of transmissibility ratio is: (A) 0.0531 (C) 0.0162 (B) 0.9922 (D) 0.0028

6.

The differential equation governing the vibrating system is: y

x k m C

(A) (B) (C) (D)

mẍ + cẋ + k (x y) = m(ẍ ÿ ) + c(ẋ ẏ ) + kx = mẍ + c(ẋ ẏ ) + kx = m(ẍ ÿ ) + c(ẋ ẏ ) + k(x y) =

ME – 2007 7. The equation of motion of a harmonic oscillator is given by ω

+

ω

+

x = 0,and the initial conditions at

=

are

( )=

( ) =

. The

amplitude of ( ) after n complete cycle is (A) Xe (B) Xe 8.

( (

)

√

(

√

)

(D) X

)

√

(C) Xe

The natural frequency of the system shown below is k/2 k m k/2

9.

(A) √

(C) √

(B) √

(D) √

For an under damped harmonic oscillator, resonance (A) occurs when excitation frequency is greater than undamped natural frequency (B) occurs when excitation frequency is less than undamped natural frequency

th

th

th



(C) occurs when excitation frequency is equal to undamped natural frequency (D) never occurs ME – 2008 10. A uniform rigid rod of mass m = 1kg and length L = 1m is hinged at its centre and laterally supported at one end by a spring of spring constant k = 300 N/m. The natural frequency in rad/s is (A) 10 (C) 30 (B) 20 (D) 40 11.

The natural frequency of the spring mass system shown in the figure is closest to m=1.4 kg

k =

(A) 8 Hz (B) 10 Hz

N/m k =

The rotor shaft of a large electric motor supported between short bearings at both the end shows a deflection of 1.8mm in the middle of the rotor. Assuming the rotor to be perfectly balanced and supported at knife edges at both the ends, the likely critical speed (in rpm) of the shaft is (A) 350 (C) 2810 (B) 705 (D) 4430

ME – 2010 15. A mass m attached to a spring is subjected to a harmonic force as shown in figure. The amplitude of the forced motion is observed to be 50 mm. The value of m (in kg) is K = 3000 N

N/m

(C) 12 Hz (D) 14 Hz

ME – 2009 12. A vehicle suspension system consists of a spring and a damper. The stiffness of the spring is 3.6kN/m and the damping constant of the damper is 400Ns/m. If the mass is 50kg, then the damping factor (d) and damped natural frequency ( f ), respectively, are (A) 0.471 and 1.19Hz (B) 0.471 and 7.48Hz (C) 0.666 and 1.35Hz (D) 0.666 and 8.50Hz 13.

14.

Theory of Machines

An automotive engine weighing 240kg is supported on four springs with linear characteristics. Each of the front two springs have a stiffness of 16MN/m while the stiffness of each rear spring is 32MN/m. The engine speed (in rpm), at which resonance is likely to occur, is (A) 6040 (C) 1424 (B) 3020 (D) 955

F(t) = 100 cos(100 t) m

(A) 0.1 (B) 1.0 16.

(C) 0.3 (D) 0.5

The natural frequency of a spring – mass system on earth is ω . The natural frequency of this system on the moon (g =g / ) is (A) ω (C) 0.204 ω (B) 0.408 ω (D) 0.167 ω

ME – 2011 17. A mass of 1 kg is attached to two identical springs each with stiffness k = kN/m as shown in the figure. Under frictionless condition, the natural frequency of the system in Hz is close to x k kg k

(A) 32 (B) 23 th

th

(C) 16 (D) 11 th



18.

A disc of mass m is attached to a spring of stiff-ness k as shown in the fig. The disc rolls without slipping on a horizontal surface. The nature frequency of vibration of the system is k

m

(A)

√

(C)

√

(B)

√

(D)

√

ME – 2012 19. A concentrated mass m is attached at the centre of a rod length 2L as shown in the figure. The rod is kept in a horizontal equilibrium position by a spring of stiffness k. For very small amplitude of vibration, neglecting the weights of the rod and spring, the un-damped natural frequency of the system is

L

L

(A) √

(C) √

(B) √

(D) √

ME – 2013 20. If two nodes are observed at a frequency of 1800 rpm during whirling of a simply supported long slender rotating shaft, the first critical speed of the shaft in rpm is (A) 200 (C) 600 (B) 450 (D) 900 21.

Theory of Machines

magnitude 5 kN for seconds. The amplitude in mm of the resulting free vibration is (A) 0.5 (C) 5.0 (B) 1.0 (D) 10.0 ME – 2014 22. Critical damping is the (A) Largest amount of damping for which no oscillation occurs in free vibration (B) Smallest amount of damping for which no oscillation occurs in free vibration (C) Largest amount of damping for which the motion is simple harmonic in free vibration (D) Smallest amount of damping for which the motion is simple harmonic in free vibration 23.

Consider a cantilever beam, having negligible mass and uniform flexural rigidity, with length 0.01 m. The frequency of vibration of the beam, with a 0.5 kg mass attached at the free tip, is 100 Hz. The flexural rigidity (in N. m ) of the beam is _______

24.

In vibration isolation, which one of the following statements is NOT correct regarding Transmissibility (T)? (A) T is nearly unity at small excitation frequencies (B) T can be always reduced by using higher damping at any excitation frequency (C) T is unity at the frequency ratio of √ (D) T is infinity at resonance for undamped system

25.

What is the natural frequency of the spring mass system shown below? The contact between the block and the inclined plane is frictionless. The mass of the block is denoted by m and the spring

A single degree of freedom system having mass 1 kg and stiffness 10 kN/m initially at rest is subjected to an impulse force of th

th

th



Theory of Machines cos (

F=

constants are denoted by k and k as shown below.

t)

m

k

m

c

k k

x

k +k (A)√ m

( )√

k +k m

(D)√

( )√

26.

k

k

29.

A point mass is executing simple harmonic motion with an amplitude of 10 mm and frequency of 4 Hz. The maximum acceleration (m/ s ) of the mass is _______

30.

A single degree of freedom system has a mass of 2 kg, stiffness 8 N/m and viscous damping ratio 0.02. The dynamic magnification factor at an excitation frequency of 1.5 rad/s is _______

31.

A rigid uniform rod AB of length L and mass m is hinged at C such that AC = L/3, CB = 2L/3. Ends A and B are supported by springs of spring constant k. The natural frequency of the system is given by

m k +k m

Consider a single degree-of-freedom system with viscous damping excited by a harmonic force. At resonance, the phase angle (in degree) of the displacement with respect to the exciting force is (A) 0 (C) 90 (B) 45 (D) 135

27.

The damping ratio of a single degree of freedom spring-mass-damper system with mass of 1 kg, stiffness 100 N/m and viscous damping coefficient of 25 N.s/m is

28.

A mass-spring-dashpot system with mass m = 10 kg, spring constant k = 6250 N/m is excited by a harmonic excitation of 10 cos(25t) N. At the steady state, the vibration amplitude of the mass is 40 mm. The damping coefficient (c, in N.s/m) of the dashpot is _______

k

k

A L/

(A)√ ( )√

th

L/

k m

k m

th

th

( )√

k m

(D)√

k m



Theory of Machines


[Ans. C] Given ω = . ω We know that ω =ω √ . ω =ω √ ∴ = . Now =

8.

[Ans. A] Equivalent K, k

2.

3.

4.

5.

. N

9.

[Ans. B] Resonance occurs when, ω

ω

=

=

M

[Ans. D] For critical damping, = √mk = √ . =223.6 Ns/m

ω⁄ω

[Ans. *] Logarithmic decrement δ =

10.

[Ans. C]

= 0.42

√

[Ans. C]

ω=

rad/s

=

rad/s O

√ +*

T= √*

+

+ + *

K = 300 N/m

+

∴ Transmissibility ratio = 0.0163

Applying torque about hinge ‘O’ mI α + k[ ] =

[Ans. C] Differential equation governing the above vibration system is given by +

[

̈+ ( ̇ 7.

/

.

√ s/m

[Ans. C] Since the natural frequency of R (128Hz) is close to the frequency produce by any instrument(144Hz) therefore sample R will show the most perceptible induced vibration.

ω =√ =

6.

/

∴ω = √

√

∴c= = 8.

= +

]+ ̇) +

̈+

k = m

Angular frequency ω = √

= =

=√

[Ans. A]

ω = 30 rad/s

th

th

th



11.

[Ans. B]

14.

keq = k1+k2 = 5600 N/m k = π m √

12.

[Ans. A] Damping factor = = =√

√

√

i.e.

Hz

=

=√ .

=√

√

= .

.

15.

16. . .

= .

[Ans. A] ω=√

∆

As g changes ∆ will also change by same factor. 17.

[Ans. A] Natural frequency of the system ω = √

k2

k2

k1

k1

ω = √

At resonance ω= ωn =√ 18.

k = k + k = 2(k +k ) = 2(32+16) =96 106 N/m

where ke = k + k = 2k

= 2 20 = 40 kN/m

f=

ω = √ rad/s

√

) m=

m = . kg

)

240 kg

N=

.8

[Ans. A] k ω m = F/ (

[Ans. A] k1=16MN/m k2=32MN/m

√ rpm

=

. .

= .8 8

N=

8

N= = 705 rpm

= √ ( . = 8. 8 √ = 7.48 Now, ω = πN Therefore N = = 13.

.

8

√ .8 = Alternately When speed of the shaft (  ) is equal to natural speed of vibration, it is called critical speed ( n )

m

∴f=

[Ans. B] . = =

k1

k1

Theory of Machines

k = π m √

Hz

[Ans. C] I ̈ + Kxr = mr ( + mr ) ̈ + K(r. )r = mr ̈ + Kr K ̈+ = m

=

=200 rad/s

=

Natural frequency =

. rpm th

th

th

√



19.

[Ans. D] Equation of motion = mL ̈ + k( L) k L ̈+ = mL k ̈ +( ) = m ∴w =√

20.

21.

25. =

k rad/sec m

[Ans. D] System in parallel. Inclination does not matter ω =√

26.

K k +k =√ m m

[Ans. C]

[Ans. A] f=n f n = node As 2 nodes are observed, n=3 1800 = f 8 f = =

π

[Ans. C] F = ma a=

=

m/s

π

= u + at = + = . m/sec Now equating the K.E of mass with P.E of spring mv =

kx

( . ) = .

x=√

=

m=

[Ans. B]

23.

[Ans. * ] Range 0.064 to 0.067 f=

π

√

g

The phase difference to when ω = ω for all value of 27.

x

22.

mm

(EI) g( EI) √ √ = = π (mg)L π mL

= 28.

EI = =

(EI) mL ( πf) mL

[ π(

)]

.

(∅)

is

equal

[Ans. *] Range 1.24 to 1.26 m =1kg k=100N/m c=25Ns/m we know = mw √

=

( πf) =

= .

[Ans. *] Range 9 to 11 m= 10 kg k=6250 N/m k =√ = rad/sec m F = N ω= rad /sec Vibration amplitude = A=40 mm = 0.040 m F /k A= √[ ( ) ]+( ) ω =√

.

= . . 24.

Theory of Machines

[Ans. B] At =√ T is always irrespective of damping

.

th

=

th

th



= . c = mω c= . c= Ns/m 29.

=

mL ̈

K ) m k ω = m ̈ =(

.( π

= . 30.

kL

[Ans. *]Range 6.3 to 6.4 x = a sin ωt ẋ = aω cos ωt ẍ = aω sin ωt Maximum value = aω = a. ( πf) =

Theory of Machines

ω=√

k m

)

m/s

[Ans. *] Range 2.0 to 2.4 s 8 ω =√ =√ = m

rad⁄s

= . rad⁄s Magnification factor

= .

= √(

(

√[

( ) ] +(

) ) +(

)

= .

.

.

)

= . 8 31.

[Ans. D] A ω /ω

F G ̈

F

m =I α L F ( )

L mL F ( )=* + m( G) + α

k

( )

k

L

L

k(

)(

( )=[ )

k(

L

+ m( ) ] ̈ L mL )( ) = Q̈

th

th

th



Machine Design

Design of Static Loading ME 1.

2006 According to Von-Mises’ distortion energy theory, the distortion energy under three dimensional stress state is represented by

M.

(A) (

)

(

)

(B)

N.

(C) (

)

(D) ( ME 2.

ME 3.

)

2007 The piston rod of diameter 20 mm and length 700 mm in a hydraulic cylinder is subjected to a compressive force of 10 kN due to the internal pressure. The end conditions for the rod may be assumed as guided at the piston end and hinged at the other end. The Young’s modulus is 200 GPa. The factor of safety for the piston rod is (A) 0.68 (C) 5.62 (B) 2.75 (D) 11.0 2011 Match the following criteria of material failure, under biaxial stresses and and yield stress , with their corresponding graphic representations: P. Maximum – normal – stress criterion Q. Maximum – distortion – energy criterion R. Maximum – shear – stress criterion

(A) (B) (C) (D) ME 4.

P – M, Q – L, R –N P – N, Q –M, R – L P – M, Q – N, R – L P – N, Q – L, R – M

2012 The homogeneous state of stress for a metal part undergoing plastic deformation is T

(

)

Where the stress component values are in MPa. Using von Mises yield criterion, the value of estimated shear yield stress, in MPa is (A) 9.50 (C) 28.52 (B) 16.07 (D) 49.41 ME 5.

L.

th

2013 For a ductile material , toughness is a measure of (A) Resistance to scratching (B) Ability to absorb energy up to fracture (C) Ability to absorb energy till elastic limit (D) Resistance to indentation

th

th



ME 6.

2014 Consider the two states of stress as shown in configurations I and II in the figure below. From the standpoint of distortion energy (von-Mises) criterion, which one of the following statements is true?

I (A) (B) (C) (D)

7.

Machine Design

II

I yields after II II Yields after I Both yield simultaneously Nothing can be said about their relative yielding

Which one of following is NOT correct? (A) Intermediate principal stress is ignored when applying the maximum principal stress theory (B) The maximum shear stress theory gives the most accurate results amongst all the failure theories (C) As per the maximum strain energy theory, failure occurs when the strain energy per unit volume exceeds a critical value (D) As per the maximum distortion energy theory, failure occurs when the distortion energy per unit volume exceeds a critical value

th

th

th



Machine Design


[Ans. C] From Von – Mises distortion theory,

∴ Estim te v lue of she r stress .

( 2.

.

)

3.

[Ans. C] P – M, Q – N, R – L.

4.

[Ans. B] Given,

[Ans. B] Area of stress-strain under curve till fracture

6.

[Ans. C] Von Mises yield criteria,

)

√(

√(

)

√(

)

( .

(

)

(

√

∴

So, ∴ (C) Both yield simultaneously

. √ According to von Mises yield criterion, permissible value of tensile yield strength,

.

)

For case - II

. √ Similarly, minimum principle stress,

√

(

√

∴

)

√(

)

For case – I

nd Maximum principle stress,

√

.

5. = 0.68 (

)

.

[Ans. A] Factor of safety =

(

7.

[Ans. B]

) (

)

.

.

( .

)

(

.

)

.

th

th

th


)


Machine Design

Design for Dynamic Loading ME 1.

ME 2.

2006 A cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycles is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will be (A) 1071 cycles (C) 281914 cycles (B) 15000 cycles (D) 928643 cycles 2007 A thin spherical pressure vessel of 200 mm diameter and 1 mm thickness is subjected to an internal pressure varying from 4 to 8 MPa. Assume that the yield, ultimate, and endurance strength of material are 600, 800 and 400 MPa respectively. The factor of safety as per Goodm n’s rel tion is (A) 2.0 (C) 1.4 (B) 1.6 (D) 1.2

ME 3.

2008 An axial residual compressive stress due to a manufacturing process is present on the outer surface of a rotating shaft subjected to bending. Under a given bending load, the fatigue life of the shaft in the presence of the residual compressive stress is (A) decreased (B) increased or decreased, depending on the external bending load (C) neither decreased nor increased (D) increased

ME 4.

2009 A forged steel link with uniform diameter of 30mm at the centre is subjected to an axial force that varies from 40kN in compression to 160kN in tension. The tensile ( ), yield ( ) and corrected endurance ( ) strengths of the steel material are 600MPa, 420MPa and

240MPa respectively. The factor of safety against fatigue endurance as per oderberg’s criterion is (A) 1.26 (C) 1.45 (B) 1.37 (D) 2.00 ME 5.

2013 A bar is subjected to fluctuating tensile load from 20 kN to 100 kN. The material has yield strength of 240 MPa and endurance limit in reversed bending is 160 MPa. According to the soderberg principle , the area of cross – section in mm of the bar for a factor of safety of 2 is (A) 400 (C) 750 (B) 600 (D) 1000

ME 6.

2014 In a structure subjected to fatigue loading, the minimum and maximum stresses developed in a cycle are 200 MPa and 400 MPa respectively. The value of stress amplitude (in MPa) is _______

th

th

th



Machine Design


[Ans. C]

3.

[Ans. D] As we c n see from the Gerber’s p r bol the endurance strength remains constant at maximum value irrespective of the compressive load. Where as for tensile load it decreases sharply with increases in and stress in the shaft, its strength in fatigue increase.

A

log

C B

log

3

6

Equation of straight line connecting ( log ) and ( log ) y . . . x ) Y– . . (x At y log . . (x ) x = 5.43 log . N = 268269 2.

Endurance strength

Compression

4.

[Ans. B] Stress induced pr t

[Ans. A] Dia .d = 30mm (Tesion) (compression) Tensile strength Yield strength Maximum stress, A

(

. Minimum stress, A Equivalent Stresses

) mm (Tensile)

( .

√

) mm (Compression)

Stress amplitude,

(

√

.

e n stress

Goodman equation

.

n

.

) (

.

)

.

Similarly,

n

strength

yield

( (

.

) )

.

et . . be n n . n n . n equ tion of soderberg

n

th

th

th



.

Machine Design

. .

5.

[Ans. D]

(

( 6.

)

(

)

)

A

mm

[Ans. *]Range 99 to 101

th

th

th



Machine Design

Design of Joints ME 1.

ME 2.

2006 A 60 mm long and 6 mm thick fillet weld carries a steady load of 15 kN along the weld. The shear strength of the weld material is equal to 200 MPa. The factor of safety is (A) 2.4 (C) 4.8 (B) 3.4 (D) 6.8

4.

The resultant shear stress on bolt P is closest to (A) 132 MPa (C) 178 MPa (B) 159 MPa (D) 195 MPa

ME 5.

2010 A bracket (shown in figure) is rigidly mounted on wall using four rivets. Each rivet is 6 mm in diameter and has an effective length of 12 mm.

2007 A bolted joint is shown below. The maximum shear stress, in MPa, in the bolts at A and B, respectively are holes of 20

.

32

32 20

1000 N

mm bolts 10

6 mm thick

C

40 75

B

25

40

F = 10 kN

A

20

12

(All dimensions in the figure are in mm)

(A) 242.6, 42.5 (B) 42.5, 242.6 ME

(C) 42.5, 42.5 (D) 242.6,242.6

2008 Statement for Linked data Question 3 & 4 A steel bar of mm is cantilevered with two M 12 bolts (P and Q) to support a static load of 4 kN as shown in the figure. 100

40

12

Direct shear stress (in MPa) in the most heavily loaded rivet is (A) 4.4 (C) 17.6 (B) 8.8 (D) 35.2

150

ME 6.

100 1.7m 400

2012 A fillet welded joints is subjected to transverse loading F as shown in the figure. Both legs of the fillets are of 10mm size and the weld length is 30mm. if the allowable shear stress of the weld is 94MPa, considering the minimum throat area of the weld, the maximum allowable transverse load in kN is F

P

Q

4 kN

3.

F

The primary and secondary shear loads on bolt P, respectively, are (A) 2 kN, 20 kN (C) 20 kN, 0 kN (B) 20 kN, 2 kN (D) 0 kN, 20 kN

(A) 14.44 (B) 17.92

th

th

(C) 19.93 (D) 22.16

th



ME

2013 Common data for questions 7 & 8: A single riveted lap joint of two similar plates as shown in the figure below has the following geometrical and material details. w

P

t

7.

8.

w

Machine Design

(A) 83 (B) 125 ME 9.

P

(C) 167 (D) 501

2014 For the three bolt system shown in the figure, the bolt material has shear yield strength of 200 MPa. For a factor of safety of 2, the minimum metric specification required for the bolt is

t

Width of the plate w =200 mm, thickness of the plate t = 5 mm, number of rivets n =3, diameter of the rivet dr = 10mm, diameter of the rivet hole dh= 11mm, allowable tensile stress of the plate allowable shear stress of the rivet and allowable bearing stress of the rivet . If the rivets are to be designed to avoid crushing failure, the maximum permissible load P in kN is (A) 7.50 (C) 22.50 (B) 15.00 (D) 30.00

(A) M8 (B) M10 10.

(C) M12 (D) M16

A bolt of major diameter 12 mm is required to clamp two steel plates. Cross sectional area of the threaded portion of the bolt is 84.3mm .The length of the threaded portion in grip is 30 mm,while the length of the unthreaded portion in grip is 8mm.Young's modulus of material is 200 GPa.The effective stiffness (inMN/m) of the bolt in the clamped zone is

If the plates are to be designed to avoid tearing failure , the maximum permissible load P in kN is

th

th

th



Machine Design


[Ans. B] Factor of safety trength of m teri l Actu l lo d or strength on m teri l (in ) (in (in . (in

2.

)

)

3.

[Ans. A] 100

[Ans. *] d . d = 0.84 (10) = 8.4 mm

1.7m

.

mm bolts 6 mm thick

(P) 4 kN e

rim ry she r due to

40

n

A P = 10 kN

Now for secondary load effect of P & T e mm s r s r e

150

rim ry force n T

r

s

B

20

100

r

C

40

400

s

holes of 20

(d )

. . x stress t A

.

)

.

o stress

r s r s

. . ( . )

. ( x stress t ) (Primary force will act equally in all both) {And in bolt B secondary shear force will not act} {so in bolt B stress is due to only primary S.F.} econd ry she r force (due to P and ) s s (r r ) e ) s ( r r s s At point ‘A’ √(

)

(

. )

r

r

e

s r (

s

)

s .

4.

[Ans. B] Resultant shear strain on bolt P = 18 kN (here d = 12 mm) (

)

. th

th

th



5.

[Ans. B] Direct shear stress will be same in all the rivets. ( )

d 6.

.

A A

[Ans. C] S = 10mm, l= 30mm,

d

Maximum allowable transverse load, (in KN), F = 0.707S

d . This is minimum diameter so nearest diameter greater is M10

. . . 7.

8.

9.

Machine Design

10.

kN

[Ans. C] Crushing failure p dt )( =3 ( = 22.5 kN

)( )

[Ans. C] Tearing of plates ( p d )t = 200 [200 3(11)]

[Ans. *]Range 460 to 470 stiffness of thre ded portion A E . l m stiffness of unthre ded portion A E l . m the connection is in series. o effective stiffness

kN .

[Ans. B] he r lo d on e ch bolt

.

. ⁄m .

A

th

th

th



Machine Design

Design of Shaft and Shaft Component ME 1.

ME 2.

2012 A solid circular shaft needs to be designed to transmit a torque of 50N.m. If the allowable shear stress of the material is 140MPa, assuming a factor of safety of 2, the minimum allowable design diameter in mm is (A) 8 (C) 24 (B) 16 (D) 32

ultimate and corrected endurance strength of the shaft material are 300 MPa, 500 MPa and 200 MPa, respectively, then the factor of safety for the shaft is _______ 3.

A shaft is subjected to pure torsional moment. The maximum shear stress developed in the shaft is 100 MPa. The yield and ultimate strengths of the shaft material in tension are 300 MPa and 450 MPa, respectively. The factor of safety using maximum distortion energy (vonMises) theory is _______

2014 A rotating steel shaft is supported at the ends. It is subjected to a point load at the center. The maximum bending stress developed is 100 MPa. If the yield,


[Ans. B] T d (Torsion of shaft Equation) T

(

Its fatigue loading case. So we will compare Alternating stain (100MPa) with endurance strength (200MPa). endur nce strength ctor of s fety Altern ting str in Factor of safety = 200/100 = 2

) d (T is in N – mm, d in mm

and in N/mm2) d d 2.

mm

3.

[Ans. *]Range 1.7 to 1.8 Yield stress (k) in pure shear (torsion) according to von-Mises criteria

[Ans. *]Range 1.9 to 2.1 Due to rotation of shaft , same point will have +100 MPa bending stress as well as bending stress. o m x Altern ting st in will be ) –( .

where √ un xi l torsion (here the yield strength) √ ∴

th

. developed

ctor of s ferty n

th

th

.

.



Machine Design

Design of Bearing ME 1.

ME 2.

2005 Which one of the following is a criterion in the design of hydrodynamic journal bearings? (A) Sommerfield number (B) Rating life (C) Specific dynamic capacity (D) Rotation factor 2007 A natural feed journal bearing of diameter 50 mm and length 50 mm operating at 20 revolution/s and carries a load of 2 kN. The viscosity of the lubricant is 20 mPa-s, the radial clearance is 50 m . The Sommerfield number for the bearing is (A) 0.062 (C) 0.250 (B) 0.125 (D) 0.785

3.

A ball bearing operating at a load F has 8000 hours of life. The life of the bearing, in hours, when the load is doubled to 2F is (A) 8000 (C) 4000 (B) 6000 (D) 1000

ME 4.

2008 A journal bearing has a shaft diameter of 40mm and a length of 40 mm. the shaft is rotating at 20 rad/s and the viscosity of the lubricant is 20mPa-s. the clearance is 0.020 mm. the loss of torque due to the viscosity of the lubricant is approximately: (A) 0.040 Nm (C) 0.400 Nm (B) 0.252 Nm (D) 0.6562 Nm

ME 5.

2010 A lightly loaded full journal bearing has journal diameter of 50 mm, bush bore of 50.05 mm and bush length of 20 mm. If rotational speed of journal is 1200 rpm and average viscosity of liquid lubricant is 0.03 Pa s, the power loss (in W) will be (A) 37 (C) 118 (B) 74 (D) 237

ME 6.

2011 Two identical ball bearing P and Q are operating at loads 30 kN and 45 kN respectively. The ratio of the life of bearing P to the life of bearing Q is (A) 81/16 (C) 9/4 (B) 27/8 (D) 3/2

ME 7.

2014 A hydrodynamic journal bearing is subject to 2000 N load at a rotational speed of 2000 rpm. Both bearing bore diameter and length are 40 mm. If radial cle r nce is m nd be ring is lubricated with an oil having viscosity 0.03 Pa.s, the Sommerfeld number of the bearing is _______

8.

Ball bearings are rated by a manufacturer for a life of revolutions. The catalogue rating of a particular bearing is 16 kN. If the design load is 2 kN, the life of the bearing will be p× revolutions, where p is equal to _______

th

th

th



Machine Design


[Ans. A] Sommerfield Number, also known as Bearing Characteristic Number, (

2.

d

.

l

(

)

[Ans. D] Life (

6.

.

p .

.

]

.

or

) ommerfeld umber ( ) w . ( mm ) d . ( ) ( ) . .

[Ans. *]Range 500 to 540 Life of bearing p revolution = p million resolution ( ) for b ll be ring ( )

.

. m.

.

[Ans. *] Range 0.75 to 0.85

8.

.

.

here ife of be ring P = Load ( o d) ( ife) ( ife) ( o d) [

.

m

( )

=1000 hrs

.

. .

[Ans. B]

)

A = 2.0096 N ( . dl) Torque = . ( torque r)

.

. . v.

( ( )

m s

. .

mm

[Ans. A] v c

force

. dv dy

7.

∴ New life = 4.

v

.

.125 3.

[Ans. A]

).( )

[Ans. B] Sommerfield number r ( ) p Where, r is radius of journal is viscosity of lubricant, N is number of revolution per second p is bearing pressure on projected Area C is radial clearance Therefore. p

5.

(

.

( )

million resolution million resolution

p

th

)

th

th



Machine Design

Design of Brakes and Clutches 2005 Statement for Linked Answer Questions 1 & 2: A band brake consists of a lever attached to one end of the band. The other end of the band is fixed to the ground. The wheel has a radius of 200 mm and the wrap angle of the band is . The braking force applied to the lever is limited to 100N, and the coefficient of friction between the band and the wheel is 0.5. No other information is given.

(A) 30 (B) 40

100N

1.

The maximum tension that can be generated in the band during braking is (A) 1200 N (C) 3224 N (B) 2110 N (D) 4420 N The maximum wheel torque that can be completely braked is (A) 200 N.m (C) 604 N.m (B) 382 N.m (D) 844 N.m

ME 3.

2006 A disk clutch is required to transmit 5 kW at 2000 rpm. The disk has a friction lining with coefficient of friction equal to 0.25. Bore radius of friction lining its equal to 25 mm. Assume uniform contact pressure of 1 MPa. The value of outside radius of the friction lining is (A) 39.4 mm (C) 97.9 mm (B) 49.5 mm (D) 142.9 mm

(C) 45 (D) 60

ME 5.

2008 A clutch has outer and inner diameters 100 mm and 40 mm respectively. Assuming a uniform pressure of 2 MPa and coefficient of friction of liner material 0.4, the torque carrying capacity of the clutch is (A) 148 Nm (C) 372 Nm (B) 196 Nm (D) 490 Nm

ME 6.

2010 A band brake having band – width of 80 mm, drum diameter of 250 mm, coefficient of friction of 0.25 and angle of wrap of 270 degrees is required to exert a friction torque of 1000 N m. The maximum tension (in kN) developed in the band is (A) 1.88 (C) 6.12 (B) 3.56 (D) 11.56

ME 7.

2012 Force of 400N is applied to the brake drum of 0.5m diameter in a band – brake system as shown in the figure, where the wrapping angle is 1800. If the coefficient of friction between the drum and the band is 0.25, the braking torque applied, in N.m is.

1m

2.

ME 4.

mm 450

200mm

1m

mm

mm

mm

ME

2007 A block-brake shown below has a face width of 300 mm and a mean co-efficient of friction of 0.25. For an activating force of 400 N, the braking torque in Nm is

400N

(A) 100.6 (B) 54.4 th

th

(C) 22.1 (D) 15.7 th



ME 8.

9.

Machine Design

2014 A disc clutch with a single friction surface has coefficient of friction equal to 0.3. The maximum pressure which can be imposed on the friction material is 1.5 MPa. The outer diameter of the clutch plate is 200 mm and its internal diameter is 100 mm. Assuming uniform wear theory for the clutch plate, the maximum torque (in N.m) that can be transmitted is _______ A drum brake is shown in the figure. The drum is rotating in anticlockwise direction. The coefficient of friction between drum and shoe is 0.2. The dimensions shown in the figure are in mm. The braking torque (in N.m) for the brake shoe is _______

rum

10.

A four-wheel vehicle of mass 1000 kg moves uniformly in a straight line with the wheels revolving at 10 rad/s. The wheels are identical, each with a radius of 0.2 m. Then a constant braking torque is applied to all the wheels and the vehicle experiences a uniform deceleration. For the vehicle to stop in 10 s, the braking torque (in N.m) on each wheel is _______


[Ans. B]

T

T

200 mm

T

.

100 N

(r

T 1m

.

1m

4.

[Ans. C]

200 mm

3.

150 mm

[Ans. B] (T T (

∴ re

A N

f

Taking moment about A,

m

[Ans. A] According to uniform pressure theory, P = 1 MPa Power transmitted

400 mm 400 N

T T e e . T T Maximum tension produced in band is 2110 N r ) .

)

. mm

T

T )

(

.

r

. Taking moment about O,

2.

)

5.

ing torque

m

[Ans. B] Using uniform pressure theory torque carrying capacity of clutch is given by r T ( ) r th

th

th



T

(

r )(

T

p(

r )

(

r )

Machine Design

) of rum

.

.

((

.

.

(

)

(

) )

)

= 195936 Nmm = 195.936 Nm. 6.

7.

T(

e

T (

e

T

.

).

T

.

10.

)

.

m

[Ans. *] Range 529 to 532 r r T ∫ ( r)dr r r ∫ rdr * ( . .

c(r r )

r )(r . m

.m

. m

[Ans. B] T e e . T T T . ∴ r ing torque pplied in .m (T T T) . ( . )

T

9.

Breaking Torque =

T (Frictional Torque)

.

. 8.

mm

[Ans. D] (T T) T e T

( .

.

r

[Ans. *] Range 9 to 11 m=mass of vehicle =1000 kg =Angular velocity of wheel = 10 rad/sec R = Radius of wheel = 0.2m linear velocity of wheel = linear velocity of vehicle u . m sec final velocity of vehicle v t = 10 sec V = u + at 0 = 2+a×10 . m sec After application of brake, vehicle moves: u s ( ) . s s m . let braking force at each wheel=B Total braking force = 4B Change in K.E of vehicle = work done against braking force

+

r )

mu ( )

)

.

50 = B Braking torque at each wheel = m

[Ans. *]Range 63 to 65 FBD of lever plus shoe: .

mm

.

.

th

th

th



Machine Design

Design of Spur Gears ME 1.

ME

2.

2006 Twenty degree full depth involute profiled 19-tooth pinion and 37-tooth gear are in mesh. If the module is 5 mm, the center distance between the gear pair will be (A) 140 mm (C) 280 mm (B) 150 mm (D) 300 mm

(A) (B) (C) (D) 6.

Match the type of gears with their most appropriate description. Type of Description gear P Helical 1 Axes non parallel and non intersecting Q Spiral 2 Axes parallel and Bevel teeth are inclined to the axis R Hypoid 3 Axes are parallel and teeth are parallel to the axis S Rack and 4 Axes are pinion perpendicular and intersecting, and teeth are inclined to the axis 5 Axes are perpendicular and used for large speed reduction 6 Axes parallel and one of the gears has infinite radius (A) (B) (C) (D)

7.

A spur gear has a module of 3 mm, number of teeth 16, a face width of 36 mm and a pressure angle of 20°. It is transmitting a power of 3 kW at 20 rev/s. Taking a velocity factor of 1.5, and a form factor of 0.3, the stress in the gear tooth is about (A) 32MPa (C) 58MPa (B) 46 MPa (D) 70 MPa

2007 Common Data Questions 2, 3 & 4 A gear set has a pinion with 20 teeth and a gear with 40 teeth. The pinion runs at 30 rev/s and transmits a power of 20 kW. The teeth are on the full – depth system and have a module of 5 mm. The length of the line of action is 19 mm. The center distance for the above gear set in mm is (A) 140 (C) 160 (B) 150 (D) 170

3.

The contact ratio of the contacting tooth is (A) 1.21 (C) 1.29 (B) 1.25 (D) 1.33

4.

The resultant force on the contacting gear tooth in N is (A) 77.23 (C) 225.80 (B) 212.20 (D) 289.43

ME 5.

2008 One tooth of a gear having 4 module and 32 teeth is shown in the figure. Assume that the gear tooth and the corresponding tooth space make equal intercepts on the pitch circumference. The dimension ‘ ’ nd ‘b’ respectively re closest to. a Pitch circle

m

6.08 mm, 4 mm 6.48 mm, 4.2 mm 6.28 mm, 4.3 mm 6.28 mm, 4.1 mm

b

th

th

th



ME

8.

9.

2009 Statement for Linked data Question 8 & 9 A full depth involute spur pinion of 4 mm module and 21 teeth is to transmit 15 kw at 960 rpm. Its face width is 25 mm The tangential force transmitted (in N ) is (A) 3552 (C) 1776 (B) 2611 (D) 1305

ME 10.

Given that the tooth geometry factor is 0.32 and the combined effect of dynamic load and allied factors intensifying the stress is 1.5; the minimum allowable stress (in MPa) for the gear material is (A) 242.0 (C) 121.0 (B) 166.5 (D) 74.0

Machine Design

2014 A spur pinion of pitch diameter 50 mm rotates at 200 rad/s and transmits 3 kW power. The pressure angle of the tooth of the pinion is 20°. Assuming that only one pair of the teeth is in contact, the total force (in newton) exerted by a tooth of the pinion on the tooth on a mating gear is _______


[Ans. A] Centre distance = =

2.

[Ans. B] Centre distance = (

3.

4.

)

5. (

)

(T

[Ans. D] Given m = 4 and T = 32 So D = m T = = 128 mm

= 140mm

T

T)

. mm But (because tooth thickness = tooth space) 2a = 12.566 a = 6.28 mm and b = 4.1 mm

= 150 mm

[Ans. C] Length of line of action = 19 mm Length of arc of contact mm . mm cos d circul r pitch mm z length of rc cont ct r tion circul r pitch . .

C

G

.

Pitch circle

F E

R

64 A

[Ans. *] (Answer key is not matching with IIT keys) P=T 3 ∴ T. T = 106.1 Nm Now, T = Ft. γ Here γ mm . m =

m

b

. ∴A

cos . .

b m A A m ( . ) b . mm Addendum = module Dedendum = . module

. .

∴ esult nt force th

th

th



6.

[Ans. A]

Machine Design

Power = Torque Angul r speed P = Tw = r P = Tangential force × Radius × Radius, r = mm P= r Helical

. 9.

[Ans. B] Tooth geometry factor, Y = 0.32 Combined effect of dynamic load and allied factor intensifying the stress is f . bYm f

Spiral bevel

Pinion

(

+

)( .

. 10.

)(

)

.


Rack

7.

[Ans. C] Lewis equation is given as ∇d b ym F can be calculated as D = mZ = 3 16 = 48 mm = 1.5, b = 36, y= 0.3, m = 3 ∴ F= =

1.5

36

ower ower

0.3 3

65 Nearest stress in gear tooth taking in ccount some . . . 8.

T (

r) d .

T ngeti l force cos

[Ans. A] Module, m = 4 mm Number of teeth, T = 21 Speed, N = 960 rpm Pressure angle, Face width, b = 25 mm Full depth involute spur pinion Transmission of power, P = 15 kW at 960 rpm

cos(

th

th

)

cos(

th

)

.



Fluid Mechanics

Fluid Properties ME 1.

ME 2.

2006 For a Newtonian fluid (A) Shear stress is proportional to shear strain (B) Rate of shear stress is proportional to shear strain (C) Shear stress is proportional to rate of shear strain (D) Rate of shear stress is proportional to rate of shear strain

3.

Match Group A with Group B: Group Group B P: Biot number Q: Grashof number R: Prandtl number S: Reynolds number

1: Ratio of buoyancy to viscous force 2: Ratio of inertia force to viscous force 3: Ratio of momentum to thermal diffusivities 4: Ratio of internal thermal resistance to boundary layer thermal resistance (A) P-4, Q-1, R-3, S-2 (B) P-4, Q-3, R-1, S-2 (C) P-3, Q-2, R-1, S-4 (D) P-2, Q-1, R-3, S-4

2014 Consider the turbulent flow of a fluid through a circular pipe of diameter, D. Identify the correct pair of statements. I. The fluid is well-mixed II. The fluid is unmixed III. < 2300 IV. > 2300 (A) I, III (C) II, III (B) II, IV (D) I, IV


[Ans. C] For a Newtonian fluid, shear stress is proportional to rate of shear strain.

3.

[Ans. A] =

=

( ) ( )

=

2.

uoy n y or s ous or v om ntum d us v ty = = rm l d us v ty n rt or = s ous ro =

[Ans. D] We know for turbulent flow of a fluid ylond’s numb r = Also, due to turbulent nature of flow, fluid will be well mixed.

th

th

th



Fluid Mechanics

Fluid Statics ME 2005 1. A U - tube manometer with a small quantity of mercury is used to measure the static pressure difference between two locations A and B in a conical section through which an incompressible fluid flows. At a particular flow rate, the mercury column appears as shown in the figure. The density of mercury is 13600 kg/ and g =9.81 m/ .Which of the following is correct? A

below. Density of water is 1000 kg/ . The minimum mass of the gate in kg per unit width (perpendicular to the plane of paper ), required to keep it closed is 5m

(A) 5000

(C) 7546

(D) 9623

ME 4.

2014 Water flows through a pipe having an inner radius of 10 mm at the rate of 36 kg/hr at 25°C. The viscosity of water at 25°C is 0.001 kg/m.s. The Reynolds number of the flow is _______

5.

For a completely submerged body with ntr o gr v ty ‘ ’ nd ntr o buoy n y ‘ ’, t ond t on o st b l ty will be (A) G is located below B (B) G is located above B (C) G and B are coincident (D) Independent of the locations of G & B

6.

A siphon is used to drain water from a large tank as shown in the figure below. Assume that the level of water is maintained constant. Ignore frictional effect due to viscosity and losses at entry and exit. At the exit of the siphon, the velocity of water is

B

150mm

(A) Flow direction is A to B and p p =20 kPa (B) Flow direction is B to A and p p =1.4 kPa (C) Flow direction is A to B and p p =20 kPa (D) Flow direction is B to A and p p =1.4 kPa

(B) 6600

ME 2010 2. For the stability of a floating body, under the influence of gravity alone, which of the following is TRUE? (A) Metacentre should be below centre of gravity. (B) Metacentre should be above centre of gravity. (C) Metacentre and centre of gravity must lie on the same horizontal line. (D) Metacentre and centre of gravity must lie on the same vertical line.

tum

(A) √ g(

ME 2013 3. A hinged gate of length 5 m, inclined at with the horizontal and with water mass on its left, is shown in the figure

)

(B) √ g (C) √ g (D) √ g th

th

th



Fluid Mechanics


[Ans. A]

x = centroid of prism = = = Taking moments about hinge) F(x mg ( ) os = m = 9612.88 Kg

B A 150m

4.

[Ans. *] Range 635 to 638 =

PA-PB = g.

=13600 9.81 0.15 = 20 kPa As pressure is decreasing from A to B, So flow direction is A to B. 2.

3.

[Ans. B] For stable equilibrium meta centre should be above centre of gravity.

ṁ =

=

=

m

=

= (

=

)

= 5.

[Ans. A]

[Ans. D] mg

t bl B has to be located above

6.

[Ans. B]

s p on mg

tum

Let us solve it by Pressure diagram (here it is triangular prism)

P and R are open to atmosphere, so gauge pressure at P and R will be zero y pply ng rnoull ’s pr n pl , t nd R: =

m

=

=

nd

=

=

g = g =√ g At exist of siphon, velocity of water = √ g

g sn

F = volume of diagram = g sn = 61312.5 th

th

th



Fluid Mechanics

Fluid Kinematics ME 2005 1. A leaf is caught in a whirlpool. At a given instant, the leaf is at a distance of 120 m from the center of the whirlpool. The whirlpool can be described by the following velocity distribution: =

(

) m/s and

=

m/s, where r (in meters) is the distance from the center of the whirlpool. What will be the distance of the leaf from the center when it has moved through half a revolution? (A) 48 m (C) 120 m (B) 64 m (D) 142 m 2.

The velocity components in the x and y directions of a two dimensional potential flow are u and v, respectively. Then

is

equal to

(A)

(C)

(B)

(D)

ME 2006 3. A two-dimensional flow field has velocities along the x and y directions given by u = x t and v = 2xyt respectively, where t is time. The equation of streamlines is: (A) = constant (B) = constant (C) = constant (D) not possible to determine 4.

In a two-dimensional velocity field with velocities u and v along the x and y directions respectively, the convective acceleration along the x-direction is given by

(A) u

v

(C) u

v

(B) u

v

(D) v

u

ME 2007 5. Which combination of the following statements about steady incompressible forced vortex flow is correct? P: Shear stress is zero at all points in the flow. Q: Vorticity is zero at all points in the flow. R: Velocity is directly proportional to the radius from the centre of the vortex. S: Total mechanical energy per unit mass is constant in the entire flow field. (A) P and Q (C) P and R (B) R and S (D) P and S ME 2008 Statement for Linked Answer Questions 6 and 7: The gap between a moving circular plate and a stationary surface is being continuously reduced, as the circular plate comes down at a uniform speed V towards the stationary bottom surface, as shown in the figure. In the process, the fluid contained between the two plates flows out radially. The fluid is assumed to be incompressible and inviscid. R r

h

Moving circular plate

V

Stationary surface

6.

The radial velocity v at any radius r, when the gap width is h, is

7.

(A) v =

(C) v =

(B) v =

(D) v =

The radial component of the fluid acceleration at r = R is

th

(A)

(C)

(B)

(D)

th

th



8.

For a continuity equation given by ⃗ ⃗ = to be valid, where ⃗ is the velocity vector, which one of the following is a necessary condition? (A) steady flow (B) irrotational flow (C) inviscid flow (D) incompressible flow

ME 2009 9. You are asked to evaluate assorted fluid flows for their suitability in a given laboratory application. The following three flow choices, expressed in terms of the two-dimensional velocity fields in the x-y plane, are made available. P. u = 2y, v = -3x Q. u = 3xy, v = 0 R. u = 2x, v = 2y Which flows(s) should be recommended when the application requires the flow to be incompressible and irrotational? (A) P and R (C) Q and R (B) Q (D) R

Fluid Mechanics

ME 2014 12. Consider the following statements regarding streamline(s): (i) It is a continuous line such that the tangent at any point on it shows the velocity vector at that point (ii) There is no flow across streamlines (iii)

=

=

s the differential

equation of a streamline, where u, v and w are velocities in directions x, y and z, respectively (iv) In an unsteady flow, the path of a particle is a streamline Which one of the following combinations of the statements is true? (A) (i), (ii), (iv) (C) (i), (iii), (iv) (B) (ii), (iii), (iv) (D) (i), (ii), (iii) 13.

Consider a velocity field ⃗ = (y ̂ x ̂ ), where K is a constant. The vorticity, , is C / (B) K (D) K /2

ME 2010 10. Velocity vector of a flow field is given as ⃗ = xy ̂ x z .̂ The vorticity vector at (1, 1, 1) is (A) ̂ ̂ (C) ̂ ̂ ̂ (B) ̂ ̂ (D) ̂ ME 2011 11. A streamline and an equipotential line in a flow field (A) Are parallel to each other (B) Are perpendicular to each other (C) Intersect at an acute angle (D) An identical

th

th

th



Fluid Mechanics


[Ans. B] Given data: Radial distance(r) = 120m Radial velocity distribution =

i.e. = of

leaf, 3.

+ m/s

*

[Ans. A] Given data: =x t = xyt

Angular velocity distribution of leaf, V0 = *

+ m/s

Let = Angle turned by a leaf from 0 For

=

= revolutions

=

(i.e. = radians). The radial distance of leaf from centre? We know, Vr =

= Integrating both sides ln x = ln y ln x y = ln x y = onst nt

= Rate of change of

radial distance of leaf =r

=r

= Rate of change of

angular displacement of leaf Given

=

*

=*

+=

-------- (1)

+=r

-------(2)

4.

[Ans. A] Two dimensional velocity field with velocities u, v and along x and y direction. Acceleration along x direction, =

Dividing Eq(1) and Eq(2) =

r

*

=

r

+

r

*

r

+

= = 2.

=

v

=

=

Convective acceleration+ temporal acceleration

dr = d r Integrating both sides, we get boundary limits dr ∫ = ∫ d r

log *

u

=

[log r ]

=

[ ]

+=

Since, therefore

Convective acceleration = u

v

5.

[Ans. B]

6.

[Ans. A] Given data: Let V = Downward velocity of circular plate (in m/sec) R = Radius of circular plate (m) h = Gap (width) between the moving plate and stationary plate. Vr = Radial v lo ty t r d us ‘r’ According to continuity Eq(law of conservation of mass) ṁ = constant

= =

= 0 for 2-dimensional field,

m

[Ans. D] For two dimensional flow, continuity equation should be satisfied

th

th

th



= =

8. n ompr ss bl r

= =

[Ans. D] The basic equation of continuity for fluid flow is given by

lu d

=

=

Radial velocity at r = R = 7.

=

Now if remains constant, then only we can write ⃗ . ⃗ =0

or

[Ans. C] let ar = Radial component of acceleration at r = R. according to law of conservation of energy m (mass flow rate) = constant (c) =C For incompressible fluid A.V = constant ( ) Volume flow rate = C For given instance, Volume of matter = constant (fixed) r = onst nt Here r and h are variable. Differentiate w.r.t time(t) d r = dt r * r += * r

r

i.e,

=

=

9.

[Ans. D] Incompressible flow satisfy continuity equation u v = x y P. u = y, v = x y x x y Incompressible flow = Q. u = xy, v = xy x y Compressible flow y R. u = x, v = y x y x y Incompressible flow = For irrotational flow condition is v u w = ( )= x y For P: u = y, v = x x y w = * + x y

+=

=

=

* +

* +

Differentiate Vr w.r.t “t” =

=

Hence incompressible flow

Where V = velocity of circular plate moving down d = dt r r = =

Fluid Mechanics

=

=

= = =

*

, *

+ =

* +=

(Rotational flow) For Q: u = xy, v =

+

=

w = *

=

= For R: u = th

th

x

xy + y

x

ot t on l low

x, v = y th



w =

y x

*

x y

=

Fluid Mechanics

+

=

rrot t on l low 10.

[Ans. D] Velocity vector =|

| xy

= [

x z x z ]

z

[ [

= = x t , , =

z

x zx

x

x z

y

xy ] xz

x

x

11.

[Ans. B]

12.

[Ans. D] (i), (ii), (iii) are correct but (iv) in an unsteady flow, the path of particle is not streamline

13.

[Ans. A] ⃗ = (y ̂

x ̂ ),

vort ty = w = =[

xy ]

y

[

x

u ] y

y ]=

th

th

th



Fluid Mechanics

Fluid Dynamics ME 2005 1. A venturimeter of 20 mm throat diameter is used to measure the velocity of water in a horizontal pipe of 40 mm diameter. If the pressure difference between the pipe and throat sections is found to be 30 kPa then, neglecting frictional losses, the flow velocity is (A) 0.2 m/s (C) 1.4 m/s (B) 1.0 m/s (D) 2.0 m/s

ME 2011 4. Figure shows the schematic for the measurement of velocity of air (density =1.2 kg/m ) through a constant - area duct using a pitot tube and a water tube manometer. The differential head of water (density = 1000 kg/m ) in the two columns of the manometer is 10 mm. take acceleration due to gravity as 9.8 m/s . The velocity of air in m/s is

ME 2007 2. In a steady flow through a nozzle, the flow velocity on the nozzle axis is given by ) , where x is the distance

10 mm

v=u (

Flow

along the axis of the nozzle from its inlet plane and L is the length of the nozzle. The time required for a fluid particle on the axis to travel from the inlet to the exit lane of the nozzle is (A) (B)

(A) 6.4 (B) 9.0

(C) ln

(D)

ME 2010 3. A smooth pipe of diameter 200 mm carries water. The pressure in the pipe at section S1 (elevation: 10m) is 50 kPa. At section S2 (elevation: 12m) the pressure is 20 kPa and velocity is 2 ms . Density of water is 1000 gm and acceleration due to gravity is 9.8 ms . Which of the following is TRUE (A) flow is from S1 to S2 and head loss is 0.53 m (B) flow is from S2 to S1 and head loss is 0.53 m (C) flow is from S1 to S2 and head loss is 1.06 m (D) flow is from S2 to S1 and head loss is 1.06 m

(C) 12.8 (D) 25.6

ME 2012 5. A large tank with a nozzle attached contains three immiscible , inviscid fluids as shown. Assuming that the changes in h1 , h2 and h3 are negligible, the instantaneous discharge velocity is

(A) √ g

(

)

(B) √ g (C) √ g (

)

(D) √ g (

th

th

)

th



ME 2013 6. Water is coming out from a tap and falls vertically downwards. At the tap opening , the stream diameter is 20mm with uniform velocity of 2 m/s. Acceleration due to gravity is 9.81 m/s . Assuming steady, inviscid flow, constant atmospheric pressure everywhere and neglecting curvature and surface tension effects, the diameter in mm of the stream 0.5 m below the tap is approximately (A) 10 (C) 20 (B) 15 (D) 25 ME 2014 7. In a simple concentric shaft-bearing arrangement, the lubricant flows in the 2 mm gap between the shaft and the bearing. The flow may be assumed to be a plane Couette flow with zero pressure gradient. The diameter of the shaft is 100 mm and its tangential speed is 10 m/s. The dynamic viscosity of the lubricant is 0.1 kg/m.s. The frictional resisting force (in Newton) per 100 mm length of the bearing is _______

Fluid Mechanics

For an incompressible flow field, ⃗ , which one of the following conditions must be satisfied? ⃗⃗⃗ = (A) (C) ⃗⃗⃗⃗ ⃗⃗⃗ =

8.

⃗⃗⃗ =

(B)

(D)

⃗⃗⃗

⃗⃗⃗⃗

⃗⃗⃗ =

9.

A fluid of dynamic viscosity 2× kg/ms and density 1 kg/m flows with an average velocity of 1 m/s through a long duct of rectangular (25 mm × 15 mm) cross-section. Assuming laminar flow, the pressure drop (in Pa) in the fully developed region per meter length of the duct is _______

10.

A flow field which has only convective acceleration is (A) a steady uniform flow (B) an unsteady uniform flow (C) a steady non-uniform flow (D) an unsteady non-uniform flow


[Ans. D] We know,

=

A1V1 =A2V2

=

=

=

=4 = m/s So velocity of flow is 2.0 m/sec

h

2.

20mm

40mm

,

V1,P1

= pply ng + =

[Ans. B]

= =

rnoull ’s Equ t on

+ z1 =

+

= =

=u (

+ z2

=u (

) )

(for horizontal pipe) th

th

th



(

Integrating ∫

(

*ln ( = 3.

z =z = =

= dt

)

)

Fluid Mechanics

= ∫ dt

g

=

g

= g

)+ = ln

g

, g

g

g

g

ln

= g

[Ans. C] pply rnoull ’s t

g g

g

or m

=

= As area is constant. Velocity head will be same

g

=√ g

g

g

g

g

[

]

= = m =1.06m Heat loss occurs in the direction 1-2 flow is from s1 to s2 and heat loss is 1.06m 4.

6.

[Ans. B]

[Ans. C] Given data Density of air, = g/m Density of water, = g/m Acceleration due to gravity (g) =9.8 m/s Manometer reading, = mm = m

z = =

Velocity of air =C = √ g Where Cv = 1 = w

(

r

= =

=

5.

v = v =

=

(

( =

,

=

d =d √ m t rs o

g

g

=

g

z

g

=v v d v v = v

√

= mm

[Ans. *]Range 15 to 16 v = = = y = = = =

m/s

g nd

v

r 7.

= =

√

[Ans. A] = g g rnoull s qu t on b tw n g

d v =

) )=

z =

= g g

),

z

8.

[Ans. A] ⃗ = th

th

d v rg n th

/m

o v lo ty v tor



9.

Fluid Mechanics

[Ans. *]Range 1.7 to 2.0 Given, = g/ms, = g/m , u = m/s Duct area = 25 mm 15 mm = Here =

= Friction factor =

= r ,

=

=

=

=

=

mm _____________(2) u

= =

rm t r

= =

______________(3) Substituting eqn. (2) & (3) in (1) = 10.

=

Pa/m.

[Ans. C] Convective acceleration is that having acceleration with respect to space Temporal on local acceleration is that having acceleration with respect to time A flow field having only convective acceleration is steady non uniform flow. Since it does not depend on time.

th

th

th



Fluid Mechanics

Boundary Layer ME 2006 Statement for Linked Answer Questions 1 & 2: A smooth flat plate with a sharp leading edge is placed along a gas stream flowing at U = 10 m/s. The thickness of the boundary layer at section r – s is 10 mm, the breadth of the plate is 1 m (into the paper) and the density of the gas = 1.0 kg/m3. Assume that the boundary layer is thin, two-dimensional, and follows a linear velocity distribution, u = U (y/ ), at the section r – s, where y is the height from plate. q

the plate to the drag force on the rear half, then (A) F < 1/2 (C) F = 1 (B) F = 1/2 (D) F > 1 ME 2010 5. Match the following P: Compressible flow Q: Free surface flow R: Boundary layer flow S: Pipe flow

r U

T: Heat convection

U

p

s

(A) (B) (C) (D)

Flat plate

1.

2.

The mass flow rate (in kg/s) across the section q – r is (A) zero (C) 0.10 (B) 0.05 (D) 0.15

4.

6.

A phenomenon is modeled using n dimensional variables with k primary dimensions. The number of non – dimensional variables is (A) (C) (B) (D)

The integrated drag force (in N) on the plate, between p – s, is: (A) 0.67 (C) 0.17 (B) 0.33 (D) zero

ME 2007 3. A model of a hydraulic turbine is tested at a head of 1/4th of that under which the full scale turbine works. The diameter of the model is half of that of the full scale turbine. If N is the RPM of the full scale turbine, then the RPM of the model will be (A) N/4 (C) N (B) N/2 (D) 2N Consider an incompressible laminar boundary layer flow over a flat plate of length L, aligned with the direction of an oncoming uniform free stream. If F is the ratio of the drag force on the front half of

U: Reynolds number V: Nusselt number W: Weber number X: Froude number Y: Mach number Z: Skin friction coefficient

ME 2012 7. An incompressible fluid flows over a flat plate with zero pressure gradient. The boundary layer thickness is 1mm at a location where the Reynolds number is 1000. If the velocity of the fluid alone is increased by a factor of 4, then boundary layer thickness at the same location, in mm will be (A) 4 (C) 0.5 (B) 2 (D) 0.25

th

th

th



Fluid Mechanics

ME 2014 8. Consider laminar flow of water over a flat plate of length 1 m. If the boundary layer thickness at a distance of 0.25 m from the leading edge of the plate is 8 mm, the boundary layer thickness (in mm), at a distance of 0.75 m, is _______


[Ans. B] ṁ = = d u = dy

3.

[Ans. C] Applying similarity condition, √ √ ( ) =( )

=

∫ ydy

=

* + =

= =

√

=

= = 2.

√

=

=

g/s

[Ans. C] Integrated drag force (FID) =∫

=∫

Where

4.

[Ans. D] Drag force

=C =

dx

√

=

=

√

Where = momentum thickness =∫

*

+ dy

=∫

*

+ dy

=*

, √ Now Drag force on front half √

/

+

=

=

√ Drag force on rear half / / =

= =

=

u

=

*

=

s y

+

=( Now

th

th

√ =

) / /

th



=

7.

√

(

√

)

=

[Ans. C]

√

5.

[Ans. D]

6.

[Ans. C] By Buckingham – pie theorem, Non–dimensional variables, = n – k.

Fluid Mechanics

=√

=√

=√ √

=

8.

=

=

mm

[Ans. *] Range 13.5 to 14.2 √x √x

th

=

th

√x

= √

th

=



Fluid Mechanics

Flow through Pipes ME 2006 1. A siphon draws water from a reservoir and discharges it out at atmospheric pressure. Assuming ideal fluid and the reservoir is large, the velocity at point P in the siphon tube is:

velocity profile at section B downstream is y y ,

u=

y y

{

P

3.

The ratio (A)

,

y

/u is (C)

/

(B) 1 4.

2.

(A) √

(C) √

(B) √

(D) √

(C)

(B)

(D)

(D)

The ratio

(where

/

and

are the

pressures at section A and B, respectively, nd s t d ns ty o t lu d s (A)

(

/

(B)

The velocity profile in fully developed laminar flow in a pipe of diameter D is given by u = u r / , where r is the radial distance from the center. If the viscosity of the fluid is , the pressure drop across a length L of the pipe is: (A)

/

ME 2007 Linked Answer Questions: Q.3 – Q.4 Consider a steady incompressible flow through a channel as shown below. y

5.

(C)

)

(D)

/

⁄

(

)

/

Consider steady laminar incompressible axi-symmetric fully developed viscous flow through a straight circular pipe of constant cross – sectional area at a Reynolds number of 5. The ratio of inertia force to viscous force on a fluid particle is (A) 5 (C) 0 (D) ∞ (B)

ME 2009 6. The velocity profile of a fully developed laminar flow in a straight circular pipe, as shown in the figure, is given by the expression u (r) = where

( )(

)

is a constant. The average

velocity of fluid in the pipe is u(r)

r

A

B

R

x

X

The velocity profile is uniform with a value of u at the inlet section A. The th

th

th



7.

(A)

( )

(C)

( )

(B)

( )

(D)

( )

Water at C is flowing through a 1.0 km long G.I pipe of 200mm diameter at the rate of 0.07 /s. If value of Darcy friction factor for this pipe is 0.02 and density of water is 1000 kg/ , the pumping power (in kW) required to maintain the flow is (A) 1.8 (C) 20.5 (B) 17.4 (D) 41.0

Fluid Mechanics

ME 2014 11. For a fully developed flow of water in a pipe having diameter 10 cm, velocity 0.1 m/s and kinematic viscosity m /s, the value of Darcy friction factor is _______ 12.

ME 2010 8. The maximum velocity of a one – dimensional incompressible fully developed viscous flow, between two fixed parallel plates, is 6 ms . The mean velocity ( in ms ) of the flow is (A) 2 (C) 4 (B) 3 (D) 5

Water flows through a 10 mm diameter and 250 m long smooth pipe at an average velocity of 0.1 m/s. The density and the viscosity of water are 997 kg/m and 855× N.s/ m respectively. Assuming fully-developed flow, the pressure drop (in Pa) in the pipe is _______

ME 2012 9. Oil flows through a 200 mm diameter horizontal cast iron pipe (friction factor, f = 0.0225) of length 500 m. The volumetric flow rate is 0.2 m /s . The head loss (in m) due to friction is (assume g = 9.81 m/s ) (A) 116.18 (C) 18.22 (B) 0.116 (D) 232.36 ME 2013 10. For steady, fully developed flow inside a straight pipe of diameter D, neglecting gr v ty ts, t pr ssur drop p ov r a length L and the wall shear stress are related by (A)

=

(C)

=

(B)

=

(D)

=

th

th

th



Fluid Mechanics


[Ans. C]

r

* u=u (

P

r

u [

p r x

]=

h1

r

p=u [

)+

][

]=

r

u

h2 3.

[Ans. A] Assume width of channel is b. applying continuity equation at A and B.

Q By energy conservation, velocity at point Q = √ g As there is a continuous and uniform flow, so velocity of liquid at point Q and P is same (i.e. Vp=VQ) i.e. kinetic energy of water = potential energy

H u

A

= √ mg =

4.

=

g

[Ans. D] By Hagen – Poiseuille law, for steady laminar flow in circular pipes wton’s l w o v s os ty ross s t on

g

b =

[Ans. C] pply ng

=√ g

=

b = u

= mg

Vm

B

u

mv = mg

2.

H 2

rnoull ’s qu t on t

=

g =

g

nd

g g

=

-----------------(1) = [

Shear stress across a section varies w t ‘r’ s =

d [u * dr

Using the value of

-----------------(2)

For fully developed laminar flow Eq (1) = Eq (2) du p r = dr x du p = [ ]r dr x r

+] =

] from first part of the

question, we get, = * 5.

+

[Ans. A] ynold’s numb r =

p r x th

th

th

=



6.

[Ans. A] u=

9. ( )(

d =u = =

[Ans. A]

)

= = = g= d=

rdr ∫

rdr

∫

(

dp )( dx

r

) rdr

=

or u 7.

r

=

=

= =

=

v=

11. =

m/s

=

= g = = 17.4 kW

)=

=

[Ans. *] Range 0.06 to 0.07 =

=

=

/

(

=

Head loss,

[Ans. C] =

p ) x

=

=

8.

[Ans. A] =(

[Ans. B]

=

=

=

10.

dp ( ) dx

m

=

Solving, we get dp = ( ) dx Now Q = Area x average velocity u

Fluid Mechanics

v

n m t

=

( ) v s os ty

= = 12.

064

[Ans. *] Range 6800 to 6900 =

m ⁄s =

=

= =

=

= m ⁄s

th

th

th



Fluid Mechanics

Hydraulic Machines ME 2006 1. A horizontal shaft centrifugal pump lifts water at 65 . The suction nozzle is one meter below pump centerline. The pressure at this point equal 200 kPa gauge and velocity is 3 m/s. Steam tables show saturation pressure at 65 is 25 kPa, and specific volume of the saturated liquid is 0.001020 m3/ kg. The pump Net Positive suction Head (NPSH) in meters is:

ME 2007 4. Match the items in columns I and II. Column I Column II P : Centrifugal 1 : Axial flow compressor Q : Centrifugal pump 2 : Surging R : Pelton wheel 3 : Priming S : Kaplan turbine 4 : Pure impulse (A) , , , (B) , , , (C) , , , (D) , , , 5.

The inlet angle of runner blades of a Francis turbine is 90°. The blades are so shaped that the tangential component of velocity at blade outlet is zero. The flow velocity remains constant throughout the blade passage and is equal to half of the blade velocity at runner inlet. The blade efficiency of the runner is (A) 25% (B) 50% (C) 80% (D) 89%

P 1m

(A) 24 2.

3.

(B) 26

(C) 28

(D) 30

A large hydraulic turbine is to generate 300 kW at 1000 rpm under a head of 40 m. For initial testing, a 1:4 scale model of the turbine operates under a head of 10 m. The power generated by the model (in kW) will be (A) 2.34 (B) 4.68 (C) 9.38 (D) 18.75 In a Pelton wheel, the bucket peripheral speed is 10 m/s, the water jet velocity is 25 m/s and volumetric flow rate of the jet is 0.1 / . If the jet deflection angle is and the flow is ideal, the output developed is: (A) 7.5 kW (C) 22.5 kW (B) 15.0 kW (D) 37.5 kW

ME 2008 6. Water, having a density of 1000 kg/m , issues from a nozzle with a velocity of 10 m/s and the jet strikes a bucket mounted on a Pelton wheel. The wheel rotates at 10 rad/s. The mean diameter of the wheel is 1 m. The jet is split into two equal streams by the bucket, such that each stream is deflected by 120°, as shown in the figure. Friction in the bucket may be neglected. Magnitude of the torque exerted by the water on the wheel, per unit mass flow rate of the incoming jet, is Deflected jet

120°

Incoming jet

120°

Deflected jet th

th

th



(A) (B) (C) (D)

0 (N.m)/(kg/s) 1.25 (N.m)/(kg/s) 2.5(N.m)/(kg/s) 3.75(N.m)/(kg/s)

ME 2009 7. Consider steady, incompressible and irrotational flow through a reducer in a horizontal pipe where the diameter is reduced from 20cm to 10cm. The pressure in the 20cm pipe just upstream of the reducer is 150 kPa. The fluid has a vapour pressure of 50kPa and a specific weight of 5 kN/m3. Neglecting frictional effects, the maximum discharge (in m3/s) that can pass through the reducer without causing cavitation is (A) 0.05 (C) 0.27 (B) 0.16 (D) 0.38 ME 2010 8. A hydraulic turbine develops 1000 kW power for a head of 40m. If the head is reduced to 20m, the power developed (in kW) is (A) 177 (C) 500 (B) 354 (D) 707 ME 2011 9. A pump handling a liquid raises its pressure from 1 bar to 30 bar. Take the density of a liquid as 990 g/m . The isentropic specific work done by the pump in kJ/kg is (A) 0.10 (C) 2.50 (B) 0.30 (D) 2.93

Fluid Mechanics V2

W1

W2

V1 U

(A) 0 (B) 1

(C) 0.5 (D) 0.25

ME 2013 11. In order to have maximum power from a pelton turbine , the bucket speed must be (A) Equal to the jet speed (B) Equal to half of the jet speed (C) Equal to twice the jet speed (D) Independent of the jet speed ME 2014 12. An ideal water jet with volume flow rate of 0.05 m /s strikes a flat plate placed normal to its path and exerts a force of 1000 N. Considering the density of water as 1000 kg/m , the diameter (in mm)of the water jet is _______ 13.

Steam at a velocity of 10 m/s enters the impulse turbine stage with symmetrical blading having blade angle 30°. The enthalpy drop in the stage is 100 kJ. The nozzle angle is 20°. The maximum blade efficiency (in percent) is _______

14.

At the inlet of an axial impulse turbine rotor, the blade linear speed is 25 m/s, the magnitude of absolute velocity is 100 m/s and the angle between them is 25°. The relative velocity and the axial component of velocity remain the same between the inlet and outlet of the blades. The blade inlet and outlet velocity triangles are shown in the figure. Assuming no losses, the specific work (in J/kg)

ME 2012 10. The velocity triangles at the inlet and exit of the rotor of a turbomachine are shown. V denotes the absolute velocity of the fluid. W denotes the relative velocity of the fluid and U denotes the blade velocity. Subscripts 1 and 2 refer to inlet and outlet respectively. If = and = , then the degree of reaction is

m/s

m/s

m/s

/ m/s

th

th

th



15.

Fluid Mechanics

Kaplan water turbine is commonly used when the flow through its runner is (A) axial and the head available is more than 100 m (B) axial and the head available is less than 10 m (C) radial and the head available is more than 100 m (D) mixed and the head available is about 50 m


[Ans. A] Given data Gauge reading, = g =

=

/m

=

Velocity,

= = = =

g ug

= =

= m / g

=

=

=

=

os os

[Ans. A] (P) Centrifugal compressor – Surging (Q) Centrifugal pump – Priming (R) Pelton wheel – Pure impulse (S) Kaplan turbine – Axial flow

5.

[Ans. C]

m

/m g/m

u

4.

= m/s

Velocity head at inlet,

v

Vw1 = u1

=

=

NPSH= Pressure at centre line (gauge reading) + static head (Hs) + Velocity

Vf1 =Vf2=

V1

Inlet Triangle

head * + = 20.387+1+0.46+2.6 = 24m 2.

Vf2

[Ans. A] For similar turbines specific power will be same. =

/

= =

Vf2 = v2

u1

/

* + * +

* + * +

Outlet Triangle

/

Given, v = v =

/

v =v =

= 3.

v = v =u

[Ans. C] = = Power developed by pelton wheel

v * + = u

Blade efficiency =

th

th

th



=

or

u u =

r=

=

orqu / g/s =

nd

1

Upstream 2

=

Reducer Q m 1

d =

8.

⁄

=

2

=

=

d / g

11.

[Ans. B] = v u u To find speed maximum power

=

= ;

m

=

v

=

v

u

u

u

= v = u u= 12. m

Also, discharge Q= ( d ) v = ( d ) v or

⁄

[Ans. C] For symmetrical velocity, Base of reaction is 0.5.

=

or

( )

10.

But w = w = 5 (incompressible flow)

=

⁄

[Ans. D] Work done = = =

=

or

( )

=

downstream

d =

= ( d )v

[Ans. B]

m⁄ g/s

=C

rg ,

m/s

= 0.16 m /s

9.

[Ans. B] Considering potential head difference= 0, i.e z =z pply rnoull ’s t or m

=

=

=

r=

= 7.

s

[Ans. D] Force exerted by the jet of water on the bucket per unit mass flow rate = v u v u os = / g/s When v = absolute velocity of jet = 10 m/s u = velocity of the plate =

= 20

or v = √

= 6.

Fluid Mechanics

=

= ( ) =( )

or v =

[Ans. *] Range 56 to 57 Force exerted by jet normal to the plate = Q= AV

=

(2)

From equation (1) and (2)

=

=

d = d= d=

=

th

th

m mm th



13.

[Ans. *]Range 85.1 to 89.9 Nozzle angle, = Maximum blade efficiency = os = os =

14.

[Ans. *]Range 3250 to 3300 = =

=

m/s

m/s

m/s

= u=

Fluid Mechanics

m/s

m/s

Given = sn = sn sn = sn = = = = now = os os = os = os = m/sec sp wor = u= = J/kg. 15.

[Ans. B] Kaplan turbine is an axial flow reaction turbine. It is of low head, high discharge and high specific speed turbine

th

th

th



Heat Transfer

Conduction ME –2005 1. A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/m.K, its density 9000 kg/ and its specific heat 385 J/kg.K. If the heat transfer coefficient is 250 W/ .K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s, (A) 8.7 (C) 17.3 (B) 13.9 (D) 27.7 2.

(B) convection decreases, while that due to conduction increases (C) convection and conduction decreases (D) convection and conduction increases 5.

In a composite slab, the temperature at the interface ( ) between two materials is equal to the average of the temperature at the two ends. Assuming steady one-dimensional heat conduction, which of the following statement is true about the respective thermal conductivities?

Heat flows through a composite slab, as shown below. The depth of the slab is 1m. The k values are in W/m.K. The overall thermal resistance in K/W is 2b

k = 0.02

k= 0.10

0.5m

k= 0.04 0.25m

q

(A) 17.2 (B) 21.9 3.

1m

In a case of one dimensional heat conduction in a medium with constant properties, T is the temperature at position

x,

at

time

t.

Then

is

proportional to

(A)

(C)

(B)

(D)

(A) (B)

0.5m

(C) 28.6 (D) 39.2

ME – 2006 4. With an increase in the thickness of insulation around a circular pipe, heat loss to surroundings due to (A) convection increases, while that due to conduction decreases

b

(C) (D)

ME – 2007 Linked Answer Questions: Q. 6 – Q. 7 Consider steady one-dimensional heat flow in a plate of 20 mm thickness with a uniform heat generation of 80 MW/m3. The left and right faces are kept at constant temperatures of 1600C and 1200C respectively. The plate has a constant thermal conductivity of 200 W/mK. 6. The location of maximum temperature within the plate from its left face is (A) 15 mm (C) 5mm (B) 10mm (D) 0mm 7.

The maximum temperature within the plate in 0C is (A) 160 (C) 200 (B) 165 (D) 250

th

th

th



ME – 2008 8. Steady two dimensional heat conduction takes place in the body shown in the figure below. The normal temperature gradients over surfaces P and Q can be considered to be uniform. The temperature gradient

of 0.1 W/m.K. The values of

d

at

surface P are y

surface Q,0 C 2m 1m Surface P, 100 C x

(A)

= 20 K/m,

(B)

= 0K/m,

(C)

= 10 K/m,

(D)

= 0K/m,

= 0 K/m =10K/m =10 K/m = 20 K/m

ME – 2009 9. Consider steady-state heat conduction across the thickness in a plane composite wall (as shown in the figure) exposed to convection conditions on both sides. h

Given: h

1

2

⁄

and Assuming negligible contact resistance between the wall surfaces, the interface temperature, T (i C), of the two walls will be (A) 0.50 (C) 3.75 (B) 2.75 (D) 4.50

at surface Q is

equal to 10 K/m. Surfaces P and Q are maintained at constant temperatures as shown in the figure, while the remaining part of the boundary is insulated. The body has a constant thermal conductivity

h

h

⁄

Heat Transfer

ME – 2010 10. A fin has 5 mm diameter and 100 mm length. The thermal conductivity of fin material is 400 W/mK. One end of the fin is i t i ed t ℃ d its re i i g surf ce is exposed to bie t ir t ℃ If the convective heat transfer coefficient is 40 W/m2K, the heat loss (in W) from the fin is (A) 0.08 (C) 7.0 (B) 5.0 (D) 7.8 ME – 2011 11. A pipe of 25 mm diameter carries steam. The heat transfer coefficient between the cylinder and surrounding is 5 W/ K. it is proposed to reduce the heat loss from the pipe by adding insulation having a thermal conductivity 0f 0.05 W/mK. Which one of the following statements are TRUE? (A) The outer radius of the pipe is equal to the critical radius. (B) The outer radius of the pipe is less than the critical radius. (C) Adding the insulation will reduce the heat loss. (D) Adding the insulation will increase the heat loss. ME – 2012 12. Which one of the following configurations has highest effectiveness? (A) Thin, closely spaced fins (B) Thin, widely spaced fins (C) Thick, widely spaced fins (D) Thick, closely spaced fins th

th

th



ME – 2013 13. Consider one – dimensional steady state heat conduction, without heat generation, in a plane wall; with boundary conditions as shown in the figure below. The conductivity of the wall is given by k= b ; where and b are positive constants , and T is temperature.

where

As x increases, the temperature gradient (dT/dx) will (A) Remain constant (B) Be zero (C) Increase (D) Decrease 14.

and specific heat c = 600 J/kgK. The time required in seconds to cool the steel ball in air from to is (A) 519 (C) 1195 (B) 931 (D) 2144 ME – 2014 16. Biot number signifies the ratio of (A) Convective resistance in the fluid to conductive resistance in the solid (B) Conductive resistance in the solid to convective resistance in the fluid (C) Inertia force to viscous force in the fluid (D) buoyancy force to viscous force in the fluid 17.

Consider one – dimensional steady state heat conduction along x – xis ( ≤ x ≤ ) through a plane wall with the boundary surface (x=0 and x = L) maintained at temperatures of and 100oC. heat is generated uniformly throughout the wall. Choose the CORRECT statement. (A) The direction of heat transfer will be from the surface at C to the surface at C. (B) The maximum temperature inside the wall must be greater than C. (C) The temperature distribution is linear within the wall. (D) The temperature distribution is symmetric about the mid – plane of the wall.

Consider one dimensional steady state heat conduction across a wall (as shown in figure below) of thickness 30 mm and thermal conductivity 15 W/m.K. At x , a constant heat flux, q" = 1× W/ is applied. On the other side of the wall, heat is removed from the wall by convection with a fluid at 25°C and heat transfer coefficient of 250 W/ .K. The temperature (in °C), at x = 0 is _______

q

x x

18. 15.

Heat Transfer

A steel ball of diameter 60mm is initially in thermal equilibrium at C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at C , with convective heat transfer coefficient h =20 W/ K. the thermo – physical properties of steel are ; density kg/ , conductivity k = 40 W/mK th

A material P of thickness 1 mm is sandwiched between two steel slabs, as shown in the figure below. A heat flux 10 kW/ is supplied to one of the steel slabs as shown. The boundary temperatures of the slabs are indicated in the figure. Assume thermal conductivity of this steel is 10 W/m.K. Considering one-dimensional steady state heat th

th



conduction for the configuration, the thermal conductivity (k, in W/m.K) of material P is _______ ⁄

21.

As the temperature increases, the thermal conductivity of a gas (A) increases (B) decreases (C) remains constant (D) increases up to a certain temperature and then decreases

22.

A plane wall has a thermal conductivity of 1.15 W/m.K. If the inner surface is at 1100°C and the outer surface is at 350°C, then the design thickness (in meter) of the wall to maintain a steady heat flux of 2500 W/ should be _______

20

in mm

20

K

All dimensions

STEEL SLAB

P

STEEL SLAB

q

19.

Consider a long cylindrical tube of inner and outer radii, r and r , respectively, length, L and thermal conductivity, k. Its inner and outer surfaces are maintained at and o , respectively( > ). Assuming one-dimensional steady state heat conduction in the radial direction, the thermal resistance in the wall of the tube is r r ( ) ( ) ( ) ( ) r r r ( ) ( ) ( ) r r

20.

Heat transfer through a composite wall is shown in figure. Both the sections of the wall have equal thickness (l). The conductivity of one section is k and that of the other is 2k. The left face of the wall is at 600 K and the right face is at 300 K.

Heat Transfer

e t f ow

The interface temperature composite wall is _______

(in K) of the

th

th

th



Heat Transfer


[Ans. C] K = 400 = 9000

3.

[Ans. D] One dimensional equation is,

h = 250

x

heat

conduction

t

∴ =8.33× T= d | dt

( h

[Ans. A] As the insulation thickness increases, the surface area exposed for H.T by convection increases and wall thickness also increases hence the convection H.T increases and conduction H.T reduces.

5.

[Ans. D]

)e (

|

)e (

) magnitude

| 2.

4.

[Ans. C] Resistance diagram

A

(

Q=

)

(

)

∴

C

B

6.

[Ans. C]

C

Now, ⁄

C

x

x =0

⁄

x=20mm

=0 By integrating twice ∴

⁄

x2 + C1x + C

… (1)

At x , T=433K ∴ 2 = 433 At x ×10-3m, T = 393K. ∴ 1 = 2000 For maximum temperature,

Equivalent resistance between B and C

⁄

=0

Total resistance = =25+3.6 =28.6 ⁄ th

∴

+ C1=0

x

× 10-3 = 5mm th

th



7.

8.

[Ans. B] = [putting x equation (1)]

11. ×

10-3

in

surface P1,

[Ans. C] r h / r

[Ans. D] Direction of heat flow is always normal to surface of constant temperature. So for

Heat Transfer

h

r r addition reduce heat loss

= 0 k/m

From energy conservation, Heat rate at P = Heat rate at Q d d ( ) ( ) dy dx

of

insulation

will

Q

= 20 K/m 9.

h

h

t

12.

[Ans. A] Fin effectiveness, he t tr sfer with fi he t tr sfer without fi And this is highest with thin closely spaced fins.

13.

[Ans. D] From energy equation d dt ( ) dx dx dt c c dx bt As x increases, T increases from

t

Under steady state conditions, ( t) t ( ) t

∴

t

t

or or 20

t =

(t

or 20 t or 3.826 t

(t

and

) )

to

decreases

14.

[Ans. B] General profile is

15.

[Ans. D]

or t 10.

r

r

[Ans. C]

[Ans. B] √

h

m=√

d

d

d

√ (t

t )t

(

h(

)

) t

h(

)

u ped

= 5.008W

th

th

ysis

r

th

r r



exp (

20.

t)

exp (

Heat Transfer


t)

Apply log, x

t

t = 2144 s 16.

[Ans. B] iot

17.

( )

[Ans. *] Range 620 to 630 By energy balance: ( ) q h ( t (1) (2) usi g ) q h( q h usi g ( ) qt qt

Heat transfer through section 1 =heat transfer through section 2 her

)

her

(3)

ce of sectio

esist

ce of sectio

(

)

–

21. 18.

esist

[Ans. A] Thermal conductivity of gas √ Hence if temperature increase, increases

[Ans. *]Range 0.09 to 0.11 q q

22.

[Ans. *] Range 0.33 to 0.35 x

19.

[Ans. C] 1D heat conduction through long cylindrical tube in radial direction is given by : ( ) e t tr sfer r te ( ) her her

resist

ce

x

(

)

x Hence thickness of wall = 0.345 m

resist ce r ( ) r

th

th

th



Heat Transfer

Convection ME – 2005 Statement for Linked Answer Questions 1 and 2: An un-insulated air conditioning duct of rectangular cross section 1m × 0.5 m, carrying air at C with a velocity of 10 m/s, is exposed to an ambient air of C. Neglect the effect of duct construction material. For air In the range of 20C, data are as follows: Thermal conductivity = 0.025 W/m.K: viscosity μ s r dt u ber density=1.2 kg/m3. The laminar flow Nusselt number is 3.4 for constant wall temperature conditions and, for turbulent flow, Nu =0.023 Re0.8Pr0.33 1. The Reynolds number for the flow is (A) 444 (C) 4.44 (B) 890 (D) 5.33 105 2.

The heat transfer per metre length of the duct , in watts, is (A) 3.8 (C) 89 (B) 5.3 (D) 769

ME – 2007 3. The average heat transfer co-efficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assume the plate temperature to be uniform at any instant of time and radiation heat exchange with the surroundings is negligible. The ambient temperature is the plate has a total surface area of 0.1 and a mass of 4 kg. The specific heat of the plate material is 2.5 kJ/kgK. The convective heat transfer co-efficient in W/ K, at the instant when the plate temperature is c and the change in plate temperature with time dT/dt = 0.02 K/s, is (A) 200 (C) 15 (B) 20 (D) 10

4.

The temperature distribution within the thermal boundary layer over a heated isothermal flat plate is given by ( )

( )

where

d

are the

temperatures of plate and free stream respectively, and y is the normal distance measured from the plate. The local Nusselt number based on the thermal boundary layer thickness is given by (A) 1.33 (C) 2.0 (B) 1.50 (D) 4.64 ME – 2008 5. For the three-dimensional object shown in the figure below, five faces are insulated. The sixth face (PQRS), which is not insulated, interacts thermally with the ambient, with a convective heat transfer coefficient of 10 W/m2.K. The ambient temperature is C. Heat is uniformly generated inside the object at the rate of 100 W/m3. Assuming the face PQRS to be at uniform temperature, its steady state temperature is Q

F

E

P

G R

2m

2m

H

S 1m

(A) (B)

th

(C) (D)

th

th



6.

For flow of fluid over a heated plate, the following fluid properties are known Viscosity = 0.001 Pa.s; specific heat at constant pressure = 1kJ/kg.K; thermal conductivity = 1 W/mK. The hydrodynamic boundary layer thickness at a specified location on the plate is 1 mm. The thermal boundary layer thickness at the same location is (A) 0.001 mm (C) 1 mm (B) 0.01 mm (D) 1000 mm

ME – 2009 7. A coolant fluid at C flows over a heated flat plate maintained at a constant temperature of C. The boundary layer temperature distribution at a given location on the plate may be approximated as T = 30 + 70 exp ( y) where y (in m) is the distance normal to the plate and T is in °C. If thermal conductivity of the fluid is 1.0W/mK, the local convective heat transfer coefficient (in W/ K) at that location will be (A) 0.2 (C) 5 (B) 1 (D) 10 ME – 2011 8. A spherical steel ball of 12 mm diameter at initially 1000 K. It is slowly cooled in surrounding 300 K. the heat transfer coefficient between steel ball and the surrounding is 5 W/ . The thermal conductivity of steel is 20 W/mK. The temperature difference between center and the surface of steel ball is (A) Large because conduction resistance is far higher than the convective resistance. (B) Large because conduction resistance is far less than the convective resistance. (C) Small because conduction resistance is far higher than convective resistance.

Heat Transfer

(D) Small because conduction resistance is far less than convective resistance. 9.

The ratios of the laminar hydrodynamic boundary layer thickness to thermal boundary layer thickness of flows of two fluids P and Q on a flat plate are ½ and 2 respectively. The Reynolds numbers are based on the plate length for both flows is . The Prandtl and Nusselt numbers for P are 1/8 and 35 respectively. The prandtl and nusselt numbers for Q are respectively. (A) 8 and 140 (C) 4 and 70 (B) 8 and 70 (D) 4 and 35

ME – 2013 Common data for questions 10 and 11: Water (specific heat , c kJ/kg K)enters a pipe at a rate of 0.01 kg/s and a temperature of the pipe , of diameter 50 mm and length 3m, is subjected to a wall heat flux q in W/ . 10. If q =2500x, where x is in m and in the direction of flow (x=0 at the inlet ), the bulk mean temperature of the water leaving the pipe in is (A) 42 (C) 74 (B) 62 (D) 104 11.

If q and the convection heat transfer coefficient at the pipe outlet is 1000 W/m2K, the temperature in at the inner surface of the pipe at the outlet is (A) 71 (C) 79 (B) 76 (D) 81

ME – 2014 12. The non dimensional fluid temperature profile near the surface of convectively cooled flat plate is given by –

b

c( )

Where y is measured perpendicular to the plate, L is the plate length and a, b and th

th

th



c are arbitrary constants. d are wall and ambient temperatures, respectively . If the thermal conductivity of the fluids is k and thewall flux is q'', the Nusselt number u

(

(A) a (B) b 13.

14.

)

is equ to (C) 2c (D) (b

Heat Transfer

𝜇=7.25× Ns/ k=0.625W/mK, Pr= 4.85. Using Nu=0.023 e r the convective heat transfer coefficient (in W/ .K) is _______ 15.

c)

For laminar forced convection over a flat plate, if the free stream velocity increases by a factor of 2, the average heat transfer coefficient (A) Remains same (B) Decreases by a factor of √ (C) Rises by a factor of √ (D) Rises by a factor of 4

Consider a two –dimensional laminar flow over a long cylinder as shown in the figure below.

The free stream velocity is and the free stream temperature is lower than the cylinder surface temperature . The local heat transfer coefficient is minimum at point (A) 1 (C) 3 (B) 2 (D) 4

Water flows through a tube of diameter 25 mm at an average velocity of 1.0 m/s. The properties of water are 𝜌=1000 kg/

Answer Keys & Explanation 1.

[Ans. C] Equivalent length of duct, b b ( b) b

3.

[Ans. D] From heat balance equation, h (T T0)=mc where, T0= Ambient temperature

e

μ

4.

[Ans. B] h u

e

2.

⁄

∴ h

We know that Reynolds number

y

|

Where

[Ans. D] Because Re> 2000, the flow is turbulent Nu = 0.023 = 0.023 (4.46 105 )0.8 (0.73)0.33 = 685.6 =

And y Where Tw is surdace temperature and T∞ is free-stream temperature. Nu ( )| y y |y

h

= 1.5

H.T /m length = hA. T = h.p.1. = 25.58 (2 ) 1( 30 20) = 769 W th

th

th



5.

[Ans. D] Heat generated throughout the volume q v q q Heat transfer through PQRS =Heat generated (

[Ans. D] Biot number (

9.

[Ans. A]

)

h (

)

∵

μ

(

)

Nusselt Number =

)

∴ ( r)

⁄

(∵ r

)

(

)

( )

∵ her bou d ry yer thickness at the same location = 1 mm. 7.

)

Prandtl Number = (∵ r

( r)

(

For P:

[Ans. C] ∵ Pr=

∵

)

Since value of biotnumber is very less hence conduction resistance is far less than convective resistance.

q

)

q(

6.

8.

Heat Transfer

∴ ∴ For Q, Prandtl Number ( ) Since value of Reynold Number is same for both the number P and Q hence: (pr) ( u) ( u) ( r) ( ) ∴ ( u)

[Ans. B] /mK

h

y

10. Under steady state conditions Heat transfer by conduction = Heat transfer by convection d ∴ – h dy d (or) – h dy d ( ) (or) – e ) h( dy d (or) – ( e ) h y ut y ( ∴ e ) h or h ut h

[Ans. B] ̇h

∫q

̇h

∫

̇h

dx

)dx

̇h

) ∫ xdx

̇h

x(

( ( )(

̇h

)

[ ]

( (

th

th

th

) )



11.

[Ans. D]

Heat Transfer

13.

[Ans. C] h √v

14.

[Ans. *] Range 4600 to 4625 μ

Convection does not matter as we take outlet section just convection in )= ̇ h ̇h q ( ) q ( ̇ (h h ) ( )( )( ) ( )( ) ( ) q h( where

) w

temperature at the outlet

q h

h

(

[Ans. B] q

d [ ] dy

(

)

d ( dy d [ ] dy

(

)

15.

[Ans. B]

e

i

r bou d ry yer

≤ ep r tor

(

by

)( b )[

(

y

cy

For e ≤ separation effect are negligible and conductors are dominated by friction drag, so the local heat transfer co-efficient is minimum at point 2

)

]

)b

(

q

)

h Hence convective heat transfer coefficient = 4613.67

Where h = heat transfer coefficient at the pipe outlet = 76 + 5 = 81 12.

h

)b

(

)b

q

b

( u u

q ( b

) )

th

th

th



Heat Transfer

Radiation radiating surfaces are diffuse and the medium in the enclosure is nonparticipating. The fraction of the thermal radiation leaving the larger surface and striking itself is

ME – 2005 1. A solid cylinder (surface 2) is located at the centre of a hollow sphere (surface 1). The diameter of the sphere is 1 m, while the cylinder has a diameter and length of 0.5 m each. The radiation configuration factor is (A) 0.375 (C) 0.75 (B) 0.625 (D) 1 2.

Surface-1

The following figure was generated from experimental data relating spectral black body emissive power to wavelength at three temperatures , and ( )

(

μ )

(μ )

The conclusion is that the measurements are (A) correct because the maxima in show the correct trend (B) correct bec use c ’s w is satisfied (C) wrong because the Stefan Boltzmann law is not satisfied (D) wrong because ie ’s disp ce e t law is not satisfied ME – 2008 3. A hollow enclosure is formed between two infinitely long concentric cylinders of radii 1m and 2m, respectively. Radiative heat exchange takes place between the inner surface of the larger cylinder (surface-2) and the outer surface of the smaller cylinder (surface-1). The

Surface-2

(A) 0.25 (B) 0.5

(C) 0.75 (D) 1

ME – 2009 Common Data Questions: 4 & 5 Radiative heat transfer is intended between the inner surfaces of two very large isothermal parallel metal plates. While the upper plate (designated as plate 1) is a black surface and the warmer one being maintained at , the lower plate ( plate 2) is a diffuse and gray surface with an emissivity of 0.7 and is kept at . Assume that the surfaces are sufficiently large to form a twosurface enclosure and steady state conditions to exist. Stefan Boltzmann constant is given as 5.67 . 4. The irradiation (in kW/ ) for the upper plate (plate 1) is (A) 2.5 (C) 17.0 (B) 3.6 (D) 19.5 5.

If plate 1 is also a diffuse and gray surface with an emissivity value of 0.8, the net radiation heat exchange (in kW/ ) between plate 1 and plate 2 is (A) 17.0 (C) 23.0 (B) 19.5 (D) 31.7

th

th

th



ME – 2012 6. For an opaque surface, the absorptivity ( ), transmissivity ( ) and reflectivity ( ) are related by the equation: (A) (C) (B) (D) 7.

Consider two infinitely long thin concentric tubes of circular cross section as shown in figure. If d are the diameters of inner and outer tubes respectively, then the view factor is given by

Heat Transfer

ME – 2014 9. A hemispherical furnace of 1 m radius has the inner surface emissivity, ( ) of its roof maintained at 800 K, while its floor ( ) is kept at 600 K. Stefan Boltzmann constant is ⁄ . The net radiative heat transfer (in kW) from the roof to the floor is _______ 10.

A solid sphere of radius r = 20 mm is placed concentrically inside a hollow sphere of radius r = 30 mm as shown in the figure.

r r

(A) ( )

(C) ( )

The view factor transfer is

for radiation heat

(B)

(D)

( )

( )

( )

( )

ero

( )

ME – 2013 8. Two large diffuse gray parallel plates separated by a small distance , have surface temperature of 400 K and 300 K. If the emissivities of the surfaces are 0.8 and the Stefan – Boltzmann constant is W/ , the net radiation heat exchange rate in kW/ between the two plates is (A) 0.66 (C) 0.99 (B) 0.79 (D) 3.96

th

th

th



Heat Transfer


[Ans. C]

Irradiation for body 1 ( ) et

1 2

= ⁄

=2.5 F21+F22 =1 Since, F22=0, therefore F21=1 Now, A1F12=A2F21 Now, F11+F12 =1 =( 1

2.

3.

5.

[Ans. D] , ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ , Net heat exchange between A and B ( )

) ) (

*(

)+

[Ans. D] ie ’s disp ce e t law is not satisfied, i.e., maxT=C Which tells lower for higher temperature. [Ans. B] F11+F12 =1 F21+F22 =1 F11 =0 due to concave surface F12 =1 Now A1F12 = A2F21

)

(

)

(

)

)

(

6.

[Ans. C]

7.

[Ans. D] f f f ∴ f Also f f And f f ( ) (f )

=

F21 = 0.5 F22=1 F21 =0.5 4.

(

)

= 31.7 kW/

F12=1 F21 =

)

(

F21 = F12 ∴

(

f

[Ans. A]

∴

( f

) (

)

= ⁄

th

th

th



8.

[Ans. A] (

) (

=0.66 9.

Heat Transfer

)

⁄

[Ans. *] Range 24.0 to 25.2 Roof (1) r Floor (2) r , From observation ∴ r r (

(

) r

) (

)

(

)

Watt ( 10.

)

[Ans. B] r r r

r

By summation rule : For current configuration, So, By Reciprocity Theorem : r r r r r (

r

)

th

th

th



Heat Transfer

Heat Exchanger enters the heat exchanger at 102°C, while the cold fluid has an inlet temperature of 15°C. The overall heat transfer coefficient for the heat exchanger is estimated to be ⁄ and the corresponding heat transfer surface area is 5 . Neglect heat transfer between the heat exchanger and the ambient. The heat exchanger is characterized by the following relation: exp( ). The exit temperature (in C) for the cold fluid is (A) 45 (C) 65 (B) 55 (D) 75

ME – 2005 1. Hot oil is cooled from 80 to C in an oil cooler which uses air as the coolant. The air temperature rises from 30 to C. The designer uses a LMTD value of C.The type of heat exchanger is (A) parallel flow (C) counter flow (B) double pipe (D) cross flow ME – 2007 2. In a counter flow heat exchanger, hot fluid e ters t ℃ d co d f uid e ves t ℃ M ss f ow r te of the hot f uid is 1 kg/s and that of the cold fluid is 2 kg/s. Specific heat of the hot fluid is 10kJ/kgK and that of the cold fluid is 5 kJ /kgK. The Log Mean Temperature Difference ( M ) for the he t exch ger i ℃ is (A) 15 (C) 35 (B) 30 (D) 45 ME – 2008 3. The logarithmic mean temperature difference (LMTD) of a counterflow heat exchanger is C. The cold fluid enters at C and the hot fluid enters at Mass flow rate of the cold fluid is twice that of the hot fluid. Specific heat at constant pressure of the hot fluid is twice that of the cold fluid. The exit temperature of the cold fluid (A) is (B) is (C) is (D) cannot be determined ME – 2009 4. In a parallel flow heat exchanger operating under steady state, the heat capacity rates (product of specific heat at constant pressure and mass flow rate) of the hot and cold fluid are equal. The hot fluid, flowing at 1 kg/s with Cp= 4kJ/kgK,

ME – 2011 5. In a condenser of a power plant, the steam condenser at a temperature of C. the cooling water enters at C and leaves at C. the logarithmic mean temperature difference (LMTD) of the condenser is (A) C (C) C (B) C (D) C ME – 2012 6. Water ( g ) at C enters a counterflow heat exchanger with a mass flow rate of g s. Air ( kJ/kg.K) enters at C with a mass flow rate of 2.09 kg/s. If the effectiveness of heat exchanger of the heat exchanger is 0.8, the LMTD (i ) is (A) 40 (C) 10 (B) 20 (D) 5 ME – 2014 7. In a concentric counter flow heat exchanger, water flows through the inner tube at 25°C and leaves at 42°C. The engine oil enters at 100°C and flows in the annular flow passage. The exit temperature of the engine oil is 50°C. th

th

th



Mass flow rate of water and the engine oil are 1.5 kg/s and 1 kg/s, respectively. The specific heat of water and oil are 4178 J/kg.K and 2130 J/kg.K, respectively. The effectiveness of this heat exchanger is ________ 8.

Heat Transfer

(A) a condenser (B) an evaporator (C) a counter flow heat exchanger (D) a parallel flow heat exchanger 9.

A double-pipe counter flow heat exchanger transfers heat between two water stream. Tube side water at 19 liter/s is heated from to . Shell side water at 25 liter/s entering at Assume constant properties of water: density is 1000 kg/ and specific heat is 4186 J/kgK. The LMTD(in ) is _________

In a heat exchanger, it is observed that = , where is the temperature difference between the two single phase f uid stre s t o e e d d is the temperature difference at the other end. This heat exchanger is


[Ans. D]

Which is greater than considered LMTD i.e Hence cross flow is better for given problem. 2.

Parallel Flow

For parallel flow ( ( )

)

(

(

) )

[Ans. B] Heat capacity of hot fluid = 1 10 = 10 kJ/K sec Heat capacity of cold fluid = 2 5 = 10 kJ/K sec Since heat capacity is same, so LMTD is difference of temperature at either end i.e. LMTD =

Which is less than consider LMTD i.e hot cold

3.

[Ans. C]

Counter Flow

For counter Flow ( ( )

) (

(

) )

( )

th

th

th



̇

̇

Heat Transfer

The heat exchanger is characterized by the following relation

c By energy balance c ( )

(

c(

)

)

We know for parallel flow [

=

(

)]

if

Let, (

M

) [

(

)

[

then

]

]

=

LMTD NTU x

(

(x i

) x (x

)( ⁄ s) exp(

⁄ (

)

) )

exp(

x

si ce

)

The maximum possible heat transfer rate is ( ) ) =(4 ⁄ )( Actual rate of heat transfer is

ow pp yi g hospit ru e give

=(0.46)(348)=160kW ( ) ⁄ ( )( 4.

)

[Ans. B] 1: Inlet 2: Outlet h

h

5.

[Ans. B]

c

c

Parallel Flow Heal Exchanger

h g⁄s ⁄ g Overall transfer coefficient U=1kW/m2K Heat transfer surface area A=5m2 Given: kJ/s.K. Neglect heat transfer between the heat exchanger and the ambient.

M

6.

[Ans. C] ( ) ∴

th

(

)

M

th

th


Page 310


Shell side water stream:Volume flow rate of water, ̇ iter s s ss f ow r te of w ter ̇ ̇

Now effectiveness

∴

7.

Heat Transfer

M

te of he t ost by she side w ter stre ) ̇ c( ̇ ̇ ) ̇ c ̇ ( ( ( ) ̇

[Ans. * ]Range 0.65 to 0.67

( c )

)

( c ) Ideal case would be oil reaching 25 outlet ( c) ( ) ( c) ( )

at

8.

[Ans. C] Can happen only in counter flow heat exchanger

9.

[Ans. * ]Range 10.8 to11.2

M

(

(

)

)

x

Tube side water stream:Volume flow rate of water, ̇ iter s s Mass flow rate of water, ̇ ̇ de sity ( ) epecific he t g Rate of heat transferred to tube side water ( ) Stream= ̇ ̇

th

th

th


Page 311



Engineering Materials ME – 2005 1. When the temperature of a solid metal increases. (A) strength of the metal decreases but ductility increases (B) both strength and ductility of the metal decreases (C) both strength and ductility of the metal increases (D) strength of the metal increases but ductility decreases ME – 2006 2. Match the items in columns I and II. Column I Column II (P) Charpy test (1) Fluidity (Q) Knoop test (2) Microhardness (R) Spiral test (3) Formability (S) Cupping test (4) Toughness (5) Permeability (A) P – 4 Q – 5 R – 3 S – 2 (B) P – 3 Q – 5 R – 1 S – 4 (C) P – 2 Q – 4 R – 3 S – 5 (D) P – 4 Q – 2 R – 1 S – 3 3.

4.

The ultimate tensile strength of a material is 400 MPa and the elongation up to maximum load is 35%. If the material obeys power law of hardening, then the true stress-true strain relation (stress in MPa) in the plastic deformation range is : (A) (B) (C) (D)

ME – 2007 5. If a particular Fe-C alloy contains less than 0.83% carbon, it is called (A) high speed steel (B) hypoeutectoid steel (C) hypereutectoid steel (D) cast iron ME – 2009 6. The effective number of lattice points in the unit cell of simple cubic, body centered cubic, and face centered cubic space lattices, respectively, are (A) 1,2,2 (C) 2,3,4 (B) 1,2,4 (D) 2,4,4 ME – 2010 7. The material property which depends only on the basic crystal structure is (A) fatigue strength (B) work hardening (C) fracture strength (D) elastic constant ME – 2011 8. The operation in which oil is permeated into the pores of a powder metallurgy product is known as (A) mixing (B) sintering (C) impregnation (D) infiltration 9.

The crystal structure of austenite is (A) body centered cubic (B) face centered cubic (C) hexagonal closed packed (D) body centered tetragonal

The main purpose of spheroidising treatment is to improve (A) hardenability of low carbon steels (B) machinability of low carbon steels (C) hardenability of high carbon steels (D) machinability of high carbon steels

th

th

th



Point on the graph P Q

ME – 2012 10. During normalizing process of steel, the specimen is heated (A) between the upper and lower critical temperature and cooled in still air (B) Above the upper critical temperature and cooled in furnace (C) Above the upper critical temperature and cooled in still air (D) Between the upper and lower critical temperature and cooled in furnace ME – 2014 11. The process of reheating the martensitic steel to reduce its brittleness without any significant loss in its hardness is (A) normalizing (C) quenching (B) annealing (D) tempering 12.

Description of the Point

1. Upper Yield Point 2. Ultimate Tensile Strength R 3. Proportionality Limit S 4. Elastic Limit T 5. Lower Yield Point U 6. Failure (A) P-1, Q-2, R-3, S-4, T-5, U-6 (B) P-3, Q-1, R-4, S-2, T-6, U-5 (C) P-3, Q-4, R-1, S-5, T-2, U-6 (D) P-4, Q-1, R-5, S-2, T-3, U-6 14.

The relationship between true strain and engineering strain in a uniaxial tension test is given as

(D) The stress-strain curve for mild steel is shown in the figure given below. Choose the correct option referring to both figure and table.

Match the heat treatment processes (Group A) and their associated effects on properties (Group B) of medium carbon steel Group Group B P: Tempering I: Strengthening and grain refinement Q: Quenching II: Inducing toughness R: Annealing III: Hardening S: Normalizing IV: Softening (A) P-III, Q-IV, R-II, S-I (B) P-II, Q-III, R-IV, S-I (C) P-III, Q-II, R-IV, S-I (D) P-II, Q-III, R-I, S-IV

tr ss

13.


tr

th

th

th





[Ans. A]

2.

[Ans. D]

3.

[Ans. B] But at UTS n= Hence n= = 0.3 Also = 400(1+0.35) = 540 MPa Now, 540 = K K = 774.97 Where,

4.

[Ans. C] During normalizing process of steel. The specimen in heated upto above the upper critical temperature and is then cooled is still air.

11.

[Ans. A]

12.

[Ans. C]

5.

[Ans. B]

6.

[Ans. B] For a simple cubic unit cell 8 corners have 8 atoms and each atom at one corner contributes to 8 unit cells. u

( ) ( [(

13.

[Ans. C] Point on the Graph P Q R S T

r

For body centred cubic, effective number = 8 corners + 1 centre

]

∫

tru str ss tru str

[Ans. D] Spheroidising improves machinability of high carbon steels.

tv

10.

U 14.

For FCC, effective number = 8 corner + 6 faces

) )

]

Description of the Point Proportionality Limit Elastic Limit Upper Yield Point Lower Yield Point Ultimate Tensile Strength Failure

[Ans. B] Tempering Toughness Quenching Hardening Annealing stress reliving thus softening Normalizing strengthening and grain refinement

Hence answer is (1, 2, 4) 7.

[Ans. C]

8.

[Ans. C]

9.

[Ans. B] Austenite has FCC Crystal structure.

th

th

th




Casting ME – 2005 1. A mould has a downsprue whose length is 20 cm and the cross sectional area at the base of the downsprue is 1 c . The downsprue feeds a horizontal runner leading into the mould cavity of volume 1000 c . The time required to fill the mould cavity will be (A) 4.05s (C) 6.05s (B) 5.05s (D) 7.25s 2.

Match the items of List I (Equipment) with the items of List II (Process) and select the correct answer using the given codes. List I (Equipment) List II (Process) P – Hot Chamber 1 – Cleaning Machine Q – Muller 2 – Core making R – Dielectric 3 – Die casting Baker S – Sand Blaster 4 – Annealing 5 – Sand mixing (A) P – 2 Q – 1 R – 4 S – 5 (B) P – 4 Q – 2 R – 3 S – 5 (C) P – 4 Q – 5 R – 1 S – 2 (D) P – 3 Q – 5 R – 2 S – 1

ME – 2006 3. In a sand casting operation, the total liquid head is maintained constant such that it is equal to the mould height. The time taken to fill the mould with a top gate is t . If the same mould is filed with a bottom gate, then the time taken is t . Ignore the time required to fill the runner and frictional effects. Assume atmospheric pressure at the top molten metal surfaces. The relation between t and t is: (A) t = √ t (C) t = √ (B) t = 2 t (D) t = √ t

4.

An expandable pattern is used in (A) slush casting (B) squeeze casting (C) centrifugal casting (D) investment casting

ME – 2007 5. A 200 mm long down sprue has an area of cross – section of 650 where the pouring basin meets the down sprue (i.e. at the beginning of the down sprue). A constant head of molten metal is maintained by the pouring basin. The molten metal flow rate is 6.5 × /s. Considering the end of down sprue to be open to atmosphere and an acceleration due to gravity of mm/s , the area of the down sprue in at its end (avoiding aspiration effect) should be Pouring Basin

650 200 mm Area of down sprue at its end

(A) 650.0 (B) 350.0 6.

(C) 290.7 (D) 190.0

Vo u o u o s ‘l’ and volume of sph r o r us ‘r’ r qu oth th cube and the sphere are solid and of same material. They are being cast. The ratio of the solidification time of the cube to the same of the sphere is

th

(A) ( ) ( )

(C) ( ) ( )

(B) ( ) ( )

(D) ( ) ( )

th

th



7.

Which of the following engineering materials is the most suitable material for hot chamber die casting? (A) low carbon steel (B) titanium (C) copper (D) tin

ME – 2008 8. While cooling, a cubical casting of side 40 mm undergoes 3%, 4% and 5% volume shrinkage during the liquid state, phase transition and solid state, respectively. The volume of metal compensated from the riser is (A) 2% (C) 8% (B) 7% (D) 9% ME – 2009 9. Match the items in Column I and Column II. Column I Column II P. Metallic Chills 1. Support for the core Q. Metallic 2. Reservoir of the Chaplets molten metal R. Riser 3. Control cooling of critical sections S. Exothermic 4. Progressive Padding solidification (A) (B) (C) (D) 10.

Two streams of liquid metal, which are not hot enough to fuse properly result into a casting defect known as (A) cold shut (C) sand wash (B) swell (D) scab


ME – 2010 11. In a gating system, the ratio 1: 2: 4 represents (A) sprue base area : runner area : ingate area (B) pouring basin area : ingate area : runner area (C) sprue basin area : ingate area : casting area (D) runner area : ingate area : casting area ME – 2011 12. A cubic casting of 50 mm side undergoes volumetric solidification shrinkage and volumetric solid contraction of 4% and 6% respectively. No riser is used. Assume uniform cooling in all directions. The side of the cube after solidification and contraction is (A) 48.32 mm (C) 49.94 mm (B) 49.90 mm (D) 49.96 mm 13.

Green sand mould indicates that (A) polymeric mould has been cured (B) mould has been totally dried (C) mould is green in colour (D) mould contains moisture

ME – 2013 14. A cube shaped casting solidifies in 5 min. The solidification time in min for a cube of the same material, which is 8 times heavier than the original casting , will be (A) 10 (C) 24 (B) 20 (D) 40 ME – 2014 15. An aluminum alloy (density 2600 kg/ ) casting is to be produced. A cylindrical hole of 100 mm diameter and 100 mm length is made in the casting using sand core (density 1600 kg/ ). The net buoyancy force (in Newton) acting on the core is _______

th

th

th



16.

Match the casting defects (Group A) with the probable causes (Group B): Group A Group B P: Hot tears 1: Improper fusion of two streams of liquid metal Q: Shrinkage 2: Low permeability of the sand mold R: Blow 3: Volumetric holes contraction both in liquid and solid stage S: Cold Shut 4: Differential cooling rate (A) P-1, Q-3, R-2, S-4 (B) P-4, Q-3, R-2, S-1 (C) P-3, Q-4, R-2, S-1 (D) P-1, Q-2, R-4, S-3

17.

The hot tearing in a metal casting is due to (A) high fluidity (B) high melt temperature (C) wide range of solidification temperature (D) low coefficient of thermal expansion

18.

A cylinder blind riser with diameter d and height h, is placed on the top of the mold cavity of a closed type sand mold as shown in the figure. If the riser is of constant volume, then the rate of solidification in the user is the least when the ratio h:d is pru

A cylindrical riser of 6 cm diameter and 6 cm height has to be designed for a sand casting would for producing a steel rectangular plate casting of 7 cm × 10 cm × 2 cm dimensions having the total solidification time of 1.36 minute. The total solidification time (in minute) of the riser is _______

s

s r o

(A) 1:2 (B) 2:1

19.


h v ty

(C) 1:4 (D) 4:1

th

th

th





[Ans. B]

5.

[Ans. C]

vat base=√ h=√ =1.9797 m.sec =197.98 cm/sec Area at base =1 cm2, and volume to be filled = 1000 cm3 Time required = 2.

3.

V h

= 5.05sec

200mm

[Ans. D] P-3, Q- 5, R -2, S-1

V

g= Here, Flow rate = 6.5 × V

[Ans. B] Time taken to fill the mould with top gate t …… √

Where r o ou H = Height of mould r o t h t h ht Given that, h so q o √h t …………… √ Time taken to fill the mould with bottom gate t t

√

(√h

√h

V

h

At down, V √ = 2236.06 mm/s V

6.

[Ans. D] V V Solidification time = ( ) ,

)

qu r

√h … … … …

[Ans. D] Expandable pattern is used in investment casting.

√

h = 50 mm

√ From equation (ii) & (iii) t t t t 4.

s

s

r

(

r to

)

(

(

)

(

)

r ) ( )

7.

[Ans. D]

8.

[Ans. B] The riser can compensate for volume shrinkange only in the liquid stage and transition stage and not in the solid state. Hence volume of metal that needs to be compensated from the riser = 3 + 4 = 7%

9.

[Ans. D]

th

th

th



10.

[Ans. A]

16.

11.

[Ans. A]

12.

[Ans. A] Volume of cube after solidification and contraction V

[Ans. D] Green sand mould indicates that mould is not baked or dried. i.e., it contains moisture

14.

[Ans. B] v t ( )

V

[Ans. C] Hot tearing : During solidification of molten metal in mold cavity due to nonuniform working different contraction stresses can be developed inside the casting and due to insufficient of collapsibility of mold sand some contraction stress can be developed in casting these stresses can be relieved by forming hot earns and cracks. So from above the most suitable option is C

18.

[Ans. A]

( )

v V

Group B Differential cooling rate Volumetric contraction both in liquid and solid stage Low permeability of the sand mold Improper fusion of two streams of liquid metal

17.

Weight is 8 times heavier

V

[Ans. B] Group A P: Hot tears Q: Shrinkage

R: Blow holes S: Cold Shut

side of cube after contraction

13.


pru

s

t t t

s r o

ur u

15.

r

o

s r

Volume (V) of Riser =

o s

V

h

[Ans. *]Range 7 to 8 Buoyancy force v

h v ty

h h

V

for least solidification time, surface area should b u or u

th

th

th




V V V

h

h h

19.

[Ans. *]Range 2.5 to 4.5 s r vo V

st st

h

vo V sur [

s r sur

r ] r h

ow t

t t (

[(

V ) ( V

)]

)

th

th

th




Forming Process interface is 0.1. The minimum possible thickness of the sheet that can be produced in a single pass is: (A) 1.0 mm (C) 2.5 mm (B) 1.5 mm (D) 3.7 mm

ME – 2005 1. A 2 mm thick metal sheet is to be bent at an angle of one radian with a bend radius of 100 mm If the stretch factor is 0.5, the bend allowance is 2mm

1 radian

(A) 99mm (B) 100mm

(C) 101mm (D) 102mm

ME – 2006 2. Match the items in columns I and II. Column I Column II (P) Wrinkling (1) Yield point elongation (Q) Orange peel (2) Anisotropy (R) Stretcher strains (3) Large grain size (S) Earing (4) Insufficient blank holding force (5) Fine grain size (6) Excessive blank holding force (A) P – 6 Q – 3 R – 1 S – 2 (B) P – 4 Q – 5 R – 6 S – 1 (C) P – 2 Q – 5 R – 3 S – 4 (D) P – 4 Q – 3 R – 1 S – 2 3.

4.

In a wire drawing operation, diameter of a steel wire is reduced from 10 mm to 8 mm. The mean flow stress of the material is 400 MPa. The ideal force required for drawing (ignoring friction and redundant work) is: (A) 4.48 kN (C) 20.11 kN (B) 8.97 kN (D) 31.41 kN

ME – 2007 5. The force requirement in a blanking operation of low carbon steel sheet is 5.0 kN. The thickness of the sheet is‘t’ and diameter of the blanked part is ‘ ’. For the same work material, if the diameter of the blanked part is increased to 1.5 d and thickness is reduced to 0.4 t, the new blanking force in kN is (A) 3.0 (C) 5.0 (B) 4.5 (D) 8.0 6.

Match the correct combination for following metal working processes. Processes Associated state of stress P : Blanking 1. Tension Q : Stretch 2. Compression Forming R : Coining 3. Shear S : Deep 4. Tension and Drawing Compression 5. Tension and Shear (A) (B) (C) (D)

7.

The thickness of a metallic sheet is reduced from an initial value of 16 mm to a final value of 10 mm in one single pass rolling with a pair of cylindrical rollers each of diameter of 400 mm. The bite angle in degree will be (A) 5.936 (C) 8.936 (B) 7.936 (D) 9.936

A 4 mm thick sheet is rolled with 300 mm diameter rolls to reduce thickness without any change in its width. The friction coefficient at the work-roll th

th

th



8.

In open – die forging, a disc of diameter 200 mm and height 60 mm is compressed without any barreling effect. The final diameter of the disc is 400 mm. The true strain is (A) 1.986 (C) 1.386 (B) 1.686 (D) 0.602

ME – 2008 9. In the deep drawing of cups, blanks show a tendency to wrinkle up around the periphery (flange). The most likely cause and remedy of the phenomenon are, respectively, (A) Buckling due to circumferential compression; Increase blank holder pressure (B) High blank holder pressure and high friction; Reduce blank holder pressure and apply lubricant (C) High temperature causing increase in circumferential length; Apply coolant to blank (D) Buckling due to circumferential compression; decrease blank holder pressure 10.

In a single pass rolling operation, a 20 mm thick plate with plate width of 100 mm, is reduced to 18 mm. The roller radius is 250 mm and rotational speed is 10 rpm. The average flow stress for the plate material is 300 MPa. The power required for the rolling operation in kW is closest to (A) 15.2 (C) 30.4 (B) 18.2 (D) 45.6

ME – 2010 Statement for Linked Answer Questions 11&12: In shear cutting operation, a sheet of 5 mm thickness is cut along a length of 200 mm. The cutting blade is 400 mm long (see figure) and zero – shear (S = 0) is provided on the edge. The ultimate


shear strength of the sheet is 100 MPa and penetration to thickness ratio is 0.2. Neglect friction. 400

S

11.

Assume force vs displacement curve to be rectangular, the work done (in J) is (A) 100 (C) 250 (B) 200 (D) 300

12.

A shear of 20 mm (S = 20 mm) is now provided on the blade. Assuming force vs displacement curve to be trapezoidal, the maximum force (in kN) exerted is (A) 5 (C) 20 (B) 10 (D) 40

ME – 2011 13. The shear strength of a sheet metal is 300MPa. The blanking force required to produce a blank of 100 mm diameter from a 1.5 mm thick sheet is close to (A) 45 kN (C) 141 kN (B) 70 kN (D) 3500 kN 14.

The maximum possible draft in cold rolling of sheet increases with the (A) increase in coefficient of friction (B) decrease in coefficient of friction (C) decrease in roll radius (D) increase in roll velocity

ME – 2012 15. Match the following metal forming processes with their associated stresses in the workpiece. Metal forming process A. Coining B. Wire Drawing C. Blanking D. Deep Drawing th

th

th



Type of Stress 1. Tensile 2. Shear 3. Tensile and compressive 4. Compressive Codes: A B C D (A) 4 1 2 3 (B) 4 1 3 2 (C) 1 2 4 3 (D) 1 3 2 4 16.

A solid cylinder of diameter 100 mm and height 50 mm is forged between two frictionless flat dies to a height of 25 mm. The percentage change in diameter is (A) 0 (C) 20.7 (B) 2.07 (D) 41.4

17.

In a single pass rolling process using 410mm diameter steel rollers, a strip of width 140mm and thickness 8mm undergoes 10% reduction of thickness. The angle of bite in radians is (A) 0.006 (C) 0.062 (B) 0.031 (D) 0.600

18.

Calculate the punch size in mm, for a circular blanking operation for which details are given below Size of blank 25mm Thickness of the sheet 2mm Radial clearance between 0.06mm punch and die Dia allowance 0.05mm (A) 24.83 (C) 25.01 (B) 24.89 (D) 25.17


ME – 2014 20. With respect to metal working, match Group A with Group B: Group A Group B P: Defect in I: alligatoring extrusion Q: Defect in rolling II: scab R: Product of skew III: fish tail rolling S: Product of rolling IV: seamless tube through cluster mill V: thin sheet with tight tolerance VI: semi-finished balls of ball bearing (A) P-II, Q-III, R-VI, S-V (B) P-III, Q-I, R-VI, S-V (C) P-III, Q-I, R-IV, S-VI (D) P-I, Q-II, R-V, S-VI

ME – 2013 19. In a rolling process, the state of stress of the material undergoing deformation is (A) pure compression (B) pure shear (C) compression and shear (D) tension and shear

th

th

th





[Ans. C] We know that Bend allowance = t Where R = inside radius of the bend =100mm bend angle = 1 radian t = thickness of metal sheet k = Location of neutral axis from the bottom = 0.33 where R < 2t = 0.50 where R > 2t Bend allowance

6.

[Ans. D] Blanking – Shear Stretch Forming – Tension Coining – Compression Deep drawing – tension and compression

7.

[Ans. D] Here, h = initial thickness = 16 mm h = final thickness = 10 mm Roll diameter = D = 400 mm t

2.

3.

,

Where angle of bite 9.826

[Ans. D] Wrinkling – Insufficient blank holding force Orange peel – large grain Stretcher strains – yield point elongation Earing – Anisotropy

8.

[Ans. C] Since volume of disc remains constant during the process, Hence

[Ans. B] Given h

Ignoring friction and redundant work means Hence

th

( )

=2 ( ) = 178.51 MPa ⁄ Ideal force = 178.51

4.

√

ru str r

9.

[Ans. A]

10.

[Ans. A]

[Ans. C] µ= √

√

t =2.5mm 5.

[Ans. A] t τ th sh r str ss F=τ t 5.0 kN w or τ τ t = 3 kN

rp √ t

th

th

th



[

√

]

⁄ro

r

ot pow r 11.


15.

[Ans. A] 1. Coining Compressive 2. Wire drawing Tensile 3. Blanking Shear 4. Deep drawing Tensile and compressive

16.

[Ans. D] Since volume remains constant V V

[Ans. A]

h

τtx

Work done =

h

h

t J 17.

12.

[Ans. C]

[Ans. B] 400

5

5

20

5

os

So 10mm punch should travel to cut the sheet of 5 mm thickness for 400 mm, shear is 20mm. so only 200mm is the effective length that a punch should travel to cut 5 mm thick sheet τ t s t p

13.

[Ans. C] Blanking Force,

τ

t

os

5

os

) ⁄

18.

[Ans. A] Diameter of punch = Diameter of blank

radian clearance die allowance

19.

[Ans. C]

20.

[Ans. B] Fish tail is an extrusion defect. (P – III) Product of skew rolling is semi – finished balls of ball bearing. (R – VI) So, P – III, R – VI

3 kN 14.

(

[Ans. A] Maximum possible draft h h

th

th

th




Joining Process each sheet is formed. The properties of the metallic sheets are given as: ambient temperature = 293 K melting temperature = 1793 K density = 7000 kg/ latent heat of fusion = 300 kJ/kg specific heat = 800 J/kg K Assume: 1) Contact resistance along sheet – sheet interface is 500 micro-ohm and along electrode – sheet interface is zero; 2) No conductive heat loss through the bulk sheet materials; and 3) The complete weld fusion zone is at the melting temperature. The melting efficiency (in %) of the process is (A) 50.37 (C) 70.37 (B) 60.37 (D) 80.37

ME – 2005 1. Spot welding of two 1 mm thick sheets of steel (density=8000kg/ ) is carried out successfully by passing a certain amount of current for 0.1 second through the electrodes. The resultant weld nugget formed is 5 mm in diameter and 1.5 mm thick. If the latent heat of fusion of steel is 1400 kJ/kg and the effective resistance in the welding operation in 200 Ω,the current passing through the electrodes is approximately (A) 1480 A (C) 4060 A (B) 3300A (D) 9400A 2.

The strength of a brazed joint (A) decreases with increase in gap between the two joining surfaces (B) increases with increase in gap between the two joining surfaces (C) decreases up to certain gap between the two joining surfaces beyond which it increases (D) increases up to certain gap between the two joining surfaces beyond which it decreases

5.

A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm. The linear voltage (E) – arc length (L) characteristic of the welding arc can be given as (where E is in Volt and L is in mm) (A) E = 20 + 2L (C) E = 80 + 2L (B) E = 20 + 8L (D) E = 80 + 8L

6.

Which one of the following is a solid state joining process? (A) gas tungsten arc welding (B) resistance spot welding (C) friction welding (D) submerged arc welding

ME – 2006 3. In an arc welding process, the voltage and current are 25 V and 300 A respectively. The arc heat transfer efficiency is 0.85 and welding speed is 8 mm/sec. The net heat input (in J/mm) is: (A) 64 (C) 1103 (B) 797 (D) 79700 ME – 2007 4. Two metallic sheets, each of 2.0 mm thickness, are welded in a lap joint configuration by resistance spot welding at a welding current of 10 kA and welding time of 10 milli second. A spherical fusion zone extending up to the full thickness of

th

th

th



ME – 2008 7. In arc welding of a butt joint, the welding speed is to be selected such that highest cooling rate is achieved. Melting efficiency and heat transfer efficiency are 0.5 and 0.7, respectively. The area of the weld cross section is 5 and the unit energy required to melt the metal is 10 J/ . If the welding power is 2 kW, the welding speed in mm/s is closest to (A) 4 (B) 14 (C) 24 (D) 34 ME – 2010 8. Two pipes of inner diameter 100 mm and outer diameter 110 mm each are joined by flash – butt welding using 30 V power supply. At the interface, 1 mm of material melts from each pipe which has a r s st o Ω th u t t energy is 64.4MJ , then time required for welding (in s) is (A) 1 (B) 5 (C) 10 (D) 20 ME – 2011 9. Which one among thefollowing welding processes used non – consumable electrode? (A) Gas metal arc welding (B) Submerged arc welding (C) Gas tungsten arc welding (D) Flux coated arc welding ME – 2012 10. In a DC arc welding operation, the voltage-arc length characteristic was obtained as V where the arc length was varied between 5 mm and 7 mm. Here V denotes the arc voltage in Volts. The arc current was varied from 400 A to 500 A. Assuming linear power source characteristic, the open circuit voltage and the short circuit current for the welding operation are (A) 45 V, 450 A (B) 75 V, 750 A (C) 95 V, 950 A (D) 150 V, 1500 A


ME – 2013 11. Match the correct pairs. Processes Characteristics / Applications P. friction 1. non – consumable welding electrode Q. gas metal arc 2. joining of thick welding plates R. Tungsten inert 3. consumable gas welding electrode wire S. electroslag 4. joining of welding cylindrical dissimilar materials (A) P – 4, Q – 3, R – 1, S – 2 (B) P – 4, Q – 2, R – 3, S – 1 (C) P – 2, Q – 3, R – 4, S – 1 (D) P – 2, Q – 4, R – 1, S – 3 ME – 2014 12. In solid-state welding, the contamination layers between the surfaces to be welded are removed by (A) alcohol (B) plastic deformation (C) water jet (D) sand blasting 13.

The major difficulty during welding of aluminum is due to its (A) high tendency of oxidation (B) high thermal conductivity (C) low melting point (D) low density

14.

For spot welding of two steel sheets (base metal) each of 3 mm thickness, welding current of 10000 A is applied for 0.2 s. The heat dissipated to the base metal is 1000 J. Assuming that the heat required for melting 1 m volume of steel is 20 J and interfacial contact resistance between sh ts s Ω th vo u ) of weld nugget is _______

th

th

th



15.

Within the Heat Affected Zone (HAZ) in a fusion welding process, the work material undergoes (A) Microstructural changes but does not melt (B) Neither melting nor microstructural changes (C) Both melting and microstructural changes after solidification (D) Melting and retains the original microstructure after solidification

16.


A butt weld joint is developed on steel plates having yield and ultimate tensile strength of 500 MPa and 700 MPa, respectively. The thickness of the plates is 8 mm and width is 20 mm. Improper selection of welding parameters caused an undercut of 3 mm depth along the weld. The maximum transverse tensile load (in kN) carrying capacity of the developed weld joint is _______


[Ans. C] H.G = H.U t Vo o

u

tot h t or = mL + m

t

3.

Efficiency

[Ans. D] Strength increases upto some certain gap and beyond this strength decreases.

= 70.38% 5.

[Ans. A] From power source characteristics,

[Ans. B] Power generated = VI t

V = 80

put

[Ans. C] Given, Current, Time, t s Resistance, R=500 × Ambient temperature,

a + bL Here, L = 5, I = 500 … But at L = 7, I = 460 … v From (iii) and (iv), we get b = 2, a = 20 Voltage charge = 20 + 2L

Ω

Melting temperature, Total heat supplied. t

ss

r

…

Arc voltage Characteristics V = a + bL … Equating (i) and (ii), we get

sp = 797 J/mm

4.

) [ ]

I = 4060 A 2.

t (

6.

[Ans. C] Friction welding is clearly solid state welding.

7.

[Ans. B] Welding power applied = Heating power needed th

th

th



14.

2 f = 14 mm/sec 8.

[Ans. C] V V Heat energy required for welding

V ( )

[Ans. *]Range 140 to 160 t t Heat dissipated to base metal= 1000J Heat requiring for melting Ω ; Volume of weld nugget =? ot h t supp t

t put to th h t put u Vo u

9.

[Ans. C] Gas tungsten arc welding uses a non – consumable electrode made of tungsten.

10.

[Ans. C] V

V

V

V

V V V

s t

t

t

t

V V

o

u

t

15.

[Ans. A] Heat affected zone (HAZ) is the area of base material, either a metal or a thermoplastic, which is not melted and has had its microstructure and properties altered by welding or heat intensive cutting operations

16.


p r u t vo t hort r u t urr t V V V V

V Vo ts p

11.

[Ans. A]

12.

[Ans. B] Contamination layer between the surface is removed by friction between the solids and due to which material will get localized plastic deformation. So best option is B.

13.


There is an undercut of 3mm r Maximum transverse load activity on this area = 700 MPa x u tr sv rs t s o

[Ans. A] The major difficulty during welding of aluminum is due to higher tendency of oxidation of aluminum. The feature of Al th ox t o st t s th t’s why has higher tendency of oxidation. th

th

th




Machining and Machine Tool Operations ME – 2005 1. A zigzag cavity in a block of high strength alloy is to be finish machined. This can be carried out by using

(A) (B) (C) (D)

(D) tool changing cost, non-productive cost, machining cost 4.

A 600mm 30mm flat surface of a plate is to be finish machined on a shaper. The plate has been fixed with the 600 mm side along the tool travel direction. If the tool over-travel at each end of the plate is 20 mm, average cutting speed is 8 m/min, feed rate is 0.3mm/stroke and the ratio of return time to cutting time of the tool is 1 : 2, the time required for machining will be (A) 8 minutes (C) 16 minutes (B) 12 minutes (D) 20 minutes

5.

Two tools P and Q have signatures 5°- 5°- 6°-6°- 8°- 30°- 0 and 5°- 5°- 7°- 7°- 8°- 15°- 0 (both ASA) respectively. They are used to turn components under the same machining conditions. If h and h denote the peak-to-valley heights of surfaces

electric discharge machining electro-chemical machining laser beam machining abrasive flow machining

2.

When 3-2-1 principle is used to support and locate a three dimensional work-piece during machining, the number of degrees of freedom that are restricted is (A) 7 (C) 9 (B) 8 (D) 10

3.

The figure below shows a graph, which qualitatively relates cutting speed and cost per piece produced.

produced by the tools P and Q, the ratio will be

Cost per piece

Total Cost

3

2 1

Cutting Speed

The three curves, 1, 2 and 3 respectively represent (A) machining cost, non-productive cost, tool changing cost (B) non-productive cost, machining cost, tool changing cost (C) tool changing cost, machining cost, non-productive cost

(A)

(C)

(B)

(D)

ME – 2006 6. If each abrasive grain is viewed as a cutting tool, then which of the following represents the cutting parameters in common grinding operations? (A) Large negative rake angle, low shear angle and high cutting speed (B) Large positive rake angle, low shear angle and high cutting speed (C) Large negative rake angle, high shear angle and high cutting speed (D) Zero rake angle, high shear angle and high cutting speed

th

th

th



7.

Arrange the processes in the increasing order of their maximum material removal rate. Electrochemical Machining (ECM) Ultrasonic Machining (USM) Electron Beam Machining (EBM) Laser Beam Machining (LBM) and Electric Discharge Machining (EDM) (A) USM, LBM, EBM, EDM, ECM (B) EBM, LBM, USM, ECM, EDM (C) LBM, EBM, USM, ECM, EDM (D) LBM, EBM, USM, EDM, ECM Common Data for Questions 8, 9, 10: In an orthogonal machining operation: Uncut thickness = 0.5mm Cutting speed = 20m/min Width of cut = 5 mm Chip thickness = 0.7 mm Thrust force = 200N Cutting force = 1200 N Rake angle = 15 ssu r h t’s th ory

8.

The values of shear angle and shear strain, respectively, are (A) 30.3 and 1.98 (B) 30.3 and 4.23 (C) 40.2 and 2.97 (D) 40.2 and 1.65

9.

The coefficient of friction at the tool-chip interface is (A) 0.23 (C) 0.85 (B) 0.46 (D) 0.95

10.

The percentage of total energy dissipated due to friction at the tool-chip interface is: (A) 30 % (C) 58 % (B) 42 % (D) 70 %


depth of cut is 2 mm. The chip thickness obtained is 0.48mm. If the orthogonal rake angle is zero and the principle cutting edge angle is 90  , the shear angle in degree is (A) 20.56 (C) 30.56 (B) 26.56 (D) 36.56 12.

In electro discharge machining (EDM), if the thermal conductivity of tool is high and the specific heat of work piece is low, then the tool wear rate and material removal rate are expected to be respectively (A) high and high (C) high and low (B) low and low (D) low and high

13.

In orthogonal turning of medium carbon steel, the specific machining energy is 2.0 J/ . The cutting velocity, feed and depth of cut are 120 m/min, 0.2 mm/rev and 2 mm respectively. The main cutting force in N is (A) 40 (C) 400 (B) 80 (D) 800

14.

In orthogonal turning of low carbon steel pipe with principle cutting edge angle of 90°, the main cutting force is 1000 N and the feed force is 800 N. The shear angle is 25° and orthogonal rake angle is zero. p oy r h t’s th ory th r t o o friction force to normal force acting on the cutting tool is (A) 1.56 (C) 0.80 (B) 1.25 (D) 0.64

15.

Match the most suitable manufacturing processes for the following parts. Parts Manufacturing Processes P : Computer chip 1. Electrochemical Machining Q : Metal forming 2. Ultrasonic Machining dies and molds R : Turbine blade 3. Electrodischarge Machining S : Glass 4. Photochemical Machining

ME – 2007 11. In orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide tool, the cutting velocity is 90 m/min. The feed is 0.24 mm/rev and the th

th

th



(A) (B) (C) (D)

16.

17.

Statement for Linked Answer Questions 16 & 17: A low carbon steel bar of 147 mm diameter with a length of 630 mm is being turned with uncoated carbide insert. The observed tool lives are 24 min and 12 min for cutting velocities of 90 m/min and 120 m/min respectively. The feed and depth of cut are 0.2 mm/rev and 2 mm respectively. Use the unmachined diameter to calculate the cutting velocity. When tool life is 20 min, the cutting velocity in m/min is (A) 87 (C) 107 (B) 97 (D) 114

20.

A researcher conducts electrochemical machining (ECM) on a binary alloy (density 6000 kg/ ) of iron (atomic weight 56, valency 2) and metal P (atomic w ht v y r y’s o st t = 96500 coulomb/mole. Volumetric material removal rate of the alloy is 50 /sec at a current of 2000 A. The percentage of the metal P in the alloy is closest to (A) 40 (C) 15 (B) 25 (D) 79

21.

The figure shows an incomplete schematic of a conventional lathe to be used for cutting threads with, different pitches. The speed gear box is shown and the feed gear box is to be placed. P, Q, R and S denote locations and have no other significance. Changes in should NOT affect the pitch of the thread being cut and changes in should NOT affect the cutting speed.

Neglect over – travel or approach of the tool. When tool life is 20 min, the machining time in min for a single pass is (A) 5 (C) 15 (B) 10 (D) 20

Spindle Work piece

Motor

P

Q

Cutting tool

R S Half Nut

E

ME – 2008 18. Internal gear cutting operation can be performed by (A) milling (B) shaping with rack cutter (C) shaping with pinion cutter (D) hobbing 19.


Lead screw

The correct connections and the correct placement of are given by (A) Q and E are connected. is placed between P and Q. (B) S and E are connected. is placed between R and S. (C) Q and E are connected. is placed between Q and E. (D) S and E are connected. is placed between S and E.

In a single point turning tool, the side rake angle and orthogonal rake angle are equal. is the principle cutting edge angle and its r s ≤ ≤ h h p ows the orthogonal plane. The value of is closest to (A) 0° (C) 60° (B) 45° (D) 90°

Statement for Linked Answer Questions 22 & 23: Orthogonal turning is performed on a cylindrical workpiece with shear strength of 250 MPa. The following conditions are used: cutting velocity is 180 m/min, feed is 0.20 mm/rev, depth of cut is 3 mm, chip th

th

th



22.

23.

thickness ratio = 0.5. The orthogonal rake s ˚ pp y r h t’s th ory or analysis. The shear plane angle (in degrees) and the shear force respectively are (A) 52; 320N (C) 28 ; 400 N (B) 52; 400N (D) 28 ; 320 N The cutting and frictional forces, respectively, are (A) 568 N ; 387 N (C) 440 N ; 342 N (B) 565 N ; 381 N (D) 480 N ; 356 N

ME – 2009 24. Friction at the tool-chip interface can be reduced by (A) decreasing the rake angle (B) increasing the depth of cut (C) decreasing the cutting speed. (D) increasing the cutting speed. 25.

26.

Minimum shear strain in orthogonal turning with a cutting tool of zero rake angle is (A) 0.0 (B) 0.5 (C) 1.0 (D) 2.0 Electrochemical machining is performed to remove material from an iron surface of 20mm × 20mm under the following conditions: Inter electrode gap= 0.2mm Supply voltage (DC)= 12V Specific resistance of electrolyte Ω Atomic weight of Iron= 55.85 Valency of Iron = 2 r y’s o st t = 96540 Coulombs The material removal rate (in g/s) is (A) 0.3471 (C) 34.71 (B) 3.471 (D) 347.1 Statement for Linked Answer Questions: 27 & 28 In a machining experiment, tool life was found to vary with the cutting speed in the following manner:


Cutting speed (m/ min)

Tool life (minutes)

60 90

81 36

27.

The exponent (n) and constant (k) of the y or’s too qu t o r (A) n = 0.5 and k = 540 (B) n = 1 and k = 4860 (C) n = 1 and k = 0.74 (D) n = 0.5 and k = 1.155

28.

What is the percentage increase in tool life when the cutting speed is halved? (A) 50% (C) 300% (B) 200% (D) 400%

ME – 2010 29. or too y or’s too xpo t s 0.45 and constant (K) is 90. Similarly for tool B, n = 0.3 and K = 60. The cutting speed (in m/min) above which tool A will have a higher tool life than tool B is (A) 26.7 (C) 80.7 (B) 42.5 (D) 142.9 ME – 2011 30. Match the following non – traditional machining processes with the corresponding material removal mechanism: Machining Mechanism of process material removal P. Chemical 1. Erosion machining Q. Electro – 2. Corrosive reaction chemical machining R. Electro 3. Ion displacement discharge machining S. Ultrasonic 4. Fusion and machining vaporization (A) P – 2, Q – 3, R – 4, S – 1 (B) P – 2, Q – 4, R – 3, S – 1 (C) P – 3, Q – 2, R – 4, S – 1 (D) P – 2, Q – 3, R – 1, S – 4 th

th

th



31.

A single – point cutting tool with 12° rake angle is used to machine a steel work – piece. The depth of cut, i.e. uncut thickness is 0.81 mm. The chip thickness under orthogonal machining condition is 1.8 mm. The shear angle is approximately (A) 22° (C) 56° (B) 26° (D) 76°

ME – 2012 32. In abrasive jet machining, as the distance between the nozzle tip and the work surface increases, the material removal rate (A) Increases continuously (B) Decreases continuously (C) Decreases, becomes stable and then increases (D) Increases, becomes stable and then decreases 33.

In a single pass drilling operation, a through hole of 15 mm diameter is to be drilled in a steel plate of 50 mm thickness. Drill spindle speed is 500 rpm, feed is 0.2 mm/rev and drill point angle is . Assuming 2 mm clearance at approach and exit, the total drill time (in seconds) is (A) 35.1 (C) 31.2 (B) 32.4 (D) 30.1

34.

Details pertaining to and orthogonal metal cutting process are given below: Chip thickness ratio 0.4 Undeformed 0.6 mm thickness Rake angle +10 Cutting speed 2.5 m/s Mean thickness of 25 microns Primary shear zone The shear strain rate in s during the process is (A) 0.1781 (C) 1.0104 (B) 0.7754 (D) 4.397


cut of 4mm. The rotational speed of the workpiece is 160 rpm. The material removal rate in /s is (A) 160 (C) 1600 (B) 167.6 (D) 1675.5 36.

Two cutting tools are being compared for a machining operation. The tool life equations are Carbide tool : HSS tool : Where V is the cutting speed in m/min and T is the tool life in min. The carbide tool will provide higher tool life if the cutting speed in m/min exceeds (A) 15.0 (C) 49.3 (B) 39.4 (D) 60.0

37.

During the electrochemical machining (ECM) of iron (atomic weight = 56, valency =2) at current of 1000A with 90% current efficiency, the material removal rate was observed to be 0.26 gm/s. If Titanium (atomic weight = 48, valency=3) is machined by the ECM precess at the current of 2000 A with 90% current efficiency , the expected material removal rate in gm/s will be (A) 0.11 (C) 0.30 (B) 0.23 (D) 0.52 Statement for linked answer questions 38 & 39: In orthogonal turning of a bar of 100 mm diameter with a feed of 0.25 mm/rev, depth of cut of 4mm and cutting velocity of 90 m/min, it is observed that the main (tangential) cutting force is perpendicular to the friction force acting at the chip – tool interface. The main (tangential) cutting force is 1500 N.

38.

The orthogonal rake angle of the cutting tool in degree is (A) Zero (C) 5 (B) 3.58 (D) 7.16

ME – 2013 35. A steel bar 200 mm in diameter is turned at a feed of 0.25 mm/ rev with a depth of th

th

th



39.

The normal force acting at the chip – tool interface in N is (A) 1000 (C) 2000 (B) 1500 (D) 2500

ME – 2014 40. Which one of the following instruments is widely used to check and calibrate geometric features of machine tools during their assembly? (A) Ultrasonic probe (B) Coordinate Measuring Machine (CMM) (C) Laser interferometer (D) Vernier calipers 41.

The main cutting force acting on a tool during the turning (orthogonal cutting) operation of a metal is 400 N. The turning was performed using 2 mm depth of cut and 0.1 mm/rev feed rate. The specific cutting pressure (in N/m ) is (A) 1000 (C) 3000 (B) 2000 (D) 4000

42.

th y or’s tool life exponent n is 0.2, and the tool changing time is 1.5 min, then the tool life(in min) for maximum production rate is _______

43.

A metal rod of initial length is subjected to a drawing process. The length of the rod at any instant is given by the expression, t t where t is the time in minutes. The true strain rate (in ) at the end of one minute is ___________

44.

During pure orthogonal turning operation of a hollow cylindrical pipe, it is found that the thickness of the chip produced is 0.5 mm. The feed given to the zero degree rake angle tool is 0.2 mm/rev. The shear strain produced during the operation is _______


45.

Match the Machine Tools (Group A) with the probable Operations (Group B): Group A Group B P:Centre Lathe 1:Slotting Q:Milling 2: Counter-boring R:Grinding 3:Knurling S:Drilling 4:Dressing (A) P-1, Q-2, R-4, S-3 (B) P-2, Q-1, R-4, S-3 (C) P-3, Q-1, R-4, S-2 (D) P-3, Q-4, R-2, S-1

46.

The following four unconventional machining processes are available in a shop floor. The most appropriate one to drill a hole of square cross section of 6 mm × 6 mm and 25 mm deep is (A) Abrasive Jet Machining (B) Plasma Arc Machining (C) Laser Beam Machining (D) Electro Discharge Machining

47.

A mild steel plate has to be rolled in one pass such that the final plate thickness is of the initial thickness, with the entrance speed of 10 m/min and roll diameter of 500 mm. If the plate widens by 2% during rolling, the exit velocity (in m/min) is ______

48.

A hole of 20 mm diameter is to be drilled in a steel block of 40 mm thickness. The drilling is performed at rotational speed of 400 rpm and feed rate of 0.1 mm/rev. The required approach and over run of the drill together is equal to the radius of drill. The drilling time (in minute) is (A) 1.00 (B) 1.25 (C) 1.50 (D) 1.75

49.

A rectangular hole of size 100 mm × 50 mm is to be made on a 5 mm thick sheet of steel having ultimate tensile strength and shear strength of 500 MPa and 300 MPa, respectively. The hole is made by punching process. Neglecting the effect of clearance, the punching force (in kN) is (A) 300 (C) 600 (B) 450 (D) 750 th

th

th



50.

51.

52.

53.

54.

The process utilizing mainly thermal energy for removing material is (A) Ultrasonic Machining (B) Electrochemical Machining (C) Abrasive Jet Machining (D) Laser Beam Machining A straight turning operation is carried out using a single point cutting tool on an AISI 1020 steel rod. The feed is 0.2 mm/rev and the depth of cut is 0.5 mm. The tool has a side cutting edge angle of 60°. The uncut chip thickness (in mm) is _______ A Cutting tool is much harder than the work piece. Yet the tool wears out during the tool-work interaction, because (A) extra hardness is imparted to the work piece due to coolant used (B) oxide layers on the work piece surface impart extra hardness to it (C) extra hardness is imparted to the work piece due to severe rate of strain (D) vibration is induced in the machine tool Which pair of following statements is correct for orthogonal cutting using a single-point cutting tool? P. Reduction in friction angle increases cutting force Q. Reduction in friction angle decreases cutting force R. Reduction in friction angle increases chip thickness S. Reduction in friction angle decreases chip thickness (A) P and R (C) Q and R (B) P and S (D) Q and S


twice that in operation 1. The other cutting parameters are identical. The ratio of maximum uncut chip thicknesses in operations 1 and 2 is ________ 55.

The principle of material removal in Electrochemical machining is (A) Fick's law (C) Kir hho '’s ws (B) Faraday's laws h ’s w

56.

Better surface finish is obtained with a large rake angle because (A) The area of shear plane decreases resulting in the decrease in shear force and cutting force (B) The tool becomes thinner and the cutting force is reduced (C) Less heat is accumulated in the cutting zone (D) The friction between the chip and the tool is less

57.

In a rolling process, the maximum possible draft, defined as the difference between the initial and the final thickness of the metal sheet, mainly depends on which pair of the following parameters? P: Strain Q: Strength of the work material R: Roll diameter S: Roll velocity T: Coefficient of friction between roll and work (A) Q, S (B) R, T (C) S, T (D) P, R

58.

A cast iron block of 200 mm length is being shaped in a shaping machine with a depth of cut of 4 mm, feed of 0.25 mm/stroke and the tool principal cutting edge angle of 30°. Number of cutting strokes per minute is 60. Using specific energy for cutting as 1.49 J/m , the average power consumption (in watt) is ______

Two separate slab milling operations, 1 and 2, are performed with identical milling cutters. The depth of cut in operation 2 is

th

th

th





2.

3.

[Ans. B] For making complex shape for higher strength alloy, electro-chemical machining is used because in it no mechanical force is exerted as well as tool left its impression surfaced.

5.

[Ans. B] We know that h

t where,

[Ans. C] 3-2-1 principle is also known as six point location of a three dimensional body. The bottom is supported against 3 point, the rear face against 2 point and the side of the block rest against single (1) point. So 3-2-1 principle. [Ans. A] Matching cost= [Matching time × Direct Labour Cost] So as cutting speed increases, matching time decreases and so matching cost decreases So as matching cost decreases.

h

h h

Cost/piea

t t t

ot ot ot

7.

[Ans. D] ECM – 104 mm3/min EDM – 5 x 103 mm3/min USM – 180 mm3/min EBM – 10 mm3/min LBM – 5 mm3/min

8.

[Ans. D] Here, cutting ratio, r r os t rs Where, = shear angle r angle= Shear strain = cot +tan

stro 9.

1.65

[Ans. B] ro r h t’s th ory Ø λ ,

⁄stro Cutting time required for 100 stroke for returning stroke

ot

[Ans. A] Large negative rake angle, low shear angle and high cutting speed.

[Ans. B] Time required for 1 stroke

utt utt

6.

Cutting speed

No of stroke

t h

Non- productive cost

4.

r t s

Given,

Tool cost Tool changing cost Machine cost

ot

wh r λ r t o λ µ t λ

⁄

total time

th

th

th



10.

[Ans. A] Total heat generated, Q1 = V = 24000 J/min V V s s V s V s

14.


[Ans. C] λ ortho o or ortho o t

Frictional force, F = Fc s os s

t

or ortho o os λ t

utt s λt

Uncut chip thickness for turning = s s t h p th ss r t o r t h r

t 12.

13.

15.

[Ans. A] Computer chip – photochemical machining Metal forming dies and mold – electro discharge machining Turbine blade – electro chemical machining Glass – ultrasonic machining

16.

[Ans. B] Here, D = 147 mm, L = 630 mm, f= 0.2 mm/rev, d = 2 mm Tool life equation, V o st t Now, V V

[Ans. B] Given λ t

t

t

s λt os s

os

Heat dissipated due to friction, V Q2 = 7193.69 J/min

11.

os λ s s os

λ

r os rs os s

[Ans. A] Material removal rate Thermal conductivity of tool Tool wear material removal rate [Ans. D] Cutting force; V= Cutting velocity Specific machining energy = 2.0 J/ V

(

)

V

(

)

V

V

(

)

V = 97.07 m / min 17.

[Ans. C] V N = 210.2 rpm Matching time, t

min.

t 18.

th

[Ans. C] Internal gear cutting is performed by shaping with a pinion cutter. th

th



19.

[Ans. D] Given, Also for orthogonal cutting t s t os t When s


Now, shear force = shear stress F =250 AB width of cut s

area

s

A

20.

[Ans. B] For Iron, Atomic weight, V yr For Metal P, Atomic weight, V yr For alloy, density,

0.2 C

B

where AC is incut chip thickness Now width depth of cut = 3 mm F= 250 0.426 3 320 N ⁄ ⁄s

⁄

23.

[Ans. B] [ ow y

os ] os r h t’s th ory

+ = 320 1.77

* Now, let % of P in alloy= x x V ( ) (

x

= 565N

V

x

Now

x 21.

22.

[Ans. D] Taking the engineering of a conventional lathe, is the back gear arrangement and is the feed gear box(a) and (b) are then completely wrong. Now if (c) is true then, if gear ratio of is changed the relative velocity of spindle with respect to load source in changed and hence pitch is changed. Hence (D) is correct.

α β-α α F

[Ans. D]

N

r os rs

t

)

Fv = 565 tan 34 = 381 N Frictional force F from the figure can be written as, F= FH sin +Fv os = 68+ 381 = 447N Hence none of the options seems to be correct.

x

)

= tan (

os s

24.

[Ans. D]

= th

th

th



25.

[Ans. D] During orthogonal machining. Shear strain, ot t Cutting tool having rake angle Then ot t For minimum value of shear strain ot

s os os

M= molecular weight v= valency M.R.R(gm/sec) =

= 0.3471

t

os

27.

s os s

s

os

os

os

os

os

ot

s os

ot

t

s

s

t

| Therefore is minimum at Therefore minimum shear strain is given by ot

26.


[Ans. A] or to y or’s qu t o V =C where , V = cutting speed in m/min T = tool life (min) C = constant (Taylors constant) V V or ⁄ or ⁄ or o o oth s s o o o o y or’s o st t V √

28.

t

[Ans. C] When cutting speed is halved 60 ×

[Ans. A] We know, R = l/A where, = resistivity = special resistance of electrolyte Ω cm R = resistance of electrolyte between electrodes l = inter electrode gap = 0.2 mm = 0.02 cm A = electrodes cross- section area

(or) ( ) (or) (or) % change in tool life = = =3× 29.

= 300 %

[Ans. A] V

Ω Current, I = V/R = 12/0.01 = 1200A Faraday law of electricity, E = M/v where, E = equivalent weight

(

V

(

( )

V

V

)

)

( )

V th

th

th



30.

31.

[Ans. A] P – 2, Q – 3, R – 4, S – 1

h r str r t V os os t os os 0104 /sec

[Ans. B] t t

r

r

t


os r s

os os

35.

[Ans. D] t r

t

32.

[Ans. D] In abrasive jet machining the variation is as shown

r

ov r t

⁄s 36.

[Ans. B] Carbide HSS

MRR

NTD

Equate tool life from both equation after eliminating cutting speed V

MRR=Metal removal rate NTD= Nozzle tip distance 33.

[Ans. A] Substitute in one of equation, x = 4.5 Drill bit

V 37.

x

[Ans. C] V

7.5

15

ou

s

Work piece

r st

st

t

ov r p 38.

ov r

[Ans. A]

=35.1 sec 34.

[Ans. C] r=0.4 t V t t

⁄s τ

r os rs th

th

th



It is observed that cutting force is perpendicular to the friction force acting at the chip-tool interface then = 0. sh r or = orthogonal rake angle sh r τ r to

39.

̇

[Ans. C] Ultrasonic probe - Used a sensor which generates a constant signal and also detect return signals Coordinate Measuring Machine (CMM) Used to measure physical geometrical characteristics of an object Laser interferometer - Used to check and calibrate geometric features Vernier calipers – used to measure dimension of an object.

41.

[Ans. B] r v p

utt

r y pr su

44.

( 43.

)

t t

t

t

ot ot

⁄r v

t t

y 45.

[Ans. C] P: Centre Lathe – knurling (diamond shape is formed on the cylindrical work piece) Q: Milling – slotting (this operation performed only in milling machine) R: Grinding – Dressing (Dressing is a procedure to remove a layer of abrasive from the surface of grinding wheel) S: Drilling – Counter boring (is like an enlarging the dia of already drilled hole)

46.

[Ans. D] The most appropriate choice is EDM as by taking a square tool it can be done with lesser time and loss.

47.

[Ans. *] Range 14.6 to 14.8

t

t

t

Volume flow rate should be equal t w v t w v

]

)

t

[Ans. *] Range 0.9 to 1.1

v

ru st

t

t

)

[Ans. *] Range 2.8 to 3.0 t r os rs

[Ans. *] Range 5.9 to 6.1 n= 0.2, T=1.5 min Tool life for maximum production [(

t

y

⁄ 42.

(

̇

[Ans. B] From the previously described diagram we can get the relation N= os s N – normal force acting at the chip – tool interface

40.


w

t

w v

r t th

th

th



48.

[Ans. B] t

t

With increase in shear angle, cutting force decreases (Q) chip thickness decreases (S) So, Q and S

r

54. 49.


[Ans. B]

[Ans. *]Range 0.70 to 0.72 Max. uncut chip thickness √

t

t

F=Feed , Z=Number of teeth N= rotational speed d= depth of cut, D= cutter dia so

Force = τ τ

50.

51.

[Ans. *] Range 0.08 to 0.12 Given r v, t

√

√

55.

[Ans. B] The principle of material removal in ECM s r y’s ws

56.

[Ans. A] As rake angle increase, lip angle decreases, shear plane area decreases and hence shear force and cutting force reduced.

57.

[Ans. B]

t

[Ans. D] Ultrasonic machining and Abrasive jet machining utilizing mechanical energy. Electrochemical machining utilizing electrochemical energy Laser Beam machining utilizing thermal energy .

t t

Maximum possible draft

h

coefficient of friction between roll and work R = Roll radius 58.

s

[Ans. *]Range 295 to 305 Material removal rate,

s s

t 52.

[Ans. C] Extra hardness is imparted to work piece due to server rate of strain

53.

[Ans. D] Reduction in friction angle increases shear angle.

s Average power consumption, s tt

th

th

th




Metrology and Inspection ME – 2005 1. In order to have interference fit, it is essential that the lower limit of the shaft should be (A) greater than the upper limit of the hole (B) lesser then the upper limit of the hole (C) greater than the lower limit of the hole (D) lesser then the lower limit of the hole

Drill spindle

Drill table

If Rp = RQ> 0, which one of the following would be consistent with the observation? (A) The drill spindle rotational axis is coincident with the drill spindle taper hole axis (B) The drill spindle rotational axis intersects the drill spindle taper hole axis at point P (C) The drill spindle rotational axis is parallel to the drill spindle taper hole axis (D) The drill spindle rotational axis intersects the drill spindle taper hole axis at point Q.

ME – 2007 A hole is specified as mm. The mating shaft has a clearance fit with minimum clearance of 0.01 mm. The tolerance on the shaft is 0.04 mm. The maximum clearance in mm between the hole and the shaft is (A) 0.04 (C) 0.10 (B) 0.05 (D) 0.11

ME – 2008 4. A displacement sensor (a dial indicator) measures the lateral displacement of a mandrel mounted on the taper hole inside a drill spindle. The mandrel axis is an extension of the drill spindle taper hole axis and the protruding portion of the mandrel surface is perfectly cylindrical. Measurements are taken with the sensor placed at two positions P and Q as shown in figure. The readings are recorded as Rx = maximum deflection minus minimum deflection, corresponding to sensor position at X, over one rotation.

sensor Q

ME – 2006 2. A ring gauges is used to measure (A) outside diameter but not roundness (B) roundness but not outside diameter (C) both outside diameter and roundness (D) only external threads

3.

P

mandrel

ME – 2009 5. What are the upper and lower limits of the shaft represented by 60 ? Use the following data: Diameter 60 lies in the diameter step of 50 -80mm. Fundamental tolerance unit, i = in m = 0.45 ⁄ , where D is the representative size in mm; Tolerance value for IT8 = 25i. u t v t o or ‘ ’ shaft = 5.5 (A) Lower limit = 59.924mm, Upper Limit = 59.970mm (B) Lower limit = 59.954mm, Upper Limit = 60.000mm (C) Lower limit = 59.970mm, Upper Limit = 60.016mm (D) Lower limit = 60.00mm, Upper Limit = 60.046mm th

th

th



ME – 2010 6. A taper hole is inspected using a CMM, with a probe of 2 mm diameter. At a height, Z = 10 mm from the bottom, 5 points are touched and a diameter of circle (not compensated for probe size) is obtained as 20 mm. Similarly, a 40 mm diameter is obtained at a height Z = 40 mm. The smaller diameter (in mm) of hole at Z = 0 is


ME – 2013 10.

Cylindrical pins of mm diameter are electroplated in a shop. Thickness of the plating is micron. Neglecting gauge tolerances, the size of the GO gage in mm to inspect the plated components is (A) 25.042 (C) 25.074 (B) 25.052 (D) 25.084

11.

A metric thread of pitch 2 mm and thread angle is inspected for its pitch diameter using 3 – wire method. The diameter of the best size wire in mm is (A) 0.866 (C) 1.154 (B) 1.000 (D) 2.000

Z = 40

Z = 20

Z=0

(A) 13.334 (B) 15.334 7.

(C) 15.442 (D) 15.542

sh t h s so ∅ . The respective values of fundamental deviation and tolerance are (A) 0.025, ±0.008 (B) 0.025, 0.016 (C) 0.009, ± 0.008 (D) 0.009, 0.016

ME – 2014 12. For the given assembly: 25 H7/g8, match Group A with Group B Group A Group B P. H I. Shaft Type Q. IT8 II. Hole Type R. IT7 III. Hole Tolerance Grade S. g IV. Shaft Tolerance Grade (A) P-I, Q-III, R-IV, S-II (B) P-I, Q-IV, R-III, S-II (C) P-II, Q-III, R-IV, S-I (D) P-II, Q-IV, R-III, S-I 13.

The flatness of a machine bed can be measured using (A) Vernier calipers (B) Auto collimator (C) Height gauge oo r’s ros op

14.

The diameter of a recessed ring was measured by using two spherical balls of diameter mmand mm as shown in the figure.

ME – 2011 8.

A hole is of dimension mm. The corresponding shaft is of dimension (A) (B) (C) (D)

mm. The resulting assembly has loose running fit close running fit transition fit interference fit

ME – 2012 9. In an interchangeable assemble, shafts of size

mm mate with holes of

size mm. The maximum interference (in microns) in the assembly is (A) 40 (C) 20 (B) 30 (D) 10

Recessed ring t r

th

th

th



The

distance and . The diameter (D, in, mm) of the ring gauge is _______________ 15.


condition, the dimension (in mm) of the GO plug gauge as per the unilateral tolerance system is (A)

A GO-No GO plug gauge is to be designed for measuring a hole of nominal diameter 25 mm with a hole tolerance of ± 0.015 mm. Considering 10% of work tolerance to be the gauge tolerance and no wear

(B) (C) (D)


[Ans. A] For making interference fit, seeing the figure it is essential that lower limit of shaft should be greater than upper limit of hole.

5.

[Ans. A] D=√ = 63.25 mm i = 0.45 (D)1/3 + 0.001Dµm = 1.856 × m= Tolerance = 25i = 46.4 × m Fundamental deviation δ (5.5) D 0.41 = 30.113 × 10-6 = 0.030 mm (absolute) High limit = Basic size – Fundamental deviation = 60 – 0.030 = 59.97 Lower limit = High limit – Tolerance = 59.77 – 0.0464 = 59.924mm

6.

[Ans. A] Diameter d = 2 mm.

Upper limit of Shaft

Lower limit of shaft Upper limit of hole

Lower limit of hole

2.

[Ans. A]

3.

[Ans. C] Minimum clearance = Minimum hole size – Maximum shaft size x u sh t s z = 40.00 0.01 = 39.99 mm Tolerance on shaft= 0.04mm Minimum shaft size= 39.99 – 0.04 = 39.95 Maximum clearance = Maximum hole size – Minimum shaft size = 40.05 39.95 = 0.10

4.

Z= 40

∅= 40

Z= 20

∅= 20

Z=0

x

10

[Ans. C] Equal deflection in both sensors means eccentricity parallel to the axis.

x

t x x th

th

th



o

t z

Z= 40


13.

[Ans. B] Auto collimator – flatness of machine bed like lathe

14.


40mm

B

30 40

20mm

Z= 10 A x

10

h suppos

Z=0

7.

8. 9.

[Ans. D] Tolerance = (35 – 0.009) – (35 – 0.025) = 0.016 Fundamental deviation is the deviation close to zero line = 0.009,

H=

h

h h h

[Ans. C] Size of shaft =

√ √

+0.030 Hole

+0.020

Shaft 0.010

Maximum interference =0.040

15.

[Ans. D] Maximum thickness of plating = 0.03 + 0.002=0.032 mm Size of Go gauge = 25.02+0.032 = 25.084mm [Ans. C] For thr

12.

[Ans. D] P. H Q. IT8 R. IT7 S. g

h

mm +0.040

11.

h

mm

Size of hole =

10.

h

so

[Ans. C] Transition fit.

25 mm

h

st w r s v

[Ans. D] o

u

Given Work tolerance = 0.015 – (0.015) Given gauge tolerance = 10% of work tolerance It is asked in unilateral tolerance system ow r to u

y

Hole Type (this symbol made for hole type shafts) Represent shaft tolerance Represents cut hole tolerance Shaft Type (small letter used for shaft). th

th

th




Computer Integrated Manufacturing (CIM) ME – 2005 1. The tool of an NC machine has to move along a circular arc from (5, 5) to (10, 10) while performing an operation. The center of the arc is at (10, 5). Which one of the following NC tool path commands performs the above mentioned operation? (A) N010 G02 X10 Y10 X5 Y5 R5 (B) N010 G03 X10 Y10 X5 Y5 R5 (C) N010 G01 X5 Y5 X10 Y10 R5 (D) N010 G02 X5 Y5 X10 Y10 R5 2.

ME – 2008 Statement for Linked Answer Questions 5 and 6: In the feed drive of a Point – to – point open loop CNC drive, a stepper motor rotating at 200 steps/rev drives a table through a gear box and lead screw – nut mechanism (pitch = 4mm, number of starts =1). The gear ratio = (

given by U = . The stepper motor (driven by voltage pulses from a pulse generator) executes 1 step/ pulse of the pulse generator. The frequency of the pulse train from the pulse generator is f = 10,000 pulses per minute.

Which among the NC operations given below are continuous path operations? Arc Welding (AW) Milling (M) Drilling (D) Laser Cutting of Sheet Metal (LC) (A) AW, LC and M (B) AW, D, LC and M (C) D, LC, P and SW (D) D, LC and SW

Punching in Sheet Metal (P) Spot Welding (SW)

ME – 2006 3. NC contouring is an example of (A) continuous path positioning (B) point-to-point positioning (C) absolute positioning (D) incremental positioning

) is

Pulse Generator

f

Table

Steeper motor

Gear Box

Nut Lead screw

5.

The Basic Length Unit (BLU), i.e. , the table movement corresponding to 1 pulse of the pulse generator, is (A) 0.5 microns (B) 5 microns (C) 50 microns (D) 500 microns

6.

A customer insists on a modification to change the BLU the CNC drive to 10 microns without changing the table speed. The modification can be accomplished by

ME – 2007 4. Which type of motor is NOT used in axis or spindle drives of CNC machine tools? (A) induction motor (B) dc servo motor (C) stepper motor (D) linear servo motor

(A) changing U to and reducing f to (B) changing U to and increasing f to 2f (C) changing

U to

and keeping f

unchanged (D) keeping U unchanged and increasing f to 2f.

th

th

th



7.

or r t oo ’s sur w require (A) a set of grid points on the surface (B) a set of grid control points (C) four bounding curves defining the surface (D) two bounding curves and a set of grid control points

ME – 2009 8. Match the following: NC Code Definition P. M05 1. Absolute coordinate system Q. G01 2. Dwell R. G04 3. Spindle stop S. G90 4. Linear interpolation (A) (B) 2 (C) (D) 9.

Which of the following is the correct data structure for solid models? (A) solid part faces edges vertices (B) solid part edges faces vertices (C) vertices edges faces solid parts (D) vertices faces edges solid parts

ME – 2010 10. In a CNC program block, N002 G02 G91 X … r r to (A) circular interpolation in counterclockwise direction and incremental dimension (B) circular interpolation in counterclockwise direction and absolute dimension (C) circular interpolation in clockwise direction and incremental dimension (D) circular interpolation in clockwise direction and absolute dimension


ME – 2012 11. A CNC vertical milling machine has to cut a straight slot of 10 mm width and 2 mm depth by a cutter of 10 mm diameter between points (0,0) and (100, 100) on the XY plane (dimensions in mm). The feed rate used for milling is 50 mm/min. Milling time for the slot (in seconds) is (A) 0 (C) 180 (B) 170 (D) 240 ME – 2014 12. For machining a rectangular island represented by coordinates P(0,0), Q(100,0), R(100,50) and S(0,50) on a casting using CNC milling machine, an end mill with a diameter of 16 mm is used. The trajectory of the cutter center to machine the island PQRS is (A) ( 8, 8), (108, 8), (108, 58), ( 8,58) , ( 8, 8) (B) (8,8), (94,8), (94,44), (8,44), (8,8) (C) ( 8,8), (94,0), (94,44), (8,44), ( 8,8) (D) (0,0), (100,0), (100,50), (50,0), (0,0) 13.

A robot arm PQ with end coordinates P(0,0) and Q(2,5) rotates counter clockwise about P in the XY plane by 90°. The new coordinate pair of the end point Q is

14.

For the CNC part programming, match Group A with Group B: Group A Group B P: circular interpolation, I: G02 counter clock wise Q: dwell II: G03 R: circular interpolation, III: G04 clock wise S: point to point countering IV: G00 (A) P-II, Q-III, R-I, S-IV (B) P-I, Q-III, R-II, S-IV (C) P-I, Q-IV, R-II, S-III (D) P-II, Q-I, R-III, S-IV

th

th

th





[Ans. D]

10.

[Ans. C]

2.

[Ans. A]

11.

[Ans. B] to

3.

[Ans. A] NC contouring is an example continuous path of positioning.

=141.42mm /mm

of rt

s

4.

[Ans. A]

5.

[Ans. B]

6.

[Ans. A] Stepper motor 200 steps /rev = 200 pulses/ rev Pitch = 4mm, no. of start =1, Gear ratio =

12.

[Ans. A] The best option is A because we have to consider the dimension of cutting tool also we have to take co-ordinate of cutting tool from (-8,-8).

13.

[Ans. B] os [s

F=10,000 pulses / mim pu s s

r v o

otor

rev of

lead screw r 1 pulse =

st = 5microns =

BLU ------- answer B Feed = pu s /min = /min For changing BLU = 10 micron = 0.01 mm GEAR ratio (U) has to be reduced to ½ pu s s /min pu s s = 5000 So F has to be reduced to F/2. 7.

[Ans. B]

8.

[Ans. C]

9.

[Ans. C] Correct data structure for solid models is given by vertices edges faces solid parts

[

X

X

s os

th

X

]

] X X X s

14.

][

os

s

s

os

w

[Ans. A] G00 – Print to point movement G01 – Liner interpolation G02 – Circular interpolation clock wise G03 – Circular Interpolation counter clock wise G04 – dwell for a specific time

th

th



Industrial Engineering

Production Planning and control ME – 2005 1. The distribution of lead-time demand for an item is as follows: Lead time demand Probability 80 0.20 100 0.25 120 0.30 140 0.25 The re-order level is 1.25 times the expected value of the lead-time demand. The service level is (A) 25% (C) 75% (B) 50% (D) 100% 2.

A welding operation is time-studied during which an operator was pace-rated as 120%. The operator took, on an average, 8 minutes for producing the weld-joint. If a total of 10% allowances are allowed for this operation, the expected standard production rate of the weld-joint (in units per 8 hour day) is (A) 45 (C) 55 (B) 50 (D) 60

3.

The sales of a product during the last four years were 860, 880, 870 and 890 units. The forecast for the fourth year was 876 units. If the forecast for the fifth year, using simple exponential smoothing, is equal to the forecast using a three period moving average, the value of the exponential smoothing constant is (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄

4.

An assembly activity is represented on a Operation Process Chart by the symbol (A) (C) D (B) A (D) O

ME – 2006 5. In an MRP system, component demand is: (A) forecasted (B) established by the master production schedule (C) calculated by the MRP system from the master production schedule (D) ignored ME – 2008 6. A moving average system is used for forecasting weekly demand. t and t are sequences of forecasts with parameters m and m , respectively, where m and m (m > m ) denote the number of weeks over which the moving averages are taken. The actual demand shows a step increase from to at a certain time. Subsequently, (A) neither F1(t) nor F2(t) will catch up with the value d2 (B) both sequences F1(t) and F2(t) will reach d2 in the same period (C) F1(t) will attain the value d2 before F2(t) (D) F2(t) will attain the value d2 before F1(t) ME – 2009 7. Which of the following forecasting methods takes a fraction of forecast error into account for the next period forecast? (A) simple average method (B) moving average method (C) weighted moving average method (D) exponential smoothening method ME – 2010 8. Vehicle manufacturing assembly line is an example of (A) product layout (C) manual layout (B) process layout (D) fixed layout

th

th

th



9.

The demand and forecast for February are 12000 and 10275, respectively. Using single exponential smoothening method (smoothening coefficient = 0.25), forecast for the month of March is (A) 431 (C) 10706 (B) 9587 (D) 11000


ME – 2014 11. In exponential smoothening method, which one of the following is true? (A) 0 ≤ ≤ an high value of is use for stable demand (B) 0 ≤ ≤ an high value of is use for unstable demand (C) ≥ an high value of is use for stable demand (D) ≤ 0 an high value of is use for unstable demand

ME – 2012 10. Which one of the following is NOT a decision taken during the aggregate production planning stage? (A) Scheduling of machines (B) Amount of labour to be committed (C) Rate at which production should happen (D) Inventory to be carried forward

12.

The actual sales of a product in different months of a particular year are given below: September 180 October 280 November 250 December 190 January 240 February ? The forecast of the sales, using the 4month moving average method, for the month of February is _______


[Ans. C] Expected value of lead time demand X

0

0

0

00 0

0

Z 0 Service level = 0.74≈ 0 2.

0

[Ans. A] Rating factor = 1.2 Normal time = 1.2 ×8 = 9.6 minute

X Reorder level

Standard time = 9.6 + 9.6 ×

= 140 √

0

%

00

0

0

√

= 10.56 minute/piece Total time available = 8×60 minute

√ 0

So, production rate = ≈

wel joint

ROL

3.

[Ans. C] 0

X

Reorder level = X 0

0

Z 4.

[Ans. D] th

th

th



5.

6.

[Ans. C] In a MRP system, component demand is calculated by the MRP system from the master Production Schedule.


Responsibility and stability

eman fore ast

[Ans. D] Since (m2 < m1), weightage of the latest demand would be more in F2 (t) Hence it will attain the actual value earlier.

7.

[Ans. D]

8.

[Ans. A]

9.

[Ans. C]

time

ta le esponsiveness And we know that i e um er of o servation is more in

Smoothed average forecast period t. Previous Period forecast Smoothing Constant Previous Period demand 0 0 25 (12000-10275) = 10706.5 10.

[Ans. A]

11.

[Ans. B] We know that

for

Responsiveness indicates that forecast have a fluctuating or swing pattern used for new product and for this value of n is kept low stability means that forecast pattern is level or flat, so best option is B. 12.

[Ans. *]Range 239 to 241 month moving average 0 0 0 0 Fore cast of Sale, For month of February = 240

Where N = number of observation

th

th

th




Inventory Control ME – 2006 1. Consider the following data for an item. Price quoted by a supplier Order quantity Unit price (units) (Rs.) < 500 10 ≥ 500 9

4.

In a machine shop, pins of 15 mm diameter are produced at a rate of 1000 per month and the same is consumed at a rate of 500 per month. The production and consumption continue simultaneously till the maximum inventory is reached. Then inventory is allowed to reduced to zero due to consumption. The lost size of production is 1000. If backlog is not allowed, the maximum inventory level is (A) 400 (C) 600 (B) 500 (D) 700

5.

The maximum level of inventory of an item is 100 and it is achieved with infinite replenishment rate. The inventory becomes zero over one and half month due to consumption at a uniform rate. This cycle continues throughout the year. Ordering cost is Rs. 100 per order and inventory carrying cost is Rs. 10 per item per month. Annual cost (in Rs.) of the plan, neglecting material cost, is (A) 800 (C) 4800 (B) 2800 (D) 6800

6.

Capacities of production of an item over 3 consecutive months in regular time are 100, 100 and 80 and in overtime are 20, 20 and 40. The demands over those 3 months are 90, 130 and 110. The cost of production in regular time and overtime are respectively Rs. 20 per item and Rs. 24 per item. Inventory carrying cost is Rs. 2 per item per month. The levels of starting and final inventory are nil. Backorder is not permitted. For minimum cost of plan, the level of planned production in overtime in the third month is (A) 40 (C) 20 (B) 30 (D) 0

Annual demand: 2500 units per year Ordering cost: Rs.100 per order Inventory holding rate: 25% of unit price. The optimum order quantity (in units) is: (A) 447 (C) 500 (B) 471 (D) ≥600 2.

A manufacturing shop processes sheet metal jobs, wherein each job must pass through two machines (M1 and M2, in that order). The processing time (in hours) for these jobs is: Machine Jobs P Q R S T U M1 15 32 8 27 11 16 M2 6 19 13 20 14 7 The optimal make-span (in hours) of the shop is: (A) 120 (C) 109 (B) 115 (D) 79

ME – 2007 3. The net requirements of an item over 5 consecutive weeks are 50 – 0 – 15 – 20 – 20. The inventory carrying cost and ordering cost are Re. 1 per item per week and Rs. 100 per order respectively. Starting inventory is zero Use “Least Unit Cost Te hnique “for developing the plan. The cost of the plan (in Rs.) is (A) 200 (C) 255 (B) 250 (D) 260

th

th

th



ME – 2008 7. The product structure of an assembly P is shown in the figure. Estimated demand for end product P is as follows week Demand 1 1000 2 1000 3 1000 4 1000 5 1200 6 1200 Ignore lead times for assembly and subassembly. Production capacity (per week) for component R is the bottleneck operation. Starting with zero inventory, the smallest capacity that will ensure a feasible production plan up to week 6 is P

Assembly Q

ME – 2009 9. A company uses 2555 units of an item annually. Delivery lead time is 8 days. The recorder point (in number of units) to achieve optimum inventory is (A) 7 (C) 56 (B) 8 (D) 60 ME – 2010 10. Annual demand for window frames is 10000. Each frame costs Rs. 200 and ordering cost is Rs. 300 per order. Inventory holding cost is Rs. 40 per frame per year. The supplier is willing to offer 2% discount if the order quantity is 1000 or more, and 4% if order quantity is 2000 or more. If the total cost is to be minimized, the retailer should (A) order 200 frames every time (B) accept 2% discount (C) accept 4% discount (D) order Economic Order Quantity

R

11. Sub-Assembly S

(A) 1000 (B) 1200 8.


T

(C) 2200 (D) 2400

A set of 5 jobs is to be processed on a single machine. The processing time (in days) table below. The holding cost for each job is Rs. K per day Job Processing Time P 5 Q 2 R 3 S 2 T 1 A schedule that minimizes the total inventory cost is (A) T – S – Q – R – P (B) P – R – S – Q – T (C) T – R – S – Q – P (D) P – Q – R – S – T

Little’s law is a relationship between (A) stock level and lead time in an inventory system (B) waiting time and length of the queue in a queuing system (C) number of machines and job due dates in a scheduling problem (D) uncertainty in the activity time and project completion time

ME – 2011 12. The word kanban is most appropriately associated with (A) economic order quantity (B) just – in – time production (C) capacity planning (D) product design ME – 2013 13. Customers arrive at a ticket counter at a rate of 50 per hr and tickets are issued in the order of their arrival. The average time taken for issuing a ticket is 1 min. th

th

th



Assuming that customer arrivals form a poisson process and service times are exponential, the average waiting time in queue in min is (A) 3 (C) 5 (B) 4 (D) 6 14.

15.

In simple exponential smoothing forecasting, to give higher weightage to recent demand information, the smoothing constant must be close to (A) – 1 (C) 0.5 (B) Zero (D) 1.0 In a CAD package, mirror image of a 2D point P(5,10) is to be obtained about a line which passes through the origin and makes an angle of counter clockwise with the X – axis. The coordinates of the transformed point will be (A) (7.5,5) (C) (7.5, - 5 ) (B) (10, 5) (D) (10, - 5)

Process I II III IV (A) I (B) II

18.

Jobs arrive at a facility at an average rate of 5 in an 8 hour shift. The arrival of the jobs follows Poisson distribution. The average service time of a job on the facility is 40 minutes. The service time follows exponential distribution. Idle time (in hours) at the facility per shift will be ⁄ C ⁄ ⁄ 0⁄ A component can be produced by any of the four processes I, II, III and IV. The fixed cost and the variable cost for each of the processes are listed below. The most economical process for producing a batch of 100 pieces is

Fixed Cost (in Rs.) 20 50 40 10 (C) (D)

Variable cost per piece in (in Rs.) 3 1 2 4 III IV

19.

Consider the following data with reference to elementary deterministic economic order quantity model Annual demand of an item 100000 Unit price of the item (in Rs.) 10 Inventory carrying cost per 1.5 unit per year (in Rs.) Unit order cost (in Rs.) 30 The total number of economic orders per year to meet the annual demand is ______

20.

A manufacturer can produce 12000 bearing per day. The manufacturer received an order of 8000 bearing per day from a customer. The cost of holding a bearing in stock in stock is Rs.0.20 per month. Setup cost per production run is Rs. 500. Assuming 300 working days in a year, the frequency of production run should be (A) 4.5 days (C) 6.8 days (B) 4.5 months (D) 6.8 months

21.

Demand during lead time with associated probabilities is shown below: Demand Probability 50 0.15 70 0.14 75 0.21 80 0.20 85 0.30 Expected demand during lead time is ____

22.

At a work station, 5 jobs arrive every minute. The mean time spent on each job in the work station is 1/8 minute. The mean steady state number of jobs in the system is _______

ME – 2014 16. The jobs arrive at a facility, for service, in a random manner. The probability distribution of number of arrivals of jobs in a fixed time interval is (A) Normal (C) Erlang (B) Poisson (D) Beta 17.


th

th

th





[Ans. C] Case I: Let EOQ is less than 500 C 0 0 √

√

3.

[Ans. B] Weeks Qty Carrying Cost 1 50 0 1 2 50 0 1 3 65 30 1 4 85 90 1 5 105 170 4 20 0 1 5 40 20 Order Total Limit Cost Cost Cost 100 100 2 100 100 2 100 130 2 100 190 ← unit ost 100 270 2.57 100 100 5 100 120 ← Least ost Weeks 1 2 3 4 Demand 50 0 15 20 Order Qty 65 0 0 40 Entering 15 15 0 20 inventory Carrying 15 15 0 20 cost Order cost 100 0 0 100 Total 115 115 130 130 250 Total ost for the plan s 0

n

Case II: Let EOQ is greater than 500 C 0 =√

√

471.40

which is against the assumption When price break occurs at Q = 447.21 at Q = 500

TIC = 1118.03 TIC = 1062.5

TIC at Q = 447.21 TC

C 0

C 0

00

= 1118.03 TIC at Q = 500 TIC = (average inventory) (unit inventory cost) + (No. of order per year) (cost of order) = =

C

C

4.

[Ans. B] t p

00

Hence, quantity to be ordered is 500. 2.

p

p

[Ans. B] Using Johnson’s algorithm M1 M2 IN OUT IN OUT R 0 8 8 21 T 8 19 21 35 S 19 46 46 66 Q 46 78 78 97 U 78 94 97 104 P 94 109 109 115 So optimum make – span = 115

5 20 0 0 0 0 250

r

r 000

p

00 = 500 units

r

t

th

th

th



5.

6.

7.

8.

[Ans. D] Maximum inventory 00 Inventory cycle = 1.5 months um er of y les require per year month y les months Ordinary cost = 8 × 100 = Rs 800 Average inventory per cycle = 100/2 = 50 Inventory carrying cost = Rs.10 per item per month 0 per item percycle Total Inventory carrying cost = 50 × 8 × 15 = Rs. 6000 Total ost 00 6000 = 6800 [Ans. B] For minimum total cost, the entire regular time production capacity has to be used Hence total over time capacity required 0 0 0 00 00 0 = 50 20 overtime units will be used up in the first 2 months hence 30 overtime units are needed to fulfill the 3rd month demand Since carrying costs are there, it is obvious that to minimize costs, these 30 units shoul e pro u e is month ‘ ’ using overtime capacity. [Ans. C] The M.R.P. for R is given by Week Demand Inventory 1 2000 P – 2000 2 2000 2P – 4000 3 2000 3P – 6000 4 2000 4P – 8000 5 2400 5P – 10400 6 2400 6P – 12800 If 6P 12800 ≥ 0 then the capacity is feasible Minimization, 6P = 12800 P 2200

9.


[Ans. C]

Reorder point

2555 unit x

8

365 days

or x= 56 units Alternately Reorder point = Safety stock + (delivery lead time 0 10.

)

(

)

[Ans. C] Total = purchase cost + holding cost + ordering cost It is a case of inventory with price breaks. √

√ = 387 units For Q = 387 T C

C 0000

C 00

0

00

= Rs. 2015492 For Q = 1000 T C

0000

00

0

0

0 00 = Rs.1983000 For Q = 2000. 0000

00

0 00 ept % is ount 11.

[Ans. A] To minimize the total inventory cost, i.e. the holding cost, the job with the least processing time should be done first.

000

0 00

[Ans. B] Little’s Law is a relationship etween L and W (also L &W ) L W L W L – Expected no.of customers in System. W Expected waiting time in system.

th

th

th



L W

Expected no. of customers in queue Expected Waiting time in queue.

12.

[Ans. B]

13.

[Ans. C]

14.

[Ans. D]

15.

[Ans. B] y

C

√

18.

19.

000

No. of orders 20.

(5, 10) y↔ x (10, 5) 0

17.

Unit order cost = 30,

√

45

16.


[Ans. B] Arrival pattern of customers: Although arrival might follow any pattern the frequency employed assumption which support many real world situation is that arrivals are poison distribution arrival is random, and random arrival is best es ri e y oisson’s istri ution [Ans. B] Average rate of providing facility 5 in 8 hrs. The average service time = 40 minute So total time required to provide service = 0 00 mins = 3.33 hrs. Ideal time = 8 3.33 = 4.667

[Ans. B] Cost 0 Cost 0 Cost 0 Cost 0

00 00 00 00

0 0 0 0

0

[Ans. C] Manufacturer can produce 12,000 bearings /day Received order from a customer is 8000 bearings/day Carrying cost is Rs.0.20/month s year Working days are 300/year Set up cost is Rs. 500/production Optimal production size is, √

√

Consideration 366 days/years D = 000 000 000 U 0 00 √

√

0 Frequency of the production run, p 0 000 days days 21.

[Ans. *]Range 74 to 75 Expected demand during lead time 0 0 0 0 0 0 0 0

22.

[Ans. *]Range 1.62 to 1.70 L Mean steady state number of jobs in the system L

[Ans. *] Range 49 to 51 Given, D = Total demand = 100000 C Inventory carrying cost per unit per year = 1.5

⁄

L th

th

th




Operations Research ME – 2005 Statement for Linked Answer Questions 1 & 2: Consider a linear programming problem with two variables and two constraints. The objective function is: Maximize X +X . The corner points of the feasible region are (0, 0), (0, 2), (2, 0) & ( 1.

2.

3.

4.

The number of activities that need to be crashed to reduce the project duration by 1 day is (A) 1 (B) 2 (C) 3 (D) 6 5.

A component can be produced by any of the four processes I, II, III and IV. Process I has a fixed cost of Rs.20 and variable cost of Rs. 3 per piece. Process II has a fixed cost of Rs. 50 and variable cost of Re. 1 per piece. Process III has a fixed cost of Rs. 40 and variable cost of Rs. 2 per piece. Process IV has a fixed cost of Rs.10 and variable cost of Rs. 4 per piece. If the company wishes to produce 100 pieces of the component, from economic point of view it should choose (A) Process I (C) Process III (B) Process II (D) Process IV

6.

Consider a single server queuing model with Poisson arrivals ( = 4 / hour) and exponential service ( = 4 / hour). The number in the system is restricted to a maximum of 10. The probability that a person who comes in leaves without joining the queue is (A) ⁄ (C) ⁄ (B) ⁄ 0 (D) ⁄

).

If an additional constraint X + X ≤ 5 is added, the optimal solution is (A) (

)

(C) (

)

(B) (

)

(D) (5,0)

Let Y and Y be the decision variables of the dual and v and v be the slack variables of the dual of the given linear programming problem. The optimum dual variables are (A) Y and Y (C) Y and v (B) Y and v (D) v and v A company has two factories S1, S2 and two warehouses D1, D2. The supplies from S1 and S2 are 50 and 40 units respectively. Warehouse D1 requires a minimum of 20 units and a maximum of 40 units. Warehouse D2 requires a minimum of 20 units and, over and above, it can take as much as can be supplied. A balanced transportation problem is to be formulated for the above situation. The number of supply points, the number of demand points, and the total supply (or total demand) in the balanced transportation problem respectively are (A) 2, 4, 90 (C) 3, 4, 90 (B) 2, 4, 110 (D) 3, 4, 110 A project has six activities (A to F) with respective activity durations 7, 5, 6, 6, 8, 4 days. The network has three paths A B, C D and E F. All the activities can be crashed with the same crash cost per day.

ME – 2006 Statement for Linked Answer Questions 7 & 8: Consider a PERT network for a project involving six tasks (a to f). Task Predecessor Expected Variance of task time the task time (in days) (in ays ) a 30 25 b a 40 64 c a 60 81 d b 25 9 e b, c 45 36 f th

d, e th

20 th

9



7.

8.

9.


The expected completion time of the project is: (A) 238 days (C) 171 days (B) 224 days (D) 155 days

Item

The standard deviation of the critical path of the project is: (C) √ 00 days (A) √ days (D) √ days (B) √ days The table gives details of an assembly line. Work station I II III IV V VI Total task time 7 9 7 10 9 6 at the workstation (in minutes) What is the line efficiency of the assembly line? (A) 70% (C) 80% (B) 75% (D) 85%

10.

A stockist wishes to optimize the number of perishable items he needs to stock in any month in his store. The demand distribution for this perishable item is: Demand (in 2 3 4 5 units) Probability 0.10 0.35 0.35 0.20 The stockist pays Rs.70 for each item and he sells each at Rs.90. If the stock is left unsold in any month, he can sell the item at Rs.50 each. There is no penalty for unfulfilled demand. To maximize the expected profit, the optimal stock level is: (A) 5 units (C) 3 units (B) 4 units (D) 2 units

11.

A firm is required to procure three items (P, Q and R). The prices quoted for these items (in Rs.) by suppliers S1, S2 and S3 are given in table. The management policy requires that each item has to be supplied by only one supplier and one supplier supply only one item. The minimum total cost (in Rs.) of procurement to the firm is

Suppliers S1

S2

S3

P

110

120

130

Q

115

140

140

R

125

145

165

(A) 350 (B) 360 12.

(C) 385 (D) 395

The number of customers arriving at a railway reservation counter is Poisson distributed with an arrival rate of eight customers per hour. The reservation clerk at this counter takes six minutes per customer on an average with an exponentially distributed service time. The average number of the customers in the queue will be (A) 3 (C) 4 (B) 3.2 (D) 4.2

ME – 2008 Common Data for Questions 13 and 14: Consider the Linear Programme (LP) Max 4x + 6y subject to x y≤ x y≤ x y≥0 13. After introducing slack variables s and t, the initial basic feasible solution is represented by the table below (basic variables are s = 6 and t = 6, and the objective function value is 0). 4 6 0 0 0 S 3 2 1 0 6 T 2 3 0 1 6 x y s t RHS After some simplex iterations, the following table is obtained 0 0 0 2 12 S 5/3 0 1 1/3 2 Y 2/3 1 0 1/3 2 x y s t RHS From this, one can conclude that

th

th

th



(A) The LP has a unique optimal solution (B) The LP has an optimal solution that is not unique (C) The LP is infeasible (D) The LP is unbounded 14.

15.

The dual for the LP is (A) Min 6u + 6v subject to u v≥ u v≥ u v≥0 (B) Max 6u + 6v subject to u v≤ 2u + 3v ≤ 6 u v≥0 (C) Max 4u + 6v subject to u v≥ u v≥ u, v ≥ 0 (D) Min 4u + 6v subject to u v≤ 2u + 3v ≤ 6 u v≥0

16.

For the standard transportation linear program with m sources and n destinations and total supply equaling total demand, an optimal solution (lowest cost) with the smallest number of nonzero xij values (amounts from source i to destination j) is desired. The best upper bound for this number is (A) mn (C) m + n (B) 2(m + n) (D) m + n 1

17.

In an M/M/1 queuing system, the number of arrivals in an interval of length T is a Poisson random variable (i.e. the probability of there being n arrivals in an interval of length T is

. The

probability density function f(t) of the inter-arrival time is given by (C) (A) ( ) (D)

(B)

ME – 2009 Common Data Questions: 18 & 19: Consider the following PERT network: 3

For the network below, the objective is to find the length of the shortest path from node P to node G. Let dij be the length of directed arc from node i to node j. Let sj be the length of the shortest path from P to node j. Which of the following equations can be used to find sG? Q P

G

R

(A) (B) (C) (D)


SG = Min(SQ, SR) SG = Min (SQ dQG, SR dRG) SG = Min(SQ+dQG, SR+dRG) SG = Min(dQG,dRG)

1

6

5

2

7

4

The optimistic time, most likely time and pessimistic time of all the activities are given in the table below. Activity Optimistic Most likely Pessimistic time (days) time (days) time (days) 1–2 1 2 3 1–3 5 6 7 1–4 3 5 7 2–5 5 7 9 3–5 2 4 6 5–6 4 5 6 4–7 4 6 8 6–7 2 3 4 th

th

th



18.

The critical path duration of the network (in days) is (A) 11 (C) 17 (B) 14 (D) 18

19.

The standard deviation of the critical path is (A) 0.33 (C) 0.77 (B) 0.55 (D) 1.66

20.

Six jobs arrived in a sequence as given below: Jobs Processing Time (days) I 4 II 9 III 5 IV 10 V 6 VI 8 Average flow time (in days) for the above jobs using Shortest Processing Time rule is (A) 20.83 (C) 125.00 (B) 23.16 (D) 139.00

21.

Consider the following Linear Programming Problem (LPP): Maximum z x x Subject to x ≤ x ≤ x x ≤ x ≥0 x ≥0 (A) The LPP has a unique optimal solution (B) The LPP is infeasible (C) The LPP is unbounded (D) The LPP has multiple optimal solutions

22.

The expected time of a PERT activity in terms of optimistic time , pessimistic time ( ) and most likely time is given by

(A)

=

(B)

=

(C)

=

(D)

=


ME – 2010 Common Data for Questions 23 and 24: Four jobs to be processed on a machine as per data listed in the table. Job Processing time (in days) Due Date 1 4 6 2 7 9 3 2 19 4 8 17 23. If the Earliest Due Date (EDD) rule is used to sequence the jobs, the number of jobs delayed is (A) 1 (C) 3 (B) 2 (D) 4 24.

Using the Shortest Processing Time (SPT) rule, total tardiness is (A) 0 (C) 6 (B) 2 (D) 8

25.

The project activities, precedence relationships and durations are described in the table. The critical path of the project is Activity Precedence Duration (in days) P 3 Q 4 R P 5 S Q 5 T R,S 7 U R,S 5 V T 2 W U 10 (A) T V (C) U W (B) T V (D) U W

26.

Simplex method of solving linear programming problem uses (A) All the points in the feasible region (B) Only the corner points of the feasible region (C) Intermediate points within the infeasible region (D) Only the interior points in the feasible region

th

th

th



ME – 2011 Common Data for Question 27 and 28: One unit of product P1 requires 3 kg of resource R1 and 1 kg of resource R2. One unit of product P2 requires 2 kg of resource R1 and 2 kg of resource R2. The profits per unit by selling product P1 and P2 are Rs.2000 and Rs.3000 respectively. The manufacturer has 90 kg of resource R1 and 100 kg of resource R2. 27. The unit worth of resource i.e. dual price of resource in Rs/kg is (A) 0 (C) 1500 (B) 1350 (D) 2000 28.

The manufacturer can make a maximum profit of Rs. (A) 60000 (C) 150000 (B) 135000 (D) 200000

29.

Cars arrive at a service station according to oisson’s istri ution with a mean rate of 5 per hour. The service time per car is exponential with a mean of 10 minutes. At steady state, the average waiting time in the queue is (A) 10 minutes (C) 25 minutes (B) 20 minutes (D) 50 minutes

ME – 2012 Linked Answer Questions Q.30 and Q.31 For a particular project, eight activities are to be carried out. Their relationships with other activities and expected duration are mentioned in the table below. Activity Predecessors Duration (days) a 3 b a 4 c a 5 d a 4 e b 2 f d 9 g c, e 6 h f, g 2


30.

The critical path for the project is (A) a-b-e-g-h (C) a-d-f-h (B) a-c-g-h (D) a-b-c-f-h

31.

If the duration of activity f alone is changed from 9 to 10 days, then the (A) Critical path remains same and the total duration to complete the project changes to 19 days. (B) Critical path and the total duration to complete the project remains the same. (C) Critical path changes but the duration to complete the project remains the same. (D) Critical path changes and the total duration to complete the project changes to 17 days.

ME – 2013 32. A linear programming problem is shown below: Maximize Subject to ≤ 0 ≤ ≥0 It has (A) An unbounded objective function (B) Exactly one optimal solution (C) Exactly two optimal solutions (D) Infinitely many optimal solutions ME – 2014 33. If there are m source and n destinations in a transportation matrix, the total number of basic variables in a basic feasible solution is (A) m n (C) m n (B) m n (D) m

th

th

th



34.

A project has four activities P, Q, R as shown below. Normal Predecessor Activity duration (days)

and S

Cost slope (Rs./d ay) P 3 500 Q 7 P 100 R 4 P 400 S 5 R 200 The normal cost of the project is Rs. 10,000/- and the overhead cost is Rs. 200/- per day. If the project duration has to be crashed down to 9 days, the total cost (in Rupees) of the project is _______ 35.

A minimal spanning tree in network flow models involves (A) All the nodes with cycle/loop allowed (B) All the nodes with cycle/loop not allowed (C) Shortest path between start and end nodes (D) All the nodes with directed arcs

36.

Consider the project network, where numbers along various activities represent the normal time. The free float on activity 4-6 and the project duration, respectively, are

(A) 2, 13 (B) 0, 13


37.

Consider an objective function Z x x x x and the constraints x x ≤ x x ≤ x ≥0 x ≥0 The maximum value of the objective function is ________

38.

The total number of decision variables in the objective function of an assignment problem of size n × n (n jobs and n machines) is C n (A) n (D) n (B) 2n

39.

The precedence relations and duration (in days) of activities of a project network are given in the table. The total float (in days) of activities e and f , respectively, are Activity Predecessor Duration (days) a 2 b 4 c a 2 d b 3 e c 2 f c 4 g d, e 5 (A) 0 and 4 (C) 2 and 3 (B) 1 and 4 (D) 3 and 1

(C) 2, 13 (D) 2, 12

th

th

th





[Ans. B]

7.

[Ans. D] b

(0, 5)

a

1 4

2

30

(0, 2)

d 25

60 c

45 e

20

5

6

f

Critical Path = a e f = 30 + 60 + 45 + 20 = 155 days

(0, 5)

Constant equation on optimal region Optimal solution, (

40

4

(4/3, 4/3)

(0, 2)

3

≤

no effect 8.

[Ans. A] Standard deviation

)

2.

[Ans. B]

3.

[Ans. C] For balance transportation problem, No. of supply points =m+n–1=2+ 1 =3 No. of demand point = 4 Total demand = 50 + 40 = 90

9.

4.

[Ans. C] Path Duration AB 7 + 5 = 12 CD 6 + 6 = 12 EF 8 + 4 = 12 There are three critical paths. So no.of activities to be crashed = 3

5.

[Ans. B] Total cost = fixed cost + (Number of pieces × variable cost) Variable cost of process II is less.

6.

[Ans. A] When

days = √

√ [Ans. C] Line efficiency

00 00

0%

Cycle time = max 10.

[Ans. B] P(d ≥ ) = probability that the demand for Q units or more arginal profit per unit sol = 0 0 s 0 =Marginal loss from each unit that is left unsold s 0 0 s 0 0 p ≥ 0 0 0 Demand Probability Cumulative that demand probability at this level 2 0.1 0.1 ≥0 3

0.35

0.45

4

0.35

0.8

5

0.20

1.00

0

th

th

th

0 ≥0 0 0 0 ≤0

0



Hence to maximize profit, optimal stock level is 4. 11.


matrix now, as the revised opportunity cost matrix

[Ans. C]

P

5

0

0

P 110 120 130

Q

0

20

0

Q 115 140 140

R

0

10

35

Thus final assignment is Item Suppliers Cost P 120 Q 140 R 125 The minimum cost = 120 + 140 +125 = Rs. 385

R 125 145 165 Step 1: Substract minimum entry in each column from all the entries on the column.

P

0

0

0

Q

5

20

10

R 15

25

3

12.

Step 2: Substract minimum entry in each row of job – opportunity cost matrix from all the entries of the row

0

0

0

0

15

5

0

10

20

Step 3: Draw minimum number of horizontal and vertical lines to cover all zeros. This can be done in 2 line, which is one less than the number of rows (which is 3). Thus the solution is yet not fixed. Hence go to steps 4

[Ans. B] Average number of customer in queue 0

0

13.

[Ans. B] Since the relative profit by varying any of the non basic variable in the second table does not show any increment, hence the solution must be optimal. But a change in the non- basic variable x, gives a relative profit of zero. So, the LPP has an alternate optima.

14.

[Ans. A] In matrix form, the dual of the primal

Max Z = (4 6) ( )

1 0

0

0

0

15

5

0

10

20

i.e, Max z = cx

2

subject to Ax < = b Min x>=0 W = yb

Step 4: From the uncovered entries, the minimum is 5. Thus, substract 5 from all the entries, which are uncovered. Add 5 at junction of lines. We get the following

subject : y A > = C is given y > by =0 The dual for the primal is th

th

th



Min 6u+6v Subject 3u+2v ≥ 4 2u+3v ≥ 6 u, v ≥ 0 15.

16.

[Ans. C] The only way to reach G is then by reaching Q and QG or reaching R and then RG. SG = Min(SQ+ dQG, SR + dRG)

Path I. 1-3-5-6- → II. 1-2-5-6- → III. 1- 4 - → Thus critical path is I Critical path duration is 18. 19.

[Ans. D] In such an L.P.P, m n variables are there and m + n equations/constraints are there (satisfying the demand supply requirements). But one constraint is removed as total supply equals total demand. The best upper bound xij values is (m+n-1).

17.

[Ans. C]

18.

[Ans. D]

Variance = (

)

Critical path 1-3 3-5 5-6 6-7 Total varian e

Variance ⁄ ⁄ ⁄ ⁄

tan ar 15 15 6

3

5 10 10

5

18 18 7

20.

6

4 3

EST = earlisest start time LFT = latest finish time Path along which EST and LFT is equal are called critical path. Critical path duration is 18 days Alternately Activity t t t t 1 5 3 5 2 4 4 2

2 6 5 7 4 5 6 3

3 7 7 9 6 6 8 4

( )

( )

erivation

( ) √

√varian e 0

5

2 3 2 2 7

3

1-2 1-3 1-4 2-5 3-5 5-6 4-7 6-7

[Ans. *] For any Poisson distribution, the probability density function f(t) is given by

( )

6 6 3 0 0 1 EST LFT


[Ans. A] Jobs using shortest Possible time rule I III V VI II IV

Processing time (days)

Cumulative time (days)

4 5 6 8 9 10

4 9 15 23 32 42

verage time flow

2 6 5 7 4 5 6 3

0

th

th

th



21.

[Ans. D] By Graphical method

25.

[Ans. D] T 4

1 Q

S

4

7 (2, 6) x2 =6

4

A = (4, 3)

3

3

5

5

5 7

V 2

U

W 6

7 10

PRUW (i) 3 + 5 + 5 + 10 = 23 QSTV (ii) 4 + 5 + 7 + 2 = 18 PRTV (iii) 3 + 5 + 7 + 2 = 17 QSUW (iv) 4 + 5 + 5 10 = 24

5 B

≤

2 1 A 1 2

R 5

3

8

6

2

P

9

x2


3

4

5

6

7

8

9 x 1

Max.z = 3x1+2x2 At point B, z = 18x1=4 At point A, z=18 This means LPP has multiple optimal solutions because both points have same value.

26.

[Ans. B]

27.

[Ans. A]

28.

[Ans. B] P1 P2 R1 3 2 R2 1 2 Profit per 2000 3000 unit X Y Zmax = 2000x + 3000y

22.

[Ans. A]

23.

[Ans. C] Job (j) Processing time (t )

1 4

2 7

4 8

3 2

Completion time(C )

4

11

19

21

Due Date ( )

6

9

17

19

Availability 90 100

y

B(0, 45)

A(30, 0)

x

Lateness(L ) ve Lateness – Job is completed after due date ve Lateness – Job is completed before due date jo s elaye 24.

→ x y →x y Subject to x y ≤ 0

≤

x

≤

y ≤ 00

x y≥0 x y≥0 Zmax = 2000x + 3000y Corner points satisfying both the constrains are A(30, 0) B(0, 45). ZA = 2000 × 30+3000 × 0 = 60,000/ZB = 2000 × 0 + 3000× 45 = 1,35,000/Solution is optimal at B; x =0; y = 45 3x + 2y + S1 = 90; 3 × 0 + 2 × 45 + S1 = 90 ⟹S1 = 0 i.e., R1 resource is fully utilized.

[Ans. D] Sequence by shortest processing Time (SPT) Job Process Job flow Due Tardiness time time date 3 2 0 19 0 1 4 6 0 2 7 6 + 7 =13 9 4 4 8 13+8=21 17 4 Total tardiness = 0 + 0 + 4 + 4 = 8 th

th

th



x

y 00 0 00 ⟹ 0 i.e., R2 resource is un utilized at optimality. Hence dual price of R2 is zero. 29.


Feasible region is inside line AD Optimal solution occurs at either A or D lets check 0 z 0 (0 ) z

[Ans. D] Given data: arrival rate servi e rate

0

( )

So exactly one optimal solution at D hour hour

verage waiting time in queue

33.

[Ans. C]

34.

[Ans. *] Range 12490 to 12510

hr 0

30.

7 days 100/day

hr 0 minutes

P

1

R 4 days 400/day

3

6

7

5

Various path A BEGH days A CGH days A DFH days a f h is critical path. 31.

[Ans. A] If the duration of activity F is changed from 9 to 10 days, critical path remains same and the project duration will increase to 19 days.

32.

[Ans. B] Plot constraints x y 0 x y (0, 10/7) 3x + 7y = 10

D (0,1.33)

35.

I II A

4 S

3

5 days 200/day

From Heuristic model, First find the critical path ays Crash the lowest cost slope from critical path Therefore, Crashing a tivity ‘ ’ y ays will be 10 days Now there are 2 critical paths i.e., P – Q = 10 days P – R – S = 10 days. Crashing 2 activities from each critical path i e rashing a tivity ‘ ’ y ay an rashing a tivity ‘ ’ y ay Now the project duration is crashed to 9 days. So, total cost of the project, 0 000 00 ays 00 ays 00 ay 00 ay 00

4

C

2

C

Q

3 days 500/day

[Ans. C]

1

2

B(10/3,0) A(2,0)

th

[Ans. B] Spanning tree cyclic property Cycle property is the basis for kruskal’s algorithm th

th



1. 2.

36.

Sort all edges in increasing weight order. Consider the edges in order. If the edges does not create a cycle , add it to the spanning tree otherwise discard it, stop , when n-1 edges have been added, because then we must have spanning tree. So best option from above property is B.

z x x onstraints x x ≤ x x ≤ x ≥0 x ≥0 Co-ordinate maximum values (0, 0) 0 (4, 0) 12 (0, 2) 18 So the maximum value of objective function is 18

[Ans. A] T T

T T

2

2 3 T T

0 0

1

2

3 T T

2 4

38.

5 3

5

6 4


[Ans. A] Jo

5

7

T T

n

T T

T T

n

For free float activity on activity 4-6 Free float = (EFT –EST) –T ̇

n

39.

n njo s an n ma hines Total number of decision variables

T T

T T T T


n

[Ans. B]

Project duration 13 37.

a hine

0 a 0

f T T

e g T T

T T

x

Total float for activity e = Total float for activity f

0

x

th

th

th


GQB_ME.pdf

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