GQB_CE.pdf

GATE QUESTION BANK for

Civil Engineering By

GATE QUESTION BANK

Contents

Contents #1.

#2.

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

Fluid Mechanics 1 2 3 4 5 6 7 8 9 10 11 12

#3.

Pressure and its Measurement Hydrostatic Forces on Plane Surfaces Kinematics of Flow Fluid Dynamics Flow Through Pipes Impulse Momentum Equation and Its Application Flow through Orifices and Mouth Pieces Boundary Layer Flow Viscous Flow Dimensional Analysis Impacts of jets and Turbines Open Channel Flow

149 150 – 151 152 – 153 154 – 156 157 – 158 159 160 161 162 – 166 167 168 169

Hydrology & Irrigation

170 - 191

1 2 3

170 – 177 178 179 – 181

4 5

#4.

149 - 169

Hydraulics & Hydraulic Machinery Irrigation Water Requirements of Crops Sediment, Transport and Design of Irrigation Channels Hydrology

182 183 – 191

4.Environmental Engineering

192 - 214

1 2 3 4 5 6

192 – 193 194 – 198 199 – 207 208 – 209 210 – 211 212 – 214

Quality Standards of Water Water Supply and its Treatment Waste Water Treatment Sludge Disposal Domestic Waste Water Treatment Air Pollution

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GATE QUESTION BANK

#5.

Structural Analysis 1 2 3 4 5 6

#6.

215 – 221 222 – 224 225 – 231 232 233 234 – 235

236 - 257 Simple Stress and Strain Relationship Bending Moment and Shear Force Diagram Thin Walled Pressure Vessel Simple Bending Theory Torsion Column and Struts Analysis of Statically Determinate Structures

236 – 241 242 – 246 247 248 – 251 252 – 253 254 – 256 257

258 - 277 Concrete Technology Basic of Mix Design Design of RCC structures Analysis of Ultimate Load Capacity Basic Elements of Pre-stressed Concrete Design of Pre-Stressed Concrete Beams Concrete Design

258 – 259 260 – 263 264 265 – 272 273 – 274 275 – 276 277

Steel 1 2 3 4 5 6

#9.

Trusses and Arches Influence Line Diagram and Rolling Loads Slope and Deflection Method Degree of Static Indeterminacy Force Energy Method Matrix Method of Structural Analysis

RCC 1 2 3 4 5 6 7

#8.

215 - 235

Mechanics 1 2 3 4 5 6 7

#7.

Contents

278 - 290 Introduction Plastic Analysis Welded Connections Design of Tension Member Compression member Beams

278 – 279 280 – 283 284 – 286 287 288 289 – 290

Geotechnical Engineering 1 2 3 4 5 6 7 8

291 – 325

Three Phase System, Fundamental Definitions and Relationship Index Properties and Soil Classification Permeability and Seepage Consolidation Compaction Stress Analysis Stress Analysis Surface Investigations th

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291 292 – 295 296 – 300 301 – 304 305 306 – 309 310 – 311 312 – 313 th

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GATE QUESTION BANK

9 10 11 12

Contents

Earth Pressure Stability of Slopes Bearing Capacity Pile Foundation

314 – 315 316 – 317 318 – 322 323 – 325

#10. Transportation & Surveying 1 2 3 4 5 6 7 8 9 10 11

326 - 358

Introduction to Transportation Geometric design of highway Traffic Characteristics Traffic Signs and Signal Design Intersection Design Testing and Specifications of Paving Materials Design of Rigid Flexible Pavements Introduction Measurement of Distance & Direction Theodolite & Traversing Leveling

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326 327 – 332 333 – 336 337 – 339 340 – 341 342 – 344 345 – 348 349 – 350 351 352 – 355 356 – 358

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 III

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 os

os sin

]

os (C) [ sin

sin os

]

sin (D) [ os

os sin

0

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

]

Eigen values of a matrix S

ME – 2007 6. The number of linearly independent

]. What is the matrix F?

os (A) [ sin (B) [

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.

(B) 1/2

(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

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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

ME – 2012 15.

*

x

√

(B) (√ )

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

1 is

0 (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

(D) ( ) √

√

+, the transpose

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

1 , one of the

(C) (√ )

(A) (√ )

ME – 2009 For a matrix,M-

For the matrix A=0

normalized Eigenvectors is given as

16.

11.

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

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

[

] 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) (B)

0 0

1

m trix 0 (C) (D)

1

0 0

The inverse of the matrix 0 is

1

is

( )

0

( )

0

( )

0

( )

0

The Eigenvalue of the matrix

CE – 2012

[P] = 0

15.

(A) (B) 11.

14.

1

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

1 is

1 are and 8 and 5

(C) (D)

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

i i i i

i

i

i

i

i

i

i

i

i

i i

i

i

i

i

1

1

1

1

i

1

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

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

Mathematics

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

[

the following does NOT ]

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 __________.

-

is __________.

16.

) 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 A= [

]. , ⁄

(A) [

th

th

⁄

⁄

th

-

is

] ⁄


GATE QUESTION BANK

⁄

⁄

(B) [

(C) [

⁄

(D) [

Let, A=0

3.

(A) 0 (B) 1

]

⁄

⁄

1 and

Then (a + b)= (A) ⁄ (B) ⁄

= 0

⁄

1

⁄ ⁄

(C) (D)

Given the matrix 0

The rank of the matrix [

⁄

] ⁄

2.

6.

]

⁄

Mathematics

Eigenvector is (C) 0

1

(B) 0 1

(D) 0

1

(C) 2 (D) 3

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. 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

]

ECE – 2006 4.

For the matrix 0 corresponding 0

to

the

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

(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

v =0

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

1

[

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)

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

(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

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,

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

Mathematics

(

)4 5

(

)h s

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

] (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.

4.

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

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

EE – 2005 1.

If R = [

] , then top row of

(A) , (B) ,

2.

-

(C) , (D) ,

5.

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

3.

(C) [

(B) [

]

(D) [ ]

]

EE – 2006 Statement for Linked Answer Questions 4 and 5. ] ,Q=[

] ,R=[

(B) [

] [

] [

]

(C) [

] [

] [

]

(D) [

] [

] [

]

(C) [ ]

(B) [

(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

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

P=[

]

(A) [ ]

] , one of

]

] [

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

-

(A) [

(A) [

is -

For the matrix p = [

Mathematics

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

] are

three vectors

th

th

th


GATE QUESTION BANK

(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 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 (B) A2 + 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 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) A A+ A = A (C) A+ A =  (B) (AA+ ) = A A+ (D) A A+ A = A+

14.

Let P be a real orthogonal matrix. x⃗ is a real vector [x x - with length x⃗ (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.

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)

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.

th

An Eigenvector of

[

(A) , (B) ,

(C) , (D) ,

th

-

th

] is -


GATE QUESTION BANK

EE – 2011 18.

The matrix[A] = 0

22.

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.

23.

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

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

1 and 0 1 and 0 1 and 0 1 and 0

EE – 2013 19.

The equation 0

1 1 1 1

x 1 0x 1

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

0 1 has

(A) No solution

x (B) Only one solution 0x 1

Eigenvector of the matrix A = 0

0 1.

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

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

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

Mathematics

2.

1 respectively. The matrix is 1

(C) 0

1

(B) 0

1

(D) 0

1

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

(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

1 and

(A) 0

1?

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 th

th

th


GATE QUESTION BANK

5.

For a given that 0

1

matrix A, it is observed 0

1 n

0

1

0

10.

1

Then matrix A is

 2 1   1 0   1 1  (A) A       1 1  0 2  1 2

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

0 2 (D) A    1 3

7.

12.

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

IN – 2009 8.

The matrix P =[

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 (B) 1 (C) 2 (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

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

IN – 2007 6. Let A = [ ] i j n with n = i. j. Then the rank of A is (A) (C) n (B) (D) n

Mathematics

] rotates a vector

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)

(

)(

)

(D) n IN – 2011 13.

The matrix M = [

Eigenvalues . An Eigenvector corresponding to the Eigenvalue 5 is , - . One of the Eigenvectors of the matrix M is (A) , (C) , √ (B) , (D) , IN – 2013 14. The dimension of the null space of the matrix [

about the axis[ ] by an angle of (A) (B) 9.

(C) (D)

] has

(A) 0 15.

] is (B) 1

(C) 2

(D) 3

One of Eigenvectors corresponding to the two Eigenvalues of the matrix 0

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

(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.

Now E × F = G [Ans. A] [

4.

[Ans. B] Given

sin os

]

]

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. 2.

os [ sin

,E-

∴

[Ans. A] For S

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. 5.

[Ans. A]

6.

[Ans. B] 0

n

in onsistent

No (

1 Eigenv lues re 2, 2 I)

(

I)

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

.

/ I)

( ⁄ ) ( ) n ( ⁄ ) ( ( ) minimum of m n) For inconsistence ( ) ( ⁄ ) ∴ he highest possi le r nk of is 3.

7.

[Ans. A] ( I) . olving for , Let the symmetric and real matrix be A = 0 Now |

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

sin os

1 |

) Which gives ( ⟹ ⟹ Hence real Eigen value.

]

]

th

th

th


GATE QUESTION BANK

8.

[Ans. B]

12.

Let

0

eigenv lues re

1

Eigenve tor is x 13.

taking K

]

[

→

[

]

[

[ [

→ →

If

14.

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

15.

[Ans. B] 1 eigenv lues v lue

0

Eigen vector will be . Norm lize ve tor

)

]

[√( )

(

) ]

]

*

[

]

16.

]

system will h ve solution

][

M x

→ MM

I

0

⁄ √ + ⁄ √

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

]

/

(

[

[Ans. A] iven M

]

√( )

[

→

]

infinite m ny solutions

[Ans. B] ( ⁄ )

11.

[

( )

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

x verify the options

[Ans. C]

→

⁄

10.

1 → Eigenv lues re

0

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

9.

[Ans. A]

n

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

Mathematics

[

| ]

1

x Equating the elements x

→

[

|

]

⁄

th

th

th


GATE QUESTION BANK

→

[

|

]

|

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

[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 Taking transpose 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) ̅ ( x x̅ re Eigenve tors they i i i 0

nnot e zero )

19.

[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. [Ans. A] |

20.

[Ans. D] 0

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 ( )  3×3 P 2×3

|

So, |

|

(Taking 2 common from each row) ( )

Hence Eigenvalue of a symmetric matrix are real 18.

Mathematics

|

th

th

th


GATE QUESTION BANK

 3×2 P( )  (2×3) (3×3) (3×2) 2×2 ( ( ) )  2×2 4.

[

[

| ]→

[

| ]→

| ]

[

[

| ]

|

[

|

] ]

[Ans. A]

]

Inverse of 0

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

|

Now for infinite solution last row must be completely zero ie –2=0 n –7=0  n 8.

→

[

→

Then by Gauss elimination procedure [

| ]

→

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

Mathematics

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

0

1

∴0

1 is (

1

)

(

) 0

9.

1

0

1 1

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

10.

0

)P

[Ans. B] A=0

1

Characteristic equation of A is 6.

7.

[Ans. B] ∑ = Trace (A) + + =1+5+1=7 Now = 2, = 6 ∴ 2+6+ =7 =3

|

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

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

|=0

| ]

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

14.

1 =(

∴ 0

i

i

,( = 15.

[

0

i

0 ) i

i)( i

i

,

-[

i

,

] -

-

[Ans. 88] The determinant of matrix is [

]

→ [

0

]

[

20.

1

i -

]

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

i

i

i

i

1

]

→

1

[

]

→

1

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

[

19.

[Ans. B] 0

]

,

1

i) i

][

K JK

[Ans. B] 0

Mathematics

[

]

Interchanging Column 1& Column 2 and taking transpose

[Ans. 0.5] 0.5

[

th

th

]

th


GATE QUESTION BANK

|

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

|

* (

)

= (

(

)+

|

)

|

( 21.

)

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

] →

[

( )

(

)

( )

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 ontr i tion o 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

( ) ]

( )

[

] ( )

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. 2] [

CS 1.

Mathematics

[Ans. B] 0

1

th

th

th


GATE QUESTION BANK

5.

6.

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

| ] →

→

[

[

) =0 = –1, 1 Only one matrix has an Eigenvalue of 1 which is 0

8.

[Ans. D] |

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

| ]

[Ans. A] Eigenvalues of 0 |

Eigenvalues of 0 |

[Ans. D] 0

1

| )( )(

(

Eigenvalues of 0

√ Eigenvalues of A are √ respectively So Eigenvalues of (√ )

)(

n

√

n ( √ ) n

1

n

| =0

(

) )

1

) =0 ) = i or 1 = 1 –i or 1 + i

|

|

(

|= 0

( (

1

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

= 0, 0

|

( )

10.

=0,1

Eigenvalues of 0

( )

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

| =0 =0

x

9.

1

| =0

1

Correct choice is (A)

| ]

Now as long as – 5 0, rank (A) =rank (A|B) =3 ∴ can be any real value except 5. Closest correct answer is (D). 7.

(

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

Mathematics

√

n

√

) =0 th

th

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

15.

[

[ ] 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

]

]

→ →

→

[

] →

→

[

] →

[

16.

]

( ) ( ) no of v ri ∴ nique solution exists 13.

14.

les

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

[ | |

ECE 1.

2.

[Ans. 0]

[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, ,

-

[Ans. A] We know,

] (

[Ans. 6] Let A =

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

Mathematics

16

0 )

0

th

=I 7=0 1

th

0

th

1 1


GATE QUESTION BANK

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

Or 2a 0.1b=0,  2a 

3.

7.

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

8.

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

[Ans. C] 0

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

 x1  1   =0  x2   x + 2x = 0 x = 2x x x  1= 2 2 1 Putting = 5, 0

Hence, 0 4.

1 is Eigenvector.

[Ans. C] 1 We know th t it is Eigenvalue

0

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

x

[Ans. A] or m trix We know

0

1

|I A|=0

|

|

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

2 –I2

 +32 =0  = 4, 8 (Eigenvalues) For

= 4, ( I

)=0

1

)=0

1

v =0 1 For

= 8, ( I

v =0 6.

1

[Ans. C] [

Mathematics

]

| [

]

| =0

( 

( ) th

)(( , th

)=0

) j th

j


GATE QUESTION BANK

11.

12.

13.

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

Then AB = [4]; BA

[

Here m = 1, n = 4 ) And et(I et of , -

]

et(I

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

1

’ will

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

I

18.

| |

[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]

) ]

| |

[Ans. B] ,

Let

-

[ ]

]

19.

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

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

[

I

I

[

]

]

→

th

th

→ →

[

[

]

]

th


GATE QUESTION BANK

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

les

∴ cof. (A) = [

0

=[

1

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

EE 1.

]

Adj (A) =, of ( )-

[Ans. B] onsi er

22.

Mathematics

Dividing by |R| = 1 gives

2.

[Ans. B] ] j( ) | |

, of tor( )| |

| |=|

=[

]

∴ 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= [

]

|

x = k

= 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).

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

cof. (1, 1) = + |

|=2+3=5

= :1:0

cof. (2, 1) =

|= 3

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

|

cof. (2, 1) = + |

i.e. x x x = k : k : 0

|= +1

th

th

th


GATE QUESTION BANK

3.

4.

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

5.

)(

7.

|

9.

= (x ) = x

x

[Ans. A] 1

|A – | = 0  |

| =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

|

x (

i

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.

[Ans. D] ⃗ k

)

)

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 ‘ ’

(

[Ans. B] xy xx | yx

|

x

)

i

)

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

|

|

 (   (

8.

L(x) = |

]

Hence Eigenvalue of M : | M

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

|

6.

M=[

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

Mathematics

)

⃗ =[ x k

x

⃗( k

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

(

x )

I)(

I) I

(

]

I) I

(

I

I)(

I) I

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GATE QUESTION BANK

I (

15.

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

16.

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

I) (

I) I

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.

Argument matrix C =*

[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

→

os

x in )

|| x⃗ || = √x

(x in

x

|| x⃗ || = || x̅|| for any vector x̅

x

+

, *

+

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

[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

Mathematics

is

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

os ) 18.

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

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GATE QUESTION BANK

19.

20.

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

p q pr qs [ ] pr qs r s ∴ hey h ve s me r nk N IN 1.

[Ans. B]

[Ans. D] Eigen value

Given:

0

1

Characteristic equation is,

Eigenvectors 0

1 n 0

Let matrix 0 x

Mathematics

|A

1

x 10

1

0

1

0

10

1

0

1

|

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

1

0

I|= |

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

1

-

Solving 0 21.

1

0

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

22.

[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.

23.

[Ans. C] p q 0 1 r s ( pplying → p q →r s element ry tr nsform tions)

2.

[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 1 X1    , X2    , 1  1, 2  2  1  2 th

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GATE QUESTION BANK

We also know that

, where

11.

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

12.

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

1 1 P   X1 X 2      1 2

 0   1 0  & D=  1    0 2   0 2 Hence

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

[Ans. B] A= [

A=[

]=[

]

[

]

Hence, rank (A) =1

13.

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

8.

[Ans. C]

9.

[Ans. C] Approach 1: (

0

1

0

10

0

0

1 1

0

]

→

[

→

[

]

1 ]

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

)=0

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

]

[

I)

| )(

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

1

|

| (

I

1

I)

Now | I

- is also vector

Apply row operations 0

(

14.

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

For given A = [

Assume, ∴A

]

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

Using elementary transformation

7.

Mathematics

15.

[Ans. A] A=|

[Ans. D]

|

Characteristics equation | th

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GATE QUESTION BANK

|

Mathematics

| j j j

[

j

x ] 0x 1

0 1

x x

j j

[

j j x

x ] 0x 1

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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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

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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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th

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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)

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th

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 ____________

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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 

 x2  exp    8 dx is 2 0 (A) 1 (C) 2 (B) (D) I

1

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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)

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

x

1

0

(B)

and respectively. What is the

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

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

1 (A)   2

CDF 1

0

1

0

11. 1

0

x

1

1 1 1

0 0 1

2

10

1 (C)   2 2

(B)

(D)

N=1

(C) M + N = 1 (D) M + N = 3

CDF

1

1

1

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 - 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

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

___________________ 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]

∫ t( t 6

t)dt t

t 6

7

[

]

∫ t(

[

t

∴ Probability = 7.

Mean (t)̅ = ∫ t f(t) dt

[Ans. A] A uniform function

t)dt

t)dt

=∫ (t =0

t )dt 1

0

7

density

Density function  1 a,x  b  f(x)   b  a a  x,x  b  0

∫ t (

t)dt

 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  Put the value of F(x), we get 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

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

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 2 2 b  a  2ab  12 (b  a)2  12 Standard deviation = √v ri nce

 

 Standard deviation =

8.

9.

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

10.

[Ans. D] Required probability =

11.

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

12.

[Ans. C]

(b  a)2 12 (b  a)

12 Given: b=1, a=0

10 12



12

X=0

X=1

(X ) is Below X (X ) has to be less than 0.5 but greater than zero 13.

( ) ( )

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

. / . /

1

[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  P (H) = 1   8 8 Alternately Probability of getting at least one head

( )

Mathematics

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

(

)

( ) ⏟

⏟

( )

he knows correct nswer

3

1 7  (3  3  1)    2 8 th

th

e does not know correct nswer so he guesses th


GATE QUESTION BANK

(

)

= probability of one employed women +probability of one employed man

( ) ( ⁄ ) ( ) ⁄ ( ) ⁄

( ⁄ )

18.

14.

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

V(x) x (

σ 15.

(x )

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

, (x)-

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

√

[Ans. A]

( ) 19.

16.

[Ans. A]

3 0.1

(x) x (x) σ

Mathematics

[Ans. *](Range 49 to 51)

[Ans. *] Range 0.25 to 0.27 p q

orm l distribution

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.

( ) ( )

( ) ( )

[Ans. *] Range 0.64 to 0.66 Let number of men = 100 Number of women = 100 No. of employed men = 80% of men = 80 No. of employed women = 50% of women = 50 Probability if the selected one person being employed

-

∫ e dz σ√ ∴ of s ving ccount holder 20.

th

[Ans. B] Mean m = np = 5.2

th

th


GATE QUESTION BANK

CE 1.

2.

(x

)

(x

)

me

e

m

p( defective in c lcul tors)

e e

[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 =

4.

[Ans. C] σ μ

5.

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

∫ x dx

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

7.

[Ans. D] ere μ cm; σ ( x 102)

=  x = 8.969 ≃ 9.0 [Ans. B] Since population is finite, hypergeometric distribution is applicable 25 Calculators

1 Defective

∫ f(x)dx

P=

Equation these two and solving, we get

23 Non-defective

5 Calculators

)

The probability expressed in percentage

=

2 Defective

x

=0 1

=

Now Z =

3.

Mathematics

4 Non-defective

=P. th

th

x

cm /

th


GATE QUESTION BANK

=P( x ) This area is shown below:

[

Mathematics

(

[ -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

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.

[Ans. D] (X )

)

(X

∴∫

( x

)

(X

6

x

)

x x x

[

e

]

15.

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

16.

[Ans. *] Range 0.25 to 0.28 ( t) e (n t) n no of vehicles veh km

)

( )

( x

{

)

)

[Ans. 6] ∫ f(x)dx f(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

( )

12.

∫ d

[Ans. C]

x otherwise

(

)

e

.

/

= 2.e = 0.2707

)dx x7

th

th

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GATE QUESTION BANK

CS 1.

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

2.

permutations of a word consisting of 10 ‘ ’ s nd ’U’s Applying formula of permutation with limited repetitions we get the answer as

[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(P/R)=

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  8   11  8C 11C = 4 5      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  which is choice (D)

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)

[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 =

(

) . /

= =

(

) (

(Binomial formula) )

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

(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 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

Mathematics

th

th

th


GATE QUESTION BANK

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

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) = )

[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

/ = P (z ≥

.z

/ = P (z ≥

(z

) = P (z ≥

) (

)

)

(

8.

)

)

(

)

(

)

P(f ce 10.

= 0.75

= 0.75 ( )

)

( )

f ulty

p

q p

not f ulty

decl red not f ulty decl red f ulty

q

q

σ =3

decl red not f ulty

From above tree (decl red f ulty)

th

=0.468

decl red f ulty

q

11.

=

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

_____(i)

[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

) (

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

(0.5263)

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

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

= 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

pq

(

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

th

th

q)(

p)

)


GATE QUESTION BANK

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

which are multiples

16.

∴ equired rob bility 12.

13.

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

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

15.

[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) (

(

( )

( )

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)

k P is no. of cars per minute travelling. 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 For k = 2 , P(2)= Hence ( ) e

)

( ∪ )

[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) ( )

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

14.

Mathematics

e ( t le st one he d) e (

TT )

( )

( )

e

e

4

5

(

⁄ ) ) e

e

So required prob bility

th

th

th


GATE QUESTION BANK

19.

[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.

20.

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

24.

Mathematics

[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 (max) ximum v lue of y

equired prob bility

(

)

x 21.

ECE 1.

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

[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) 

(function)

p 22.

23.

2.

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

[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( ∪ ∪ ) ( ) ( ) ( ) ( ) ( ( ) ( ) ∴ equired prob bility is ( ∪ ( )

[Ans. A] I

∫ e

)

√ omp ring with

σ√ ut μ

∫

∫ rom x σ σ Put σ

)

(

∫

√

e

(

)

σ

√

dx

dx

e

dx

…

nd x

in

equ tion

e

∪ )

th

th

th


GATE QUESTION BANK

3.

[Ans. C]

Probability of failing in paper 1, when A student has failed in paper 2, P    0.6 B



 P x  .dx  1



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



 ax  Ke .dx  1





or ∫

e dx

Mathematics

e

∫

(

∴

dx

)

( )

= 0.6 0.2 = 0.12

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



( )

7.

[Ans. A] CDF: F  x  



x

  PDF 

dx



4.

[Ans. D] . / ( )

P (Y/D) =

. / ( )

. / ( )

= 5.

For x<0, F  x   . / ( )

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



i



i

8.

[Ans. A] Given: Px  x   Me2|x|  Ne3|x|

i



 P  x  dx  1

i

x

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

 

 Me



2|x|









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

By simplifying 2 M N 1 3

= E[X2] [ ,X-] [Ans. C] Probability of failing in paper 1, P (A) = 0.3 Probability of failing in paper 2, P (B) = 0.2

1 2

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

  xiP xi 

6.

F  0 

conc ve upw rds

0

x



1

2

 xf  x  dx

 x P x    x  x dx

0

x

 

dx

For x>0, F  x   F 0    x  1 dx





  x  1

1

=0.4

E[X] =

x

9.

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

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 11.

Mathematics

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

[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 exp 4

x

∫ x exp 4

√ ,

-

x x

∫ x exp 4

√ [ exp (

24.

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.

[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

4/5 1/5

S

∴ Required probability = R

3.

parcel

Parcel is lost

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)

Parcel is lost

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)

x

|

[Ans. D]

=52 weeks and 2 days are extra. Out of

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= (

̅)

(

̅)

(

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

9.

[Ans. A]

̅)

V (closest answer is 0.2) 4.

8. =5.9 V.

√

S

[Ans. C] ( )

P(x)=

=

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

) ≃

∴ ( ) 12. 7.

(

(

)

nd (

)

[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

= (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

10.

= (x

1.

Consider

)

(D) x

= (x

)

he error in

 xe dx

=

1

f(x)|

(

for a continuous

)

(

)

is 2

3.

CE – 2012 The estimate of ∫ .

1 to an accuracy of at least  106 3

The Newton-Raphson iteration 1 R x n 1   x n   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 .

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.

=

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

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

series

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

)

= (x

the

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

2

(C) x

0) dx

CS – 2007

x

(B) 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) (B) (C) (D) 9.

Mathematics

Only I Only II Both I and II Neither I nor II

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

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GATE QUESTION BANK

Mathematics

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

ECE– 2014 6. The Taylor expansion of is

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

7.

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

8.

The series ∑

(C) X n 1  1  X n  (D) X n 1 

e X n 1  e X n

X2n  e Xn 1  X n   1 X n -e Xn

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 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

sin x

(A) 2

x

x

.

..

(B) 2

x

x

.

..

(C) 2

x

x

.

..

(D) 2

x

x

.

..

2 cos x

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

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GATE QUESTION BANK

Mathematics

EE– 2008 2. Equation e = 0 is required to be solved using ewton’s method with an initial guess x = . Then, after one 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

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

3.

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 ___________

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

)

(x

)

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.

IN– 2006 1. For k = 0 2 . the steps of Newton-Raphson method for solving a non-linear equation is given as 2 5 x k 1  x k  xK 2 . 3 3 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 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) th

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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 ( ) =

(0.5) = .5

=

y = sin ( 2.

[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

)=0

f(x)dx = [(y

∫ y ∫

6.

/

[(0

0)

0.70 0

[(0

2(0.70 0

0.70 0

0=0

y )]

7.

[Ans. C] x y= ∫

x

x

1 1 h

dx = (y

= ( = . 8.

( 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

y

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

= )

2(y

)]

sinx dx =

0.70 0

y )

2(y

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

5.

0.70 0

Trapezoidal rule

= 50 units

3.

0.70 0

2

2

0

55)

2 )]

3

2 y 2

y )

)

[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

h= x y

2(y

iven in question 0 1 1 2 1 0.5

∫ dx = [y x 2

y

.

.

t Δx = 2

t|

Δx = 0.0

0. = 0.

= 2t

t|

To calculate using N-R method Set up the equation as x =

rapezoidal rule y

)dt

[Ans. C]

∫ dx by trapezoidal rule x ∫ f(x)dx = [y

[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

i.e. = a

)]

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.

3.

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

2f

x

x

f )

and α β

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

4.

βγ

α βγ = 5

Now exact value ∫ f(x)dx

=

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

∫ f(x)dx = (1 + 2(4) + 15) = 12

x

x

=

(3 points Trapezoidal Rule) Here h = 1

=∫ (

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

=x

=x

5.

Mathematics

6.

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

th

∑y = 2 ∑x = 14

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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

=

= 0. = 0.7 5

f

2f

0.25(

0.

0.

0.5)

)

2x

x

=

2x

2

[x

x

]

11.

[Ans. D] Exact value of ∫ .

| 2]

.

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 ∫ (x

f ) 2

x

[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 )

2

f

(

10.

[Ans. A] I = h(f

f(x ) f(x )

=x

=x

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

= [( 0

2

4 266

0)dx 2

)

2(2 )

(

)]

= 2 5. The value of integral th

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GATE QUESTION BANK

∫ (x

0) dx = *

=

0

=2

x 5

=

0x+ 3.

.

x

=

[Ans. A] =

+

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)

x

= .5

5.

(b – a) max |f ( )| 1

=

(2 – 1) [e (2 + 2)]

=

e (

)

=

= 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 =(

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,

.

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

=

=

.

= . 07

is number of subintervals max |f ( )|

)

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

max |f ( )|

=

(α

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

α=

Now putting

)

2α =

=x =α

α= +

Where

(x

2α=α+

, x = 0.5

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

= 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  x 0 

x

e x n 1  e x n

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 ( )

4=0

f  x0 

f ' x 0 

x  x  4x0  4 3x02  2x0  4 3 0

e e

[Ans. C]

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

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

2 0

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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( –

.

2 x

as e = EE 1.

Now put x 2

)

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.

=* 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 =

( )

f(x ) = e f’(x ) = e

=

=[

( )

–(

.

[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 =

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

th

th


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.

17.

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

21.

Let

(D) ∫

What is

(C) 1 (D) 1/ln2

ME – 2009 22. The area enclosed between the curves

y 2  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  2y 2  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

(B) √ ⁄ 24.

0 (A) (B)

traversed in a counter-clockwise sense is Y B X

x

The value of (A) (B)

A path AB in the form of one quarter of a circle of unit radius is shown in the figure. 2

x+2y=2

⁄ ⁄

⁄ ⁄

(C) √ (D) 2

Integration of  x  y  on path AB

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

P

at x=2, y=1?

(A) 0 (B) ln2

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

1

18.

20.

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

A

2

(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) (D)

̂

√

37.

̂

̂

√

√

√

has

/

√

̂

√

(C) ̂

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

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

CE – 2012 19. For the parallelogram OPQR shown in the sketch, ̅̅̅̅ ̂ ̂ and ̅̅̅̅ R ̂ .̂ The area of the parallelogram is Q

(C) 1 (D)

24.

π

magnitudes a and b respectively. |⃗ ⃗ | will be equal to (A) – (⃗ ⃗ ) (C) + (⃗ ⃗ ) (B) ab ⃗ ⃗ (D) ab + ⃗ ⃗

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) ⁄

If ⃗ and ⃗ are two arbitrary vectors with

(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

(A) 0 (B) 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)x i 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 (C) (B) 1 (D) Let P=∑

7.

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

4 3 2 curve 3x  16x  24x  37 is (A) 0 (C) 2 (B) 1 (D) 3

∫ (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

.

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

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

dx = π, then the value of k

is equal to_______. 15.

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

e 1  ex monotonically increases monotonically decreases increases to a maximum value and then decreases decreases to a minimum value and then increases

function f  x  

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

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.

2 (A) P  P   P 2 (B)  P   P

th

th

th

(C) 2P      P (D)     P  2P


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   (D) 7.

8.

9.

10.

11.

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

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

13.

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? (A) (C) x 2 x (B) e (D)

lim

sin   / 2

 (A) 0.5 (B) 1 0

is (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 ECE – 2009 18. The Taylor series expansion of sinx at x   is given by x

 x  

2

(A) 1 

 .....

3!

 x  

21.

the value of the integral ∯

 .....

3!

 x  

(A) 3V (B) 5V

2

(C) 1 

3!

 .....

 x  

2

(D) 1  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,

2

(B) 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

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) 31.

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

32.

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

33.

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 =

in the interval

, magnitude

of the gradient at the point (1, 3) is (C) √ (D) ⁄

⁄

(A) √ (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 = ∫

, then S has the value (C) ⁄ (D) 1

⁄ ⁄

(A) (B)

/ is (C) (D)

Mathematics

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.

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

. The

2.

)

( ) ∫ (∫

)

( ) ∫ (∫

)

( ) ∫ (∫

)

(A)

√

(B)

√

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) (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) R 22.

(C)

Given a real-valued continuous function

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.

dx dy is. (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

√

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

+ 2x + 10, the value of

is equal to (A) 0 (B) 4

|

(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) . / (B) .

IN – 2007 9. The value of the integral ∫ ∫

(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


=

[Ans. A]

=

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

[Ans. D] o

h

π h

π o

(

0 o

1 √

0

9.

[Ans. B]

10.

[Ans. B] (

)

For V to be max

Applying L hospital rule (

[Ans. A] (

)

)

This is of the form . /

Hence, h 3.

√

1

)

. /

= |

|

|

|

= = 11.

4.

[Ans. A] (

After changing order ∫ ∫ 5.

[Ans. A] I= ∫ (

)

=2∫

[ ∫

) ( )(⃗ )(⃗ )

]

= 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.

Applying ’ Hospital rule, we get I= [Ans. A] ∫

[Ans. B] √

Given:

I=

8.

[Ans. B] Let the vectors be

(

)

(

)

√

√

√

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

∫√

L =∫ (√

(

(

)

)

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 ̂

(

Let x= a3 ⇒ a=2

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

16.

[Ans. B] L=

)

=1.22 15.

) 1

).dx

= (

)

= 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.

y 2  4x x2  4y

(0,0)

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)

24.

x4   4x 16

[Ans. B] Y

or x4  64x

B

or x(x  64)  0 3

or x3  64 or x  4  y 4

x = cos  y=sin 

 Required area = ∫ .√

Path is x2  y 2  1

/

R  e

4

 2 x3   2  x3 2    3 12 0 

(x  y)2  1  2sin  cos  2

2

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

4 64   (4)3 2  3 12 32 16 16    3 3 3 Alternately For point where both parabolas cut each other

 1 1     1 2 2 2 2 Alternately =

Given: x2  y 2  1 Put x=cos  , and y=sin 

y2  4x, x2  4y

x  y

2

 x  4 4x 2

or x4  64x

∫ (

or x3  64

 x   4,0 ,(4,0)

) 2

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

 Required area 4

x2 dx 0 4

     1 2 

  4x   0

 cos2   sin2   2sin cos  = 1  sin2

or x2  8 x

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. C]

⇒

1

1

differentiable for all x since at

o

29.

[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] o

R, except at

o

,

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

’ h 35.

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

36.

[Ans. A]

[Ans. D] ,

∫

[Ans. D]

R, and

value towards the left and right side of

28.

∫ ( )

32.

2

y is continuous for all x

( )

Standard limit formulae

̂ ̂

-

[Ans. D]

π

)

[Ans. D] If f(x) even function ∫

||

o

π* +

π(

[Ans. D] ⃗ ̂

Mathematics

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

[Ans. B] G

/

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

o

( )

o

o

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

o 46.

|

.

=

’

. /

41.

Mathematics

|

[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

= 126 kmph Where v(t) is the velocity of the rail engine. 3.

∫

= 2y

=∫ 2.

)

)

’ h o I= ∫

)

(

 = xy 

⇒

R

Here I = ∮ ( )

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

4.

[Ans. C] f = 2 +3

(

)⃗

= 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 …( ) =

(

)

=

0

=

.

( α

h

)

√ ) [ from eqn(i)]

=

∫ Then Integrating by parts we get,

(

√

h

(

)

o )1

/

√

(

√

(

(

= dI = Integrating, I = ( )

α o

h

h

)

√

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,

∫ (

= 8h

)

∫ (

Since and y = h at x =

)

* ( ) *

4h … … ( )

(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,

Also since, r=0 (

)

o

1

.

/

= 8 > 0, the point

(

o

|

minimum equal to

√ √

Since ∫ ( ) I=∫

√ √

…( )

√

√

∫ (

)

(

)

|

̅̅̅̅ ̅̅̅̅ R ̅̅̅̅ ̅̅̅̅ R

[Ans. B] Let I = ∫

o -

[Ans. A] Area = |̅ ̅ | ̅̅̅̅ ̅̅̅̅ R (

So the optimal value of f(x, y) is a

16.

)

, 19.

) o

o

(

)

o ()

[Ans. A]

The minimum value is (

o

⇒λ 18.

/

o

, we can use

π

⇒λ

Since, We have either a maxima or minima at .

Mathematics

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

[Ans. B] 1

1  x 

n



n 1  r

r 0

Cr x

r



1  x 

2





r 0

r 0

 r 0

Cr x





r 0

i 0

5.

)

(

[Ans. D] y = 3 – 16 = 12

)

– 24 48

+ 37

– 48 x = 0

x (12 – 48x 48 ) = 0 x = 0 or 12 – 48x – 48 = 0 4x – 4 = 0

∑ () g(i) =i+1

√

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.

k

⇒

(since r is a dummy variable, r can be replaced by i)

(

k

r

   r  1 x r    i  1 x i

)

–1)

)

w h (

2 1  r

  r 1 Cr x r   r 1 C1 x r

(

( …

‘ ’

1

Putting n=2,

2.

)

= Q=∑



(

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.

( (

) )



∫

∫

At At

= ln ( sec ) – ln (sec 0) = ln (√ ) = ln (

. / . /

value 10.

)–0=

[Ans. B] (

8.

)

*

(

(

) [

[Ans. A]

*(

) +

) + .

/

]

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

( ) For x =

7.

gives maximum

value

)-

(

π ]

( )



=,

[Ans. B] f(x) = sin x ⇒ ( ) o ( ) ⇒ o π π π [

π

)

th

th

th


GATE QUESTION BANK

14.

[Ans. 4]

Mathematics

∫

∫

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

∫(

2x

2

x



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 

4.

0 dx  2x 1=0 1  x 2

[Ans. A] o ’ h o ∬(

)

∮

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  

 x  x0  f ' x0   x  x 0 

2



1

2

2

 e  (x  2)(e ) 

 x  2 2

f '' x0 

2 2

For strictly bounded, 0  limy   x 0

 ......

or 0  lim y   x  2

So, y  e x is strictly bounded

 e ...... 2

2    x  2  e2  3  x   ......    2   (Neglecting higher power of x)

7.

9.

10.

lim 0

=

ex  e x ex  e x

x x 2 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  y1  y  2 x  c  x 2  x1  y  x c

x x 2 x3    .......... 1 2 3

ex  e x ex  e 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]

x2 x 4   .......... 2 4  x3 x5 x    .......... 3 5 1

y  x 1 … ( ) 2

2

5 I   ydx    x  1 dx   2.5 2 1 1

1 x (Neglecting x2 and higher order)

or cot h (x)=

(Since at x=1,y=2)

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

0

0





4 I   4x3  10 2x   dx   4x3  160x 4 dx   1

 4x 4 160 5  =   x   33 5  4 0

(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 ’

f  x   f  a    x  a  f ' a  

x  a

2

2!

f '' a 

Q

where, a=

 f  x   f      x    f '    Coefficient of (x )2 is

 x  

2

2!

f ''  

f ''  

P

2

f ''   e  sinx |at x   e



x

∫(

 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.

∬⃗ ⃗

̂

̂

̂

̂

∯

⃗

∭

23.

[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

∮ ⃗ ⃗⃗⃗

( )∭ and is the position vector)

(

⃗⃗⃗ ⃗⃗⃗

has a maximum.

[Ans. D] Apply the divergence theorem

[Ans. C]

= [

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 0(

⇒ ( )

⇒ o

⇒ ( ) Since ( ) is negative, maximum value of f(x) will be where ( )

⇒

)

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, ̂

(

√

√

)̂

(̂

√

√

=3

,

-

/

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 ) ∫ ( ) ∫ ( 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.

.

-

[Ans. D]

̂ ) ̂ )

,

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.

[Ans. A]

13.

Mathematics

[Ans. B] P=∫ = , =, ( = (

1 0

9.

10.

(

∫

[Ans. D] ̅=( )̂ ( ( ) = (0, 2) ( ) = (2, 0) Equation of starting line

[Ans. C] f(x) = ( ) ( ) So the equation f(x) having only maxima at x = 1

16.

[Ans. B] ̂ ̂

=

11.

∫ (

17.

)

C

o

.(

)̂

[Ans. A] ̂ Div ( ) =.

̂

= 1+1+1= 3

̂

)̂ )̂ /

̂

̂

||

/(

̂

̂

̂)

||

( ̂ ( =0

⃗ ̂

(

̂

̂ ̂

‘ ’

(

is undefined

Discontinuous 12.

o ( )

)

But at

̂ ̂

̂

[Ans. D]

[Ans. B] (

̇̂

̂ ̂ ∫ ∫ Along x axis ,y=0,z=0 The integral reduces to zero.

⇒ y = 2 – x , dy = – dx ̅ ̅ =( ( ) Putting ∫̅ ̅ ∫

) =1

15.

)̂

) y = 2 – x and dy = – dx

(

[Ans. B] Dot product of two vectors =1+a+ =0 So orthogonal

)

[Ans. C] ( ) ( ) ( ) ( ) ( )

)

14. √

∫

)-

18.

th

[Ans. A] ( ) o ⇒ M th

) ( ) ̂ (

( ( )

th

) ( ) ̂ (

)

(

) (

)


) )

GATE QUESTION BANK

19.

[Ans. B]

23.

[Ans. A] ( ⃗) ⃗ ⃗ ( ) ( ) ⃗

24.

[Ans. B] ( ) ( ) ⇒ ⇒ )( ⇒( ⇒ ( ) ( ) ( )

∫⃗ ∫

[

(

o

o π

)( )(

∫ ( ∫

⇒

o

) )

[

]

]

π 20.

[Ans. C] ( )

(

)

( )

( ) For number of values of ) o ( ) ( ( ) ( ( )

⁄

)

IN 1.

(

o

h

̂

o ⇒

G

o ∫ .∫

)

Unit vector along y = x is

o

o

√ o

/

/.

√

∫

√

.

) o . /

√

̂ /

√

√

√ √

. / 2.

[Ans. D] Using L Hospital Rule., numerator becomes =

From the graph, distance at π 0 . /1

π

o ̂

/

[Ans. 2] ( ∫

-

⇒

G

22.

,

w

[Ans. A]

[Ans. B]

∫ .∫

)

( ) ( )

( )

M 21.

Mathematics

3.

()

= ( )

[Ans. B] Given integral is, I=∫ Let f(x) = so curve of 1/

th

th

th

will be


GATE QUESTION BANK f(x)

Mathematics

( )

(

)

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) ̂ ̅ |R( ) =K (constant)

⇒ ⇒

i.e., (t) + (t) + (t) = constant. On analyzing the given (A) option, we find ̅(t) that R

̅( )

[Ans. C] Given : f= + where,

⇒

√ √

G o

…… + (i=0 to n) are constant. +(n 1)

…… ⇒

=0+

+

(n 1)

⇒

+n +

+

[Ans. B] ( )

(

)

(

8.

[Ans. B] ()

(

)

(

)

)

⇒ ⇒

( )

∫

∫

When ( )

(

…( )

Differentiating the above equation () , ()

When ( )

√

⇒

-

When ( )

√

……

+

6.

)

will give constant magnitude,

+

= , = nf

(

1

= and

)

o

so first differentiation of the integration will be zero. 5.

o

( ) -

) From equation (i)

And when th

th

th


GATE QUESTION BANK

()

()

()

⇒

()

()

This is Leibnitz linear equation Integrating factor I.F = ∫ the solution is ()

15.

[Ans. C] By definition

16.

[Ans. A] Unit vector=

and

∫

Mathematics

17.

=xi+yj+zk

[Ans. C] R: Y

( )

,

1 1

- o

9.

[Ans. D]

10.

[Ans. A] This is a standard question of differentiability & continuity

11.

Area =

[Ans. C] y= =(

=

).(cos x + sin x) = 0

Or x = y will be maximum at x = =

=

[Ans. C] y(2) = y(5) =

√

( )

=

18.

y=

19.

coulomb.

[Ans. B] We know that ∫

() (

∫

.

) π

14.

π . /

/

........ ( ) ( )

[Ans. C] y= y= [Ans. B] Given y = x2 + 2x + 10

20.

[Ans. B] (

)

For a G.P to converge (

⇒ 21.

= 2x + 2 |

( )wh

[Ans. B] Expansion of sin x

In a G.P 13.

X

Total charge = σ

⇒ tan x = 1

12.

+1

22.

(

⇒

)

⇒ [Ans. D] .E=0 is not irrational (it is solenoidal) [Ans. 1] From Gauss divergence theorem, we have ∫ ̅

th

)

̅ th

∫

̅

∫ th

̅


GATE QUESTION BANK

∫.

/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

7.

what is y (A) e (B) 1

2.

3.

then

(D) degree 2 order 2 ME – 2007 The solution of

⁄ ⁄

(C) (D)

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

For

+4

+ 3y =

, the particular

integral is: (C) (D)

(A) (B) 5.

6.

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.

(B) (1+ x)

(D) (1

)

(

)= 0 has

x

with the

s

(A) y

(C) y

(B) y

(D) y

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

x)

The partial differential equation

y

ME – 2010

+

(C) (1 x)

The solution of x condition y

11.

(A) (1+ x)

(D) 2  x  2

(B)   x  1

The solution of the differential equation 2 dy  2xy  e x with y (0) = 1 is: dx

(

with initial value

y (0) = 1 is bounded in the interval (C) x  1,x  1 (A)   x  

ME – 2006 4.

y

ME – 2011 12. Consider

the

differential

equation

y x. The general solution with

(A) degree 1 order 2 (B) degree 1 order 1 (C) degree 2 order 1

constant c is (A) y

th

th

t n

t n

th


GATE QUESTION BANK

(B) y

t n (

(C) y

t n ( )

(D) y

t n(

17.

x

)

the

x

differential y

with the boundary

(B) s n ( )

(D)

u (A) (B) (C) (D) 15.

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

t t

y and

x ,y-

*

x + ,y-

x ,y-

*

x + ,y-

*

x + ,y-

*

x + ,y-

x ,yt x ,yt

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

(

x

y is

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 20.

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

__________

The wronskian W(t) =|

where k is a constant,

(

t

The general solution of the differential

x

The solution to the differential equation

(B) u

with

s n( ) 19.

partial

at x

equation

equation

conditions of y(0) =0 and y(1) = 1. The complete solution of the differential equation is (A) x (C) s n( )

ME – 2013 14. The

If y = f(x) is the solution of the boundary conditions y

18.

ME – 2012 13. Consider x

)

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

th

(A)

+ (1 n)pv = (1 n)q

(B)

+ (1 n)pv = (1+n)q

(C)

+ (1+n)pv = (1 n)q

(D)

+ (1+n)pv = (1+n)q

th

th


GATE QUESTION BANK

2.

The solution of

7.

y in the range (A) (B)

y x

( os x ( os x

(C)

( os x

(D)

( os x

,

xy

x

= x y with the condition that y = 1 at x = 0 is

is given by

(B) In(y) =

s n x) s n x)

8.

s n x)

given that at x = 1,

(A)

(C)

(B)

(D)


(D) y =

CE – 2008 9.

The general solution of (A) (B) (C) (D)

10.

+ y = 0 is

y = P cos x + Q sin x y = P cos x y = P sin x y=Psn x

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

= 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

+4

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

y = 0 is

5.

(C) In(y) =

(A) y =

s n x)

The solution of the differential equation x

The solution for the differential equation

( )

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 4.

Mathematics

+ 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

CE – 2007 6. The degree of the differential equation

y

= 0 are

(C) 3 and 3 (D) 3 and 1

+ 2x = 0 is (A) 0 (B) 1

(C) 2 (D) 3

13.

The solution to the ordinary differential equation (A) y = (B) y = (C) y = (D) y =

th

th

+ + + + +

6y = 0 is

th


GATE QUESTION BANK

14.

The partial differential equation that can be formed from z = ax + by + ab has the form (w t p (A) (B) (C) (D)

n q

)

Mathematics

ECE – 2006

3.

For the differential equation

d2 y  k2y  0 dx2

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

z = px + qy z = px + pq z = px + qy + pq z = qx + pq

CE – 2011 15. The solution of the differential equation + = x, with the condition that y = 1 at

y

∑

y

∑

y

∑

os

m x

x = 1, is (A) y =

+

(C) y = +

(B) y = +

(D) y =

+

CE – 2012 16. The solution of the ordinary differential equation

y=0 for the boundary

ECE – 2007 4. The solution of the differential equation d2 y  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

condition, y=5 at x = 1 is (A) y (C) y (B) y (D) y CE – 2014 17. The

integrating

for

the

equation

k2

differential s

(A) (B)

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

(C) (D)

ECE – 2005 1. A solution of the following differential equation is given by

2x 3x (C) y  e  e

2x 3x (B) y  e  e

2x 3x (D) y  e  e

The following differential equation has

y y y y

xp( x⁄ ) y xp x⁄ y s n x⁄ y xp x⁄ y

(A) t (B) x t

x t

x t

(C) x t (D) x t

t t

ECE – 2009 6. The order of the differential equation 3

3

d2 y  dy  4 t    y  e is dt 2  dt 

d2 y  dy  2  4   y 2 x 2 dt  dt 

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

y y y y

differential equation

2x 3x (A) y  e  e

3

y y y y

ECE – 2008 5. Which of the following is a solution to the

d2 y dy  5  6y  0 2 dx dx

2.

x

(A) 1 (B) 2

degree=2, order=1 degree=1, order=2 degree=4, order=3 degree=2, order=3 th

th

(C) 3 (D) 4

th


GATE QUESTION BANK

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 2. Straight Lines dy y  Q. dx x 3. Hyperbolas dy x R.  dx y S. (A) (B) (C) (D)

dy x  dx y

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

x with the initial condition

A function n x satisfies the differential equation

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 ECE– 2011 10. The solution of the differential equation y y (A) x (B) x

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 | l v lu o

(A) (B)

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

9.

ECE\EE\IN – 2012 11. With initial condition x(1) = 0.5, the solution of the differential equation,

num r

ECE – 2010 8. Consider a differential equation y x

Mathematics

is (C) y (D) y

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) ECE – 2014 14. If the characteristic equation of the differential equation y

has two equal roots,

t n t v lu s o (A) ±1 (B) 0,0 th

th

r (C) ±j (D) ±1/2 th


GATE QUESTION BANK

15.

16.

17.

18.

Which ONE of the following is a linear non-homogeneous differential equation, where x and y are the independent and dependent variables respectively? (A)

xy

(C)

(B)

xy

(D)

If z

EE – 2011 3. With K as a constant, the possible solution for the first order differential equation is

xy

(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) With initial values y(0) = y (0) = 1, the solution of the differential equation tx

s

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 – 2013 4. A function y x x is defined over an open interval x = (1,2). At least at one point in this interval , (A) 20 (B) 25

x (A) (B) (C) (D)

n

6.

x x x x

t t t t

, the solution is

(C) 30 (D) 35

(B) s n t

os t

(C) s n t

os t

(D) os t

t

Consider

the

x

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

with initial conditions | t

is exactly

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

EE – 2010 2. For the differential equation x

(C) (D)

(A) (B)

xy ln xy then

y

Mathematics

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 (B) C1e2x + 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 d2 y 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

xy x

lnx x

y

lnx x

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 Since, y (1 + x) e

y 6.

,c=1  x2

[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] v n y

y x

y x

y

nt gr t ng y nx

x

y

x

∫ y

∫

and x

[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

Given condition y m ns t x

y

₂

y

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

[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 x x

x

yx

r or yx

9.

∫

x

olut on y x x

8.

n

x

x

x x

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 y

5.

–x

8. x

[Ans. B] =

x

y +1=0

θ θ0) (Newton’s law of cooling)

θ θ θ

t =∫

∫

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 y  0.25hy y +y =0 Putting k = 0 in above equation 0.25h y y +y =0 Since, y = 1 and h = 1

0C

= Now at t = 30 minutes Θ

±√

y =2

6.

y x

y

[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 y y y

0.25 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

x

17.

m1 = 

m x

1 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] d2 y dy  5  6y  0 2 dx dx

A.E. is D2  5D  6  0  D=2,3 2x 3x Hence, solution is y  e  e

2.

sn

[Ans. D] k2D2y= y y2 y 2  2 1 D  2 y  2 k  k 

[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

3

d2 y  dy  2  4   y 2 x dt dt 2  

So, x  t   ke3t , Hence x  t   2e3t is one solution (for

Order of highest derivative=2 Hence, most appropriate answer is (B) 3.

[Ans. A] Given, Differential equation,

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]

d2 y  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

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.

12. x

y

[Ans. D]

t ∫t t

xt

t

[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

The solution is, n x

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

[Ans. C]

y t

n x m

Auxiliary equation m olut on n x Since, n

Given y ln y When y y

±



Since, n  must be zero) Therefore

10.

∫

xt

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

and x

(hence

y

t t

x

y t

y

t

t

t

t t t

y y t | t th

th

th


GATE QUESTION BANK

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.

15.

[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

17.

18.

EE 1.

xy ln xy z x

ox

xy

z y

y

[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. *] 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 i.e.

.

…

x (t) x

[Ans. A]

t xy

is a first order linear

equation non xy

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 ∫

omog n ous

0 is a first order linear

equation (homogeneous) r non l n r qu t ons 16.

z y

Mathematics

[Ans. C] z xy ln xy z y ln xy xy y y ln xy y x xy z x ln xy xy x x ln xy x y xy z ox xy ln xy xy x

2.

[Ans. B] x x t t th

th

x th


GATE QUESTION BANK

Auxiliary equation m m m m (m+4)(m+2)=0 m= 2, 4 x(t) = x(0) = 1 1=

m

w subst tut y t

IN 1.

… (1)

 … On solving (1) & (2), we get

[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

sn t

2.

[Ans. C] x y xy y y x

x

s

[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 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]

6.

[Ans. A] Given

6.

s ts

)

m m m Since there is double root at 2, so general solution of the given differential equation would be +

and

y

5.

x

x

x

Integrate on both sides

4.

y

x(

y

[Ans. A] y

x

x

[Ans. C]

x(t)= 2 3.

nx n



|

Mathematics

= 1 + y2

Integrating ∫

y y x

x

=∫ x

Or t n y = x + c Or y = t n x th

th

th


GATE QUESTION BANK

7.

[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

√ 11.

[Ans. A] Given partial differential equation is x

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

9.

± os x

ẏ ẏ

y

8.

[Ans. D] y t ÿ t

y y

y=

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 Hyperbolic if El ps Compare the given differential equation with standard from A = 1, B = 0, C = 0 Parabolic

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

x  y 

2

x

2 2

y 2

2

k

 k

y (C)

2

 x2 2

x  y

 k

2

(D)

2

k

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

f(z) = u(x, y) +iv(x, y) where i = 1 . If u = xy, the expression of v should be

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| |

+ +

z z

z being a complex variable. The value of I will be (A) I = 0: singularities set = ϕ (B) I = 0: singularities set =,

πn

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

-

7.

z

+

i i

ECE – 2006 1. The value

of

∮|

2.

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

π/2: singul riti s s t { nπ n 2 } (D) None of the above 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)

(C) I

∮

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

dz

2

(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 1 f z  at z=2 is 2 2  z  2  z  2 

5.

(A)

(C)

(B)

(D)

ECE – 2014 11. C is a closed path in the z-plane given by |z|=3. The value of the integral ∮(

12.

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

+

is given by (A) 2π (B) 2π +

z

, then ∮

+

ECE – 2010 7. The residues of a complex function z and 1

(B)

and

(C)

where C is the

∮

contour |z-i/2| = 1 is (A) 2πi (C) t n z (B) π (D) πi t n z EE – 2011 2. A point z has been plotted in the complex plane, as shown in figure below. nit ir l

(D)

z

and and

ECE – 2011 8. The value of the integral ∮ where is the circle |z| is given by (A) 0 (C) 4/5 (B) 1/10 (D) 1

Given f (z)

nit ir l

lm

z

nit ir l

lm

nit ir l

y lmlm

nit ir l

y

. If C is a

counterclockwise path in the z – plane such that |z+1| =1, the value of (A) (B)

lm

y

ECE\EE\IN – 2012 9. If x = √ then the value of x is ⁄ (C) x (A) ⁄ (D) 1 (B)

∮

+ j2 j2

at its poles are

(A)

10.

(C) 4π (D) 4π

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

The value of

z

(C) 2πj (D) 2πj

+ j2 j2

EE – 2007 1.

If f(z) =

) z is

(A) 4π (B) 4π

ECE – 2009 6.

Mathematics

y

z z is 2

(C) (D) 2 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 (

)

6.

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 – 2006 3. The value of the integral of the complex function

) + 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 the counter clockwise

Mathematics

f(s)

3s  4 (s  1)(s  2)

Along the path s  3 is (A) 2j (B) 4j

(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.

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

m gin ry xis

m gin ry xis The value of ∮

j

l xis

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 (D)

√

)

x

IN – 2009

+j

√ y

pl n

j

√

(Note:

j

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

5.

[Ans. A] +i 2 2 i + 2i

6.



7.

∫y x

∮ ∮ 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

 (x2  y 2  2ixy)    m  i  2   or v 

i

x y

y = r sin x y r os

or

Integrating, w  i

i

[Ans. C]

u u i x y

dw  y  ix dz Replacing x by z and y by 0, we get dw  0  iz dz Where, z = x + iy dw = izdz

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 dw u v  i dz x x

+ 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

∫

2

th

th

∫

os 2πz 2z z *

+ *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 )

]

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.

x +

x

x

y

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 ∮

z z +4 |z j| 2 2j  I  2j  2j 2 2.

[Ans. B] iv n

log

1 y  u  iv  loge  x  iy   log x2  y 2  i tan 1   2 x



x v + x 2 i.e. v = 3x2 + f(x) iii Now applying equation (iii) we get u v y x [ x+

[Ans. D] Given ,

∫ x y

y

Mathematics



Since, u is constant, therefore 1 log x2  y 2  c 2 x +y Which is represented set of concentric circles.



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 1 1 lim  s  1 2  S 1 S 1 2



n

th

∮

th

s



s

th

2πj ( ) 2

πj


GATE QUESTION BANK

4.

[Ans. A] Residue of z=2 is d 2 lim  z  2 f  z   z 2 dz 

7.

)

X(z) = Poles are Z= 0, Z =1, Z=2 Residue at Z=0 is lim Residue at Z =1 is lim Residue z =2 is lim

2 i )

2 i

8.

[Ans. A] z+4 ∮ z + 4z +

2 im 2 i

2 i

m

i

iz

log

2 i

√

4

2 2 i + i√ 2

√ 2 √

i

z

i √

i

i

i

√

i√

9.

log i + log( √ π iz log + i ( 2nπ) 2 +log √ π iz i ( 2nπ) + log 2 π z ( 2nπ) ilog( 2

i

)

∮ ∮(

+

+ z

x

)

10.

z+

z F z 2 π j r si u o Residue at z = 0 ( 2- order )

i log i π i i 2

x log x i log π 2

[Ans. C] z z

∫ 2πj

∫

z

*∫

∫

z+

z

z+

where f (z) =1

11.

z

log y

⟹y

√

z +

x

√ ty log y

( 2 infinite number of complex solutions sin z has infinite no. of complex solutions [Ans. D] f(z) = + + z z z

z + 4z + z+2 + 2 j will be outside the unit circle o th t int gr tion v lu is ‘z ro’

⟹ log y

√

z

[Ans. A]

i

iz

6.

+

[Ans. C]

2 i

m

)

z

+ F(z 2πj

2i

ut m m

z+

+

[Ans. D] sin z

(

+

(z

z

d 1  2 1   lim lim   2 3 z 2 dz  z  2  32  z  2   z  2 

5.

Mathematics

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

+

+

wh n

i

√2 wh n

+ o

+i in

qu

r nt wh n

| | √ in

2

√(

) +(

+

+

√ √

2

i

+

√

Mathematics

+

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

5.

x

π π os ( ) + i sin ( ) 4 4

[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

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) y(n) dz

2 + i

By Formula, y y ( ( Since, both n n contour, ) )

xy

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

3

 z  a  2 j

the poles are enclosed by

therefore Value of integral=2πj + 2 2πj 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.

x

j

x +

x +

j

|

2πj

2

+

[sum o r si

2 o pol

4πj

∮

⁄

z

j j x +

lim { x +

j ption

⟹z j j ⟹ 2[ j

∮( + + + + ) 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

z

[Ans. C] 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

ME – 2014 6. Laplace transform of

is

.

The Laplace transform of

t

 f (t) dt is 0

(A)

F (s)

(B)

F (s) f (0)

(C) sF (s) – f (0) (D)

ME – 2009 2.

The inverse Laplace transform of is (A) (B)

s

1 2

s



(C) 1 – (D)

ME – 2010 3. The Laplace transform of a function . The function

(B) 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)

ME – 2013 5. The function equation

satisfies the differential and the auxiliary

conditions, Laplace transform of

. The is given by

(A)

(C)

(B)

(D)

CE – 2009 1. Laplace transfrm of the function f(x) = cosh(ax) is (A) (C) (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 3.

Consider the matrix P = *

+ . The

value of eP is (A) *

+

(B) [

]

(C) [

]

(D) [

th

th

]

th


GATE QUESTION BANK

ECE - 2010 4. The trigonometric Fourier series for the waveform f(t) shown below contains

Mathematics

The numerical value of (A) (B)

|

is

(C) (D)

ECE – 2013 9. A system is described by the differential equation

=x(t). Let x(t)

be a rectangular pulse given by ,

(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.

Assuming that y(0) = 0 and the Laplace transform of y(t) is

Given [

]

then the value of K is (A) 1 (C) 3 (B) 2 (D) 4

10.

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

|

|

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) th

th

th


GATE QUESTION BANK

EE – 2014 13.

14.

Let

be

the

transform of signal x(t). Then, (A) 0 (C) 5 (B) 3 (D) 21

Laplace is

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    (s2  s) s(s  1) s (s  1) (

)

( )

(

[

)

(

) (

[ 3.

)

]

)

]

[Ans. A] [

]

6.

[Ans. D] [

Matching coefficient of in numerator we get,

[

]

s and constant CE 1.

4.

(

]

[Ans. B] It is the standard result that L (cosh at) =

ECE 1.

[Ans. A] [

[Ans. D] {

]

} {

}

th

th

th


GATE QUESTION BANK

2.

3.

[Ans. A] ̇ (t) + 2x (t) = (t) Taking Laplace transform of both sides , we get sX(s) X(0) + 2X(s) = 1 1 X(s) = s2 From Inverse Laplace transform gives, we get [ ] [Ans. D ] eP= [

[

Therefore, the trigonometric Fourier series for the waveform f(t) contains only cosine terms and a negative value of the dc component. 5.

[Ans. D]

[

]

[

0 1  and P =    2  3



]

]

]

[

s 1  =   2 s+3

Where

Mathematics

]

1

s  3 1   s  1 s  2 2 s 1

s 3    s  1 s  2 =   2    s  1 s  2

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  

=[ 4.

⁄

]

=2

[Ans. C] Since f(t) is an even function, its trigonometric Fourier series contains only cosine terms ∫ *∫ [

7.

∫ ∫ (

[Ans. D]

+

t

)]

th

th

th


GATE QUESTION BANK

8.

Mathematics

[Ans. D] For t = Taking Laplace transform on both the sides. We have,

+ sin 11.

√

[Ans. D] By Laplace transform property,

[

]

[

(

]

[

9.

] [

|

]

12.

[Ans. B] Writing in terms of Laplace transform

(

)

[Ans. B]

=5 By taking Laplace transform ( ⁄

X(s) =

13.

[Ans. B]

(

)

( ( 14. 10.

)

[

)

]

(

)

)

[Ans. 0.5]

[Ans. D]

∫

∫

|

Value of constant term = 0.5

th

th

th


. GATE QUESTION BANK

Fluid Mechanics

Pressure and its Measurement CE– 2005 1. The reading of differential manometer of a Venturimeter, placed at to the horizontal is 11cm. If the venturimeter is turned to horizontal position, the manometer reading will be (A) Zero (C) 11cm (D) 11√ cm (B) cm √

Answer Keys & Explanations 1.

[Ans. C] The manometer reading of a venturimeter does not depend upon the inclination of venturimeter with the horizontal.

th

th

th



Fluid Mechanics

Hydrostatic Forces on Plane Surfaces CE – 2005 1. Cross – section of an object (having same section normal to the paper) submerged into a fluid consists of a square of sides 2 m and triangle as shown in the figure. The object is hinged at point P that is one meter below the fluid free surface. If the object is to be kept in the position as shown in the figure, the value of ‘x’ should be 2m

x

CE– 2014 4. Three rigid buckets, shown as in the figures (1), (2) and (3), are of identical heights and base areas. Further, assume that each of these buckets have negligible mass and are full of water. The weights of water in these buckets are denoted as d respectively. Also, let the force of water on the base of the bucket be denoted as F F d F respectively. The option giving an accurate description of the system physics is

1m 2m

(A) 2√ (B) 4√

p

h

(C) 4 m (D) 8 m

h

h

ALL THREE BUCKETS HAVE THE SAME BASE AREA (1)

CE– 2009 2. Water ( ) flows with a flow rate of 0.3 /sec through a pipe AB of 10m length and of uniform cross – section. The end ‘B’ is above and ‘A’ and the pipe makes an angle of to the horizontal for a pressureof 12 kN/ at the end B, the corresponding pressure at the end A is (A) 12.0 kN/ (C) 56.4 kN/ (B) 17.0 kN/ (D) 61.4 kN/

(3)

(2)

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

dF dF dF dF

F F F F

F F F F

CE– 2012 3. If a small concrete cube is submerged deep in still water in such a way that the pressure exerted on all faces of the cube is p, then the maximum shear stress developed inside the cube is (A) 0 (C) p (D) 2p (B)

th

th

th



Fluid Mechanics


[Ans. A] The vertical force on the surface bounded by square and triangle would be respectively, ( ) …… F for squ re edge d ti g t fro ) F x( … … for i li ed edge of triangle and acting at x/3 from P Taking moment of both the forces about P, we get x F F x ( ) ) x( x x

2.

3.

[Ans. A] P

P P

Maximum shear stress,

4.

[Ans. D]

Force on base of Bucket, F hA Base area of all buckets is same So, F F F Weight of water, i e o

√ m

[Ans. D] Applyi g Ber oulli’s equ tio betwee A and B, we get B

10m z

A

300

z = 10 sin 300 z = 5 cm

p

v

p

g    

p p p p

v

z g =p + z = 12 + 9.879 x 5 = 12 + 49.4 = 61.4 kN/m2

th

th

th



Fluid Mechanics

Kinematics of Flow CE– 2005 1. An inert tracer is Injected continuously from a point in an unsteady flow field. The locus of locations of all the tracer particles at an instance of time represents (A) Streamline (B) Pathline (C) Stream tube (D) Streakline 2.

A stream function is given by: Ψ x y + (x + 1) y The flow rate across a line joining point A(3,0) and B(0,2) is (A) 0.4 units (C) 4 units (B) 1.1 units (D) 5 units

3.

The ir ul tio ‘G’ rou d ir le of radius 2 units for the velocity field u = 2x + 3y and v = 2y is (A) u its (C) u its (B) u its (D) u its

6.

A particle moves along a curve whose parametric equations are: x t t y e 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 s ) at t = 0 is __________

CE– 2006 4. The velocity field for Flow is given by: ⃗ = (5x + 6y + 7z) ̂ + (6x + 5y + 9z) ̂ + ( x y λz) ̂ and the density varies as ρ ρ exp( 2t). In order that the mass is overed the v lue of λ should be (A) 12 (C) 8 (B) 10 (D) 10 CE– 2014 5. A plane flow has velocity components, u v= and w = 0 along x, y and z direction respectively, where T ( ) and T( ) are constants having the dimension of time. The given flow is incompressible if (A) T T (C) T (B) T

(D) T

T

th

th

th



Fluid Mechanics


[Ans. D]

2.

[Ans. C] The rate of flow is the difference in value of stream function at points A and B | ∴Q = | = 2 × ( ) × 0 + (3 +1) × ( ) = 0 = 2 × ( ) × 2 + (0 +1) × ( ) = 4 | Q | = 4 units

3.

4.

[Ans. 12] x t t y e z si ( t) dx t dt d x t dt dy e ( dt e d dt dz os ( dt d z dt ⃗ ̂ ̂

[Ans. B] Circulation = Vorticity × Area v u ( ) Are x y ( ) ( ) u its

)

e t) si

t

̂

̂ ⃗ tt ̂ ̂ ⃗ ̂ Magnitude of acceleration at t = 0 s

[Ans. C] The continuity equation is given by (ρu) (ρ ) (ρw) ρ ow ρ ρ e ρ ρ e ρ t u v w ρ ρ ρ ρ ρ x y z ∴ ρ ρ ρ λρ λ λ

5.

6.

λ

[Ans. D] For a flow to exist u v x y T T

T T

th

th

th



Fluid Mechanics

Fluid Dynamics CE– 2006 1. The necessary and sufficient condition for surf e to be lled s ‘free surf e’ is (A) No stress should be acting on it (B) Tensile stress acting on it must be zero (C) Shear stress acting on it must be zero (D) No point on it should be under any stress

CE– 2014 5. An incompressible homogeneous fluid is flowing steadily in a variable diameter pipe having the large and small diameters as 15 cm and 5 cm, respectively. If the velocity at a section at the 15 cm diameter portion of the pipe is 2.5 m/s, the velocity of the fluid (in m/s) at a section falling in 5 cm portion of the pipe is ___________

CE– 2007 2. At two points 1 and 2 in a pipeline the velocities are V and 2V, respectively. Both the points are at the same elevation. The fluid de sity is ρ The flow be assumed to be incompressible, inviscid, steady and irrotational. The difference in pressures and at points 1 and 2 is (A) ρ (C) ρ (B) ρ (D) ρ

6.

A venturimeter, having a diameter of 7.5 cm at the throat and 15 cm at the enlarged end, is installed in a horizontal pipeline of 15 cm diameter. The pipe carries an incompressible fluid at a steady rate of 30 litres per second. The difference of pressure head measured in terms of the moving fluid in between the enlarged and the throat of the venturimeter is observed to be 2.45 m. Taking the acceleration due to gravity as 9.81 s , the coefficient of discharge of the venturimeter (correct up to two places of decimal) is ______________

7.

Group I lists a few devices while Group II provides information about their uses. Match the devices with the corresponding use. Group I Group II P. Anemometer 1. Capillary potential of soil water Q. Hygrometer 2. Fluid velocity at a specific point in the flow stream P. Pitot Tube 3. water vapour content of air S. Tensiometer 4. Wind speed (A) (B) (C) (D)

CE– 2011 3. For a body completely submerged in a fluid, the centre of gravity (G) and centre of Buoyancy (O) are known. The body is considered to be in stable equilibrium is (A) O does not coincide with the centre of mass of the displaced fluid (B) G coincides with the centre of mass of the displaced fluid (C) O lies below G (D) O lies above G CE– 2013 4. For a two dimensional flow field, the stre fu tio is give s =

(

) . The magnitude of

discharge occurring between the stream lines passing through points (0,3) and (3,4) is : (A) 6 (C) 1.5 (B) 3 (D) 2

th

th

th



8.

Fluid Mechanics

A venturimeter having a throat diameter of 0.1 m is used to estimate the flow rate of a horizontal pipe having a diameter of 0.2 m. For an observed pressure difference of 2 m of water head and coefficient of discharge equal to unity, assuming that the energy losses are negligible, the flow rate (in /s) through the pipe is approximately equal to (A) 0.500 (C) 0.050 (B) 0.150 (D) 0.015


[Ans. C]

2.

[Ans. B] Applying Ber oulli’s equ tio we get +

( )

+z=

+

(

)

FB O

Balloon

G W

+z

G

FB

W

= = ρ

3.

Fig (ii)

Fig (i)

ρ

he vily lo ded so th t its ‘G’ is lower th its ‘O’ s show i Figure (i) let weight of balloon acting down and FB is buoyant force action vertically up. For equilibrium of the balloon FB = W. if the balloon is distributed in clockwise dire tio s show i figure (ii) the ‘FB’ d ‘ ’ o stitute ouple ti g i the anti – clock wise direction and brings the system in the original position. Conclusion: When FB =W and centre of buoyancy (O) is above centre of gravity (G), the body completely submerged in a fluid is said to be in stable equilibrium.

[Ans. D] A submerged body is said to be in stable equilibrium, if it comes back to its original position after a slight disturbance. The relation position of the centre of gravity (G) and centre of buoyancy (O) of a determines the stability of a sub – merged body. The position of G and O in case of completely sub – merged body are fixed. Figure below shows a balloon submerged in air with bottom portion.

4.

[Ans. B] Ψ Ψ (

)

(

)

u its th

th

th



5.

Fluid Mechanics

[Ans. *] (Range 22 to 23) Using equation of continuity A A

s 6.

[Ans. *] Range 0.93 to 0.96

l⁄ s

D

D

C

√ gh

√ (

)

(

C

)

)

√(

(

)

√ C 7.

[Ans. D]

8.

[Ans. C] Given Dia of throat d = 0.1 m Dia of pipe D = 0.2 m Pressure difference Coefficient of discharge C √

Discharge A

(

√

) and A

(

) √

√(

)

(

)

se

th

th

th



Fluid Mechanics

Flow Through Pipes CE– 2012 1. The circular water pipes shown in the sketch are flowing full. The velocity of flow (i s) i the br h pipe “ ” is dia = 4m P Q V = 6 m/s

V = 5 m/s R V= ?

(A) 3 (B) 4

dia = 2 m

(C) 5 (D) 6

CE– 2013 2. A 2 km long pipe of 0.2 m diameter connects two reservoirs. The difference between water levels in the reservoirs is 8 m. The Darcy – Weisbachfriction factor of the pipe is 0.04. Accounting for frictional, entry and exit losses, the velocity in the pipe (in m/s) is: (A) 0.63 (C) 2.52 (B) 0.35 (D) 1.25

CE– 2014 3. An incompressible fluid is flowing at a steady rate in a horizontal pipe. From a section, the pipe divides into two horizontal parallel pipes of diameters d d d (where d d ) that run for a distance of L each and then again join back to a pipe of the original size. For both the parallel pipes, assume the head loss due to friction only and the Darcy Weisbach friction factor to be the same. The velocity ratio between the bigger and the smaller branched pipes is _________ 4.

A straight 100 m long raw water gravity main is to carry water from an intake structure to the jack well of a water treatment plant. The required flow through this water main is 0.21 s. Allowable velocity through the main is 0.75 m/s. assume f = 0.01, g = 9.81 m/ . The minimum gradient (in cm/100 m length) to be given to this gravity main so that the required amount of water flows without any difficulty is _________

th

th

th



Fluid Mechanics


[Ans. B]

3.

[Ans. 2] d

4 mm

4 mm V = 5 m/s

V = 6 m/s

A

d

B

d

d

Since pipes are in parallel So, head loss will be same f f d g d g d d d d

2 mm

A

A

A

( )

( )

( )

s ∴ 2.

s

4.

[Ans. *] Range 4.7 to 4.9 s Allowable velocity = 0.75 m/s f = 0.01 g = 9.81 d

[Ans. A] Applyi g Ber oulli’s theore t e try d exit. p p losses ρ g ρ g Hence both reservoirs are at atmospheric pressure, and mean velocity is same at entry and exit. | | h fl | | g g gd

d flv gd

i i u g

g

(

h

gr die t

Hence, answer is 4.8.

)

h l

g Upon solving u s

th

th

th



Fluid Mechanics

Impulse Momentum Equation and Its Application CE– 2005 1. A tank and a deflector are placed on a frictionless trolley. The tank issues water jet (mass density of water = 1000 kg/m3), which strikes the deflector and turns by 45°. if the velocity of jet leaving the deflector s 4 m/s and discharge is 0.1 /s, the force recorded by the spring will be

CE– 2007 2. A horizontal water jet with a velocity of 10m/s and cross sectional area of 10 mm2 strikes a flat plate held normal to the flow direction. The density of water is 1000 kg/ . The total force on the plate due to the jet is (A) 100 N (C) 1 N (B) 10 N (D) 0.1 N

Spring

Tank

Jet of water

Deflector Trolley

(A) 100N (B) 100 √ N

(C) 200 N (D) 200 √ N


[Ans. D] Force in spring will be the force in horizontal direction. F ρ os θ os √

2.

[Ans. C] The force is given by F ρ F x x F

x(

)

√

th

th

th



Fluid Mechanics

Flow through Orifices and Mouth Pieces CE– 2010 1. Match List – 1( Devices) with List – II (Uses) and select the answer using the codes given below the lists: List – I A. Pitot tube B. Manometer C. Venturimeter D. Anemometer List – II 1. Measuring pressure in a pipe 2. Measuring velocity of flow in a pipe 3. Measuring air and gas velocity 4. Measuring discharge in a pipe Codes: A B C D (A) 1 2 4 3 (B) 2 1 3 4 (C) 2 1 4 3 (D) 4 1 3 2


[Ans. C]

th

th

th



Fluid Mechanics

Boundary Layer Flow CE– 2006 1. The thickness of the laminar boundary layer on a flat plate at a point A is 2cm and at a point B, 1m downstreams of A, is 3cm. What is the distance of A from the leading edge of the plate? (A) 0.50 m (C) 1.00 m (B) 0.80 m (D) 1.25m

CE– 2008 2. The flow of water (mass density = 1000kg/ and kinematic viscosity = /s) in a commercial pipe, having equivalent roughness as 0.12 mm, yields an average shear stress at the pipe boundary = 600 N/ . The value of δ’ (δ’ bei g the thi ess of l i r sub-layer) for this pipe is (A) 0.25 (C) 6.0 (B) 0.50 (D) 8.0


[Ans. B] Laminar boundary layer thickness,

2.

[Ans. D] We know that v δ

√

δ δ δ

But

√

√

x

√

ρ v

∴δ √

x

√

δ x

δ

x

∴

th

δ

th

th



Fluid Mechanics

Viscous Flow CE– 2006 1. The flow of glycerin (kinematic viscosity v=5× s) in an open channel is to be modeled in a laboratory flume using water (v = /s) as the flowing fluid. If both gravity and viscosity are important. What should be the length scale (i.e ratio of prototype to model dimensions) for maintaining dynamic similarity? (A) 1 (C) 63 (B) 22 (D) 500 Common Data for Question 2 and 3 An upwards flow of oil (mass density 800 kg/ , dynamic viscosity 0.8 kg/m-s) takes place under laminar conditions in an inclined pipe of 0.1 m diameter as shown in the figure. The pressure at sections 1 and 2 are measured as p = 435 kN/ and p = 200 kN/ 2 5m

2.

3.

The discharge in the pipe is equal to (A) 0.100 /s (C) 0.144 /s (B) 0.127 /s (D) 0.161 /s If the flow is reversed, keeping the same discharge, and the pressure at section 1 is maintained as 435 kN/ , the pressure at section 2 is equal to (A) 488 kN/ (C) 586 kN/ (B) 549 kN/ (D) 614 kN/

CE– 2007 4. Flow rate of a fluid (density = 1000 kg/ ) in a small diameter tube is 800 m /s. The length and the diameter of the tube are 2 m and 0.5 mm, respectively. The pressure

drop in 2 m length is equal to 2.0 MPa. The viscosity of the fluid is (A) 0.025 N-s/ (C) 0.00192 N-s/ (B) 0.012 N-s/ (D) 0.00102 N-s/ CE– 2008 Statements for Linked answer questions 5&6 An automobile with projected area 2.6 is running on a road with a speed of 120 km per hour. The mass density and the kinematic viscosity of air are 1.2 kg/ and 1.5 × /s respectively. The drag coefficient is 0.30 5. The drag force on the automobile is (A) 620N (B) 600N (C) 580N (D) 520N 6.

The metric horse power required to overcome the drag force is (A) 33.23 (C) 23.23 (B) 31.23 (D) 20.23

CE– 2009 7. Water flows through a 100 mm diameter pipe with a velocity of 0.015 m/sec. If the kinematic viscosity of water is 1.13 × /sec the friction factor of the pipe materials is (A) 0.0015 (C) 0.037 (B) 0.032 (D) 0.048 CE– 2011 8. A single pipe of length 1500 m and diameter 60 cm connect two reservoirs having a difference of 20m in their water levels. The pipe is to be replaced by two pipes of the same length and equal di eter ‘d’ to o vey % ore discharge under the same head loss. If the friction factor is assumed to be the same for all the pipes the v lue of ‘d’ is approximately equal to which of the following options ? (A) 37.5 cm (C) 45.0 cm (B) 40.0 cm (D) 50.0 cm th

th

th



CE– 2012 9. A trapezoidal channel is 10.0 m wide at the base and has a side slope of 4 horizontal to 3 vertical. The bed slope is 0.002. the channel is lined with smooth Concrete ( i g’s ) The hydraulic radius (in m) for a depth of flow of 3.0 m is (A) 20.0 (C) 3.0 (B) 3.5 (D) 2.1 10.

11.

13.

A rectangular open channel of width 5.0m is carrying a discharge of 100 /s. The Froude number of the flow is 0.8. The depth of flow ( in m) in the channel is (A) 4 (C) 16 (B) 5 (D) 20

Fluid Mechanics

With reference to a standard Cartesian (x, y) plane, the parabolic velocity distribution profile of fully developed laminar flow in x-direction between two parallel, stationary and identical plates that are separated by distance, h, is given by the expression h dp y u [ ( ) ] dx h I this equ tio the y = 0 axis lies equidistant between the plates at a distance h/2 from the two plates, p is the pressure variable and is the dynamic viscosity term. The maximum and average velocities are, respectively (A) u

du

(B) u

The top width and the depth of flow in a triangular channel were measured as 4m and 1m, respectively. The measured velocities on the centre line at the water surface, 0.2m and 0.8m below the surface are 0.7 m/s, 0.6 m/s and 0.4m/s, respectively. Using two – point method of velocity measurement, the discharge ( in /s) in the channel is (A) 1.4 (C) 1.0 (B) 1.2 (D) 0.8

du

(C) u

u u

du

(D) u

du

u u

CE– 2014 12. The dimension for kinematic viscosity is (A)

(C)

(B)

(D)

th

th

th



Fluid Mechanics


[Ans. C] Equating Reynolds number and Froude number, we get = ∴

= (v ) =

But

Here, CD = 0.30, A = 2.6m2 ρ V = 120 kmph ( F

=2× =( = 0.0159

)

=

6.

≈

+ 5sin

H

=

(

(

)

7.

Reynolds number, Re =

16.19 = 0.127

=

/s = 1327.43 < 2000 Hence flow is laminar

+ 5 sin

+

(

ρ = 614.48 kN/ 4.

f

)

=

+

√

= 614 kN/

8.

A

e

[Ans. D] Figure blow shows a single pipe which connects two reservoirs.

[Ans. C] Applying Hazen-Poiseuille equation, the drop in pressure is given by

But

H

[Ans. D]

[Ans. D] When the flow is reversed, then +

)

)

V = 16.19 m/s ∴ Q = AV =

) (

w tt

+ √

3,

[Ans. C] Power required to overcome the drag = Drag force × velocity

[Ans. B] Applyi g Ber oulli’s equ tio betwee (1) and (2) =

g

 FD = 520 N

= =

3.

[Ans. D] The drag force on the automobile may be given as ρ F C A

√

v =

2.

5.

(1) 20m

D

60 cm D

A

(2)

datum

Q1 V1

L=1500m

(

) s ≈

∴

(

Let V1 = velocity of water flow in single pipe (m/s) Q1 = Discharge in single pipe ( /S) hf = Head loss due to friction in single pipe (m)

s )

s⁄ s th

th

th



f

h

[where f

Substitute (1) & (3) in (2),

Fri tio f tor]

gD From continuity equation

Fluid Mechanics

) d ( Obtain d = 0.497 m = 49.7 cm say 50 cm

A

epl e Fro

9.

Ber oulli’s equ tio h

g

[Ans. D] 3V

ρg

g

10m

h for uniform diameter pipe ]

[ ∴h f

4H

A A

* ⁄ D +

Where D = 0.6 m f

3v

4H

(B

y)y

(

)

D y√

B (

√( ⁄ )

)

∴

( ) √f In case of two pipes of the same length d equ l di eter ‘d’ (p r llel pipe system), discharge in each pipe will be the same. As the discharge in each parallel pipe is same, hence velocity will also be dame in parallel pipe, given 25%

10.

[Ans. A]

5m 20m

B = 5m Q = 100 Fr = 0.8 y=?

d

2Q2

Q2 d

Q2

2Q2

F

/s

√gy A

L1=L2= 1500m

A

( ) h

f

A√gy

gd epl e

h

f

y√ y

d (

)

11.

gd d d √f

[Ans. C] Area of flow =

f

∴

y

Aver ge velo ity ( ) th

th

th



Fluid Mechanics

s A 12.

s

[Ans. C] Di e sio s of ν re ν

13.

since in SI unit

s since ν

[Ans. A] h/2 h/2

Velocity for a laminar flow between two parallel plates is given as h dp y u ( )[ ( ) ] dx h E d o ditio u u ty h dp u ( ) dx Discharge, dQ = Area velo ity dQ =

( )[ h h

h

dp ( ) ∫ ( dx (

( ) ] (dy

)

y ) dy h

dp ) dx

A dp ) ( ) (h dx h dp ( ) dx ( ) ( )

th

th

th



Fluid Mechanics

Dimensional Analysis CE – 2007 1. A 1 : 50 scale model of a spillway is to be tested in the laboratory. The discharge in the prototype is 1000 /s. The discharge to be maintained in the model test is (A) 0.057 /s (C) 0.57 /s (B) 0.08 /s (D) 5.7 /s

CE – 2013 3. Group – I contains dimensionless parameters and Group – II contains the ratios. Group – I Mach Number

Group – II P. 1. Ratio of inertial force and gravitational force Q. Reynolds 2. Ration of fluid Number velocity and velocity of sound R. Weber 3. Ratio of inertial Number force and viscous force S. Froude 4. Ratio of inertial Number force and surface tension force The correct match of dimensionless parameters in Group – I with ratios in Group – II is: (A) P – 3, Q – 2, R – 4, S – 1 (B) P – 3, Q – 4, R – 2, S – 1 (C) P – 2, Q – 3, R – 4, S – 1 (D) P – 1, Q – 3, R – 2, S – 4

CE – 2008 2. A river reach of 2.0 km long with maximum flood discharge of 10000 /s is to be physically modeled in the laboratory where maximum available discharge is 0.20 /s. For a geometrically similar model based on equality of Froude number, the length of the river reach (m) in the model is (A) 26.4 (C) 20.5 (B) 25.0 (D) 18.0


[Ans. A] Froude model law will be applicable in this case. =(

= 0.057

3.

/sec

=1

[ gr = 1]

∴ Vr = √ Now, we have x 



(

)

Lm = 26.4 m

[Ans. C] 1. Froude number lined inertial force to gravitational force 2. Reynolds number is ratio of inertial forces to viscous force 3. Mach number classifier force as subsonic sonic are supersonic depending upon ratio of velocity of fluid velocity and velocity of sound 4. Weber number is ratio of inertial force to surface tension force

[Ans. A] According to Froude Model law √

( )



) = 1000 x( )

2.



x√ th

th

th



Fluid Mechanics

Impacts of jets and Turbines CE – 2007 1. A horizontal water jet with a velocity of 10 m/s and cross sectional area of 10 strikes a flat plate held normal to the flow direction. The density of water is 1000 kg . The total force on the plate due to the jet is (A) 100 N (C) 1 N (B) 10 N (D) 0.1 N


[Ans. C] The force is given by F=ρ F F

(

)

th

th

th



Fluid Mechanics

Open Channel Flow CE – 2007 1. The flow rate in a wide rectangular open channel is 2.0 s per meter width. The h el bed slope is The i g’s roughness coefficient is 0.012. The slope of the channel is classified (A) Critical (C) Mild (B) Horizontal (D) Steep CE – 2010 2. A mild – sloped channel is followed by steep-sloped channel. The profiles of gradually varied flow in the channel are (A) (C) (B) (D)

CE – 2011 3. The flow in a horizontal, frictionless rectangular open channel is supercritical. A smooth hump is built on the channel floor. As the height of hump is increased, choked condition is attained. With further increase in the height of the hump, the water surface will (A) Rise at a section upstream of the hump (B) Drop at a section upstream of the hump (C) Drop at the hump (D) Rise at the hump


[Ans. D] The critical depth for a rectangular channel is given by y

q

( ) q ( ) ( ) g si g i g’s equ tio we get (y )

√ (y )

y For steep slopes y

√ y

2.

[Ans. D]

3.

[Ans. B] The height of the hump attains its maximum value at the choked condition. Increasing the height of hump after chocked condition results in hydraulic jump formation in case of super critical flow. Thus height of water surface will drastically changes upstream and downstream of hump.

th

th

th




Hydraulics & Hydraulic Machinery CE – 2005 1. A horizontal bed channel is followed by a steep bed channel as shown in the figure. The gradually-varied profiles over the horizontal and steep beds are

5.

A launching apron is to be designed at downstream of a weir for discharge intensity of 6.5 m ⁄s⁄m. For the design of launching aprons the scour depth is taken two times of Lacey scour depth. The silt factor of the bed material is unity. If the tailwater depth is 4.4m, the length of launching apron in the launched position is (C) 5 m (A) √5m (B) 4.7 m (D) 5√5 m

Horizontal bed Steep bed

(A) (B) (C) (D) 2.

3.

H H H H

and S and S and S and S

respectively respectively respectively respectively

Critical depth at a section of a rectangular channel is 1.5 m. The specific energy at that section is (A) 0.75 m (C) 1.5 m (B) 1.0 m (D) 2.25 m

7.

Identify the FALSE statement from the following the specific speed of the pump increase with (A) Increase in shaft speed (B) Increase in discharge (C) Decrease in gravitational acceleration (D) Increase in head

8.

A hydraulic jump occurs in a rectangular horizontal, frictionless channel. What would be the pre-jump depth if the discharge per unit width is 2 m /s/m and the energy loss is 1m? (A) 0.2 (C) 0.8m (B) 0.3m (D) 0.9m

9.

A very wide rectangular channel is designed to carry a discharge of 5 m /s per meter width. The design is based on the manning’s equations with the roughness coefficient obtained from the grain size using Strickler’s equation and results in a normal depth of 1.0m by

A partially open sluice gate discharge water into a rectangular channel. The tail water depth in the channel is 3m and Froude number is

√

. If a free hydraulic

jump is to be formed at downstream of the sluice gate after the vena contracta of the jet coming out from the sluice gate, the sluice gate opening should be (co-efficient of contraction C = 0.9) (A) 0.3 m (C) 0.69 m (B) 0.4 m (D) 0.9 m 4.

CE – 2006 6. A channel with a mild slope is followed by a horizontal channel and then by a steep channel. What gradually varied flow profiles will occur? (A) M , H , S (C) M , H , S (B) M , H , S (D) M , H , S

A triangular irrigation lined canal carries a discharge of 25 m /s at bed slope = If the side slopes of the canal are 1 : 1 and Manning’s coefficient is 0.018, the central depth of flow is equal to (A) 2.98 m (C) 4.91 m (B) 3.62 m (D) 5.61 m

th

th

th



mistake, however, the engineer used the grain diameter in mm in the Stricker’s equation instead of in meter, What should be the correct normal depth? (A) 0.32m (C) 2.00m (B) 0.50m (D) 3.20m CE – 2007 10. There is a free overfall at the end of a long open channel. For a given flow rate, the critical depth is less than normal depth. What gradually varied flow profile will occur in the channel for this flow rate? (A) M (C) M (B) M (D) S 11.

A triangular open channel has a vertex angle of 90 and carries flow at a critical depth of 0.30m. The discharge in the channel is (A) 0.08 m /s (C) 0.15m /s (B) 0.11 m /s (D) 0.2m /s Statement for linked answer questions 12 & 13 A rectangular open channel needs to be designed to carry a flow of 2.0 m /s under uniform flow conditions. The Manning’s roughness coefficient is 0.018. The channel should be such that the flow depth is equal to half width and the Froude number is equal to 0.5

12.

The bed slope of the channel to be provided is (A) 0.0012 (C) 0.0025 (B) 0.0021 (D) 0.0052

13.

Keeping the width, flow depth and roughness the same. If the bed slope of the above channel is doubled. The average boundary shear stress under uniform flow conditions is (A) 5.6 N/m (C) 12.3N/m (B) 10.8 N/m (D) 17.2N/m


CE– 2008 14. A person standing on the bank of a canal drops a stone on the water surface. He notices that the disturbance on the water surface is not traveling upstreams. This is because the flow in the canal is (A) Sub-critical (C) Steady (B) Super-critical (D) Uniform

15.

Common data questions 15, 16 and 17 A rectangular channel 6.0 m wide carries a discharge of 16.0 m /s under uniform flow condition with normal depth of 1.60 m. Manning's 'n' is 0.015. The longitudinal slope of the channel is (A) 0.000585 (C) 0.000385 (B) 0.000485 (D) 0.000285

16.

A hump is to be provided on the channel bed. The maximum height of the hump without affecting the upstream flow condition is (A) 0.50 m (C) 0.30 m (B) 0.40 m (D) 0.20 m

17.

The channel width is to be contracted. The minimum width to which the channel can be contracted without affecting the upstream flow condition is (A) 3.0 m (C) 4.1 m (B) 3.8 m (D) 4.5 m

18.

A weir on a permeable foundation with downstream sheet pile is show in the figure below. The exit gradient as per Khosla’s method is Weir 5m

Floor

10 m Downstream Sheet Pile

(A) 1 in 6.0 (B) 1 in 5.0 th

th

4m

(C) 1 in 3.4 (D) 1 in 2.5 th



19.

Water emerges from an ogee spillway with velocity = 13.72 m/s and depth = 0.3 m at its toe. The tail water depth required to form a hydraulic jump at the toe is (A) 6.48 m (C) 3.24 m (B) 5.24 m (D) 2.24 m

20.

The base width of an elementary profile of a gravity dam of height H is b. The specific gravity of the material of the dam is G and uplift pressure coefficient is K. The correct relationship for no tension at the heel is given by (A) (B)

√

√G

For a rectangular channel section, match List – I (Geometrical elements) with List – II (Proportions for hydraulically efficient section) and select the correct answer using the codes given below the lists: List – I List – II A. Top width 1. y /2 B. Perimeter 2. y C. Hydraulic Radius 3. 2y D. Hydraulic Depth 4. 4y y is the flow corresponding to hydraulically efficient section Codes: A B C D (A) 2 4 1 3 (B) 3 1 4 2 (C) 3 4 1 2 (D) 3 4 2 1

25.

The froude number of flow in a rectangular channel is 0.8 if the depth is flow is 1.5 m, the critical depth is (A) 1.80 m (C) 1.36 m (B) 1.56 m (D) 1.29 m

√

CE – 2009 21. Direct step method of computation for gradually varied flow is (A) Applicable to non-prismatic channels (B) Applicable to prismatic – channels (C) Applicable to both prismatic and non-prismatic channels (D) Not applicable to both prismatic and non-prismatic channels 22.

24.

(C) (D)

A rectangular open channel of width 4.5m is carrying a discharge of 100m /sec. The critical depth of the channel is (A) 7.09 m (C) 2.16 m (B) 3.96 m (D) 1.31 m

CE – 2010 23. The flow in a rectangular channel is subcritical. If width of the channel is reduced at a certain section, the water surface under no-choke condition will (A) Drop at a downstream section (B) Rise at a downstream section (C) Rise at an upstream section (D) Not undergo any change


CE – 2011 26. A spill way discharges flood flow at a rate of 9m /s per metre width. If the depth of flow on the horizontal apron at the toe of the spillway is 46 cm, the tail water depth needed to form a hydraulic jump is approximately given by which of the following options? (A) 2.54 m (C) 5.77 m (B) 4.90 m (D) 6.23 m 27.

th

For given discharge, the critical flow depth in an open channel depends on (A) Channel geometry only (B) Channel geometry and bed slope (C) Channel geometry, bed slope and roughness (D) Channel geometry, bed slope, roughness and Reynolds number

th

th



CE – 2013 28. For subcritical flow in an open channel, the control section for gradually varied flow profiles is (A) At the downstream end (B) At the upstream end (C) At both upstream and downstream ends (D) At any intermediate section 29.

The normal depth in a wide rectangular channel is increased by 10%. The percentage increase in the discharge in the channel is : (A) 20.1 (C) 10.5 (B) 15.4 (D) 17.2

CE – 2014 30. A horizontal jet of water with its crosssectional area of 0.0028 m hits a fixed vertical plate with a velocity of 5 m/s. After impact the jet splits symmetrically in a plane parallel to the plane of the plate. The force of impact (in N) of the jet on the plate is (A) 90 (C) 70 (B) 80 (D) 60


31.

A rectangular channel having a bed slope of 0.0001, width 3.0 m and Manning’s coefficient ‘n’ 0.015, carries a discharge of 1.0 m s. Given that the normal depth of flow ranges between 0.76 m and 0.8 m. The minimum width of a throat (in m) that is possible at a given section, while ensuring that the prevailing normal depth is not exceeded along the reach upstream of the contraction, is approximately equal to (assume negligible losses) (A) 0.64 (C) 1.04 (B) 0.84 (D) 1.24

32.

A horizontal nozzle of 30 mm diameter discharges a steady jet of water into the atmosphere at a rate of 15 litres per second. The diameter of inlet to the nozzle is 100 mm. The jet impinges normal to a flat stationary plate held close to the nozzle end. Neglecting air friction and considering the density of water as 1000 kg/m , the force exerted by the jet (in N) on the plate is _________

33.

A rectangular channel of 2.5 m width is carrying a discharge of 4 m /s. Considering that acceleration due to gravity as 9.81 m/s , the velocity of flow (in m/s) corresponding to the critical depth (at which the specific energy is minimum) is _______


[Ans. A]

2.

[Ans. D] For rectangular channel, E = 1.5y = 1.5 × 1.5 = 2.25m

3.

[Ans. C] For hydraulic jump, y 1 * 1 √1 8F + y 2 Given y = 3m

F =

√

√1

∴y = * 1

8

+

⇒ y = 0.62m Sluice gate opening = 4.

=

. .

= 0.69m

[Ans. C]

th

By Manning’s equations, Q

R

S

⇒ 25 =

)

× y

th

.

×(

√

) th

×(

A



⇒ 25 =

× y

.

×

⇒ ∆E = 1m But energy loss ∆E given in problem is also 1m. Hence the assumed value of y , i.e. 0.3m is correct.

.

⇒ y = 4.91 m 5.

[Ans. C] Lacey’s scour depth, q 6.5 R 1.35 ( ) 1.35 ( ) f 1 Scour depth =2R = 9.4m ∴ D 2R 4.4 5m

9.

[Ans. D]

7.

[Ans. D] The specific speed of a pump is given by

8.

[Ans. B]

4.7 m

6.

N


As per Stickler’s equation. n ∴

=(

Now, as per Mannings equation, 1 q y √S n ∴ Correct depth of flow, n ( n

y

N√Q H

[Ans. B] We know that a rectangular channel, y

…… 1

∴y

*

⇒y 0.74m We know that hydraulic jump is formed when supercritical flow changes to subcritical flow, if the prejump depth and post jump depth are y and y respectively then y
)

= [(

)

y

]

1 = 0.50m

10.

[Ans. B] All free fall curves are zone II profiles. Since the critical depth (y ) is less than the normal depth (y ), therefore the slope is mild. Thus, the gradually varied profile is M

11.

[Ans. B] The side slope of triangular open channel given as z horizontal to 1 vertical. When vertex angle is 90 , z = 1

+

.

)

= ⇒

.

=

1

90

Solving the above equation by taking y = 0.3m which is option (b) and thus calculating the value of y 8 ∴ 0.3y 0.3 y 9.81 ⇒ y + 0.3y 2.72 = 0 ⇒ y = 1.51 m We also known that

Now, y = *

.

⇒ ∆E =

.

. .

.

12.

. .

for a triangular

+

channel

∆E = ⇒ ∆E =

Z

.

th

⇒

Q =

⇒

Q =

⇒

Q = 0.11 m /s

.

.

[Ans. B] Given Q = 2 m3/s, n = 0.018, B = 2y, F = 0.5 We know that th

th



F=

16.

√

Where V is velocity of flow, D is hydraulic radius Now ∴

D=

F=

y

√

∴ 1.6

Squaring both sides

√

 0.25 = .

 1.74 .

+

 ∆Z  ∆Z

 y = 0.836m Using Manning’s equation, we get 1 Q A R √S n 1 2 2 0.836 0.018 0.836 ( ) √S 2  √S = 0.0461  S = 0.0021

17.

[Ans. D] The average shear stress, τ0 γRS0 Bed slope is doubled i.e. S0 = 2 0.0021 = 0.0042 τ0 = 9810

.

0.0042

 τ0 = 17.22 N/m2 14.

[Ans. A]

15.

[Ans. A] The velocity as per Manning’s equation is given by

18.

V= R Here,

But

Q = AV

∴

Q=A



16 = 9.6

. .

R .

S 1.04

∆Z

9.6

3 y 2

3 16 6 * + 2 9.81 1.74 – 1.35 0.39 ≈ 0.4m ∆Z

[Ans. C] If the channel is contracted, the depth of flow at this location decreases. Thus width of the channel can be decreased till the specific energy becomes minimum. If the channel is contracted further, the depth of flow starts decreasing. ∴

y+

E



1.6 +



1.74 =

(



1.74 =

*

 

b

.

y

.

) .

+

= 2.557 b = 4.1m

[Ans. C] The exit gradient may be given as H 1 G d √ 1

√

and

Here b = 10 m, d = 4 m, H = 5 m b 10 ∴ 2.5 d 4 1 √1 2.5 and 1.85 2 5 1 ∴ G 4 √1.85 1 ⇒ G 3.4

1.04m

.

the this the the

E

16 9.81

2

Where

S

Q = 16m3/s. n = 0.015, B =6.0m A = 6.0 1.60 = 9.6 m2 R=



∆Z

√

 0.5 =

13.

[Ans. B] When a hump is introduced on channel bed the depth of flow at location will reduce. If ∆Z is maximum height of hump so that depth of flow remains same, then y+

 0.5 =

 y=*


S

S = 0.00059

th

th

th



19.

[Ans. C] Water depth required to form a hydraulic jump may be given as y [ 1 2

y

0.3 2

√1

[ 1

8 13.72 ] 9.81 0.3 25.

3.24 m 20.

A 2y P 2y 2y y 2 Top width, T B 2y Perimeter, P B 2y 2y 2y A 2y Hydraulic depth, D y T 2y Hydraulic radius, R

8V ] gy √1

[Ans. A] Tension will not be developed at the heel with full reservoir, when H b √G

21.

[Ans. B]

22.

[Ans. B] The critical depth for a rectangular channel is given by q y g


4y

[Ans. D] Froude Number for a rectangular channel is given by F=

√

V = 0.8 √9.81 1.5 = 3.07 m/s ∴ Discharge per unit width of rectangular channel is given by q = Vy = 3.07 1.5 = 4.6 m3/s/m ∴ Critical depth, yc = ( )

But q = ∴

y

 y

.

( 26.

4.6 ) 9.81

1.29m

[Ans. C]

22.22 m sec y2

. .

3.69m y1

23.

[Ans. A]

24.

[Ans. C] Let the width of rectangular channel be B and depth of flow be y. Area of flow. A = By Wetted perimeter, P = B + 2y =

2y

For hydraulically most efficient section, P should be minimum dP ∴ 0 dy A  2 0 y  A 2y  B 2y  B 2y

q y

9m s m 0.46m y y y

2 9 0.46 y 9.81 y 5.77 m 27.

y 0.46

y

[Ans. A] y

(

) Q

y

( )

Constant

Depends on geometry only. 28.

th

[Ans. A]

th

th



29.

[Ans. D] Discharge Q = R S


So, E

A

( )

S

B

(

y

∴

y

∆

3 2

⇒ .

100

)

So, [

[for wide channels R = y/2] ⇒Q

.

.

0.789 m

manning’s formula ∴Q

( ⁄ )

0.78

100

]

0.789 ⇒ B

1

0.7893

g B 0.84m

⇒B

17.2 32. 30.

31.

[Ans. C] Force exerted in x direction = rate of change of momentum in x direction i.e. F A v v v Herev 0ms x direction ⇒F 1000 0.0028 5 5 70 N [Ans. B] n 0.015, Q 1m s, B 3.0m Normal depth of flow ranges between 0.76m to 0.8 m If prevailing normal depth of flow is not exceeded, there must not be choking of the section or it should be at boundary condition of choking. So, width of section should be such that there should be critical flow corresponding to the prevailing specific energy. 3 q i. e. , ( ) 2 g q

E

100 mm 15

Force on plate, F a. v 1000

4 318.29 N 33.

E

0.03

21.22

[Ans. *] Range 2.45 to 2.55 Q = 4m s B = 2.5 m 4 q 1.6 m s m 2.5 q y ⇒y 0.639 g V 2g

E

V

y

Q

0.639 2

√0.639 g = 2.504 m/s

. A. R s ⇒ 1 . B. y . (

.

⇒y

30 mm

At critical depth velocity head =

) 3 ( So [ ] 2 g

⇒1

21.22 m s

.

E

[Ans. *] (Range 318 to 319) Q 15 10 m s Velocity of jet, v A 0.03

.

.

. 3y . (

).S ) . 0.0001

0.78 m th

th

th




Irrigation CE – 2005 1. On which of the canal systems, R.G. Kennedy, executive engineer in the Punjab Irrigation Department made his observations for proposing his theory on stable channels? (A) Krishna Western Delta canals (B) Lower Bari Doab canals (C) Lower Chenab canals (D) Upper Bari Doab canals

2.

Uplift pressure at points E and D (figure A) of a straight horizontal floor of negligible thickness with a sheet pile at downstream end are 28% and 20%, respectively. If the sheet pile is at upstream end of the floor (figure-B), the uplift pressures at points D and C are E C

C d

b

d

b D

D

Figure-A Figure-B (A) 68% and 60% respectively (B) 80% and 72% respectively (C) 88% and 70% respectively (D) 100% and zero respectively

Answer keys & Explanations 1.

[Ans. D]

2.

[Ans. B] Uplift pressure at D , Uplift pressure at C ,

100 100 100 100

20

80

28

72

th

th

th




Water Requirements of Crops CE – 2005 1. The culturable commanded area for a distributary is 2 10 m . The intensity of irrigation for a crop is 40%. If kor water depth and kor period for the crop are 14 cm and 4 weeks, respectively, the peak demand discharge is (A) 2.63 m ⁄s (C) 8.58 m ⁄s (B) 4.63 m ⁄s (D) 11.58 m ⁄s CE – 2006 2. In a cultivated area, the soil has porosity of 45% and field capacity of 38%. For a particular crop, the root zone depth is 1.0 m, the permanent wilting point is 10% and the consumptive use is 15 mm/d. If the irrigation efficiency is 60%, what should be the frequency of irrigation such that the moisture content does not fall below 50% of the maximum available moisture? (A) 5d (C) 9d (B) 6d (D) 15d CE – 2007 3. The consumptive use of water for a crop during a particular stage of growth is 2.0 mm/day. The maximum depth of available water in the root zone is 60 mm. Irrigation is required when the amount of available water is 50% of the maximum available water in the root zone. Frequency of irrigation should be (A) 10 days (C) 20 days (B) 15 days (D) 25 days 4.

The culturable command area for a distributed channel is 20,000 hectares. Wheat is grown in the entire area and the intensity of irrigation is 50%. The kor period for wheat is 30 days and the kor water depth is 120 mm. The outlet discharge for the distributary should be (A) 2.85 m ⁄s (C) 4.63 m ⁄s (B) 3.21 m ⁄s (D) 5.23 m ⁄s

CE – 2009 5. An agricultural land of 437 ha is to be irrigated for a particular crop. The base period of the crop is 90 days and the total depth of water required by the crop is 105 cm. If a rainfall of 15 cm occurs during the base period, the duty of irrigation water is (A) 437 ha/cumec (C) 741 ha/cumec (B) 486 ha/cumec (D) 864 ha/cumec CE – 2010 Common Data for Question 6 and 7 The moisture holding capacity of the soil in a 100 hectare farm is 18 cm/m. The field is to be irrigated when 50 per cent of the available moisture in the root zone is depleted. The irrigation water is to be supplied by a pump working for 10 hours a day, and water application efficiency is 75 per cent. Details of crops planned for cultivation are as follows: Crop Root zone Peak rate of moisture depth (m) use (mm/day) X 1.0 5.0 Y 0.8 4.0 6. The capacity of irrigation system required to irrigate crop ‘X’ in 36 hectares is (A) 83 litres/sec (C) 57 litres/sec (B) 67 litres/sec (D) 53 litres/sec 7.

The area of crop ‘Y’ that can be irrigated when the available capacity of irrigation system is 40 liters/sec is (A) 40 hectares (C) 30 hectares (B) 36 hectares (D) 27 hectares

CE – 2012 8. Wheat crop requires 55 cm of water during 120 days of base period. The total rainfall during this period is 100 mm. Assume the irrigation efficiency to be 60%. The area (in ha) of the land which th

th

th



can be irrigated with a canal flow of 0.01 m ⁄s is (A) 13.82 (C) 23.04 (B) 18.85 (D) 230.40 CE – 2013 9. The transplantation of rice requires 10 days and total depth of water required during transplantation is 48 cm. During transplantation, there is an effective rainfall ( useful for irrigation) of 8 cm. the duty of irrigation water (in hectares/cumec) is : (A) 612 (C) 300 (B) 216 (D) 108


CE – 2014 10. Irrigation water is to be provided to a crop in a field to bring the moisture content of the soil from the existing 18% to the field capacity of the soil at 28%. The effective root zone of the crop is 70 cm. If the densities of the soil and water are 1.3 g/ cm and 1.0 g/ cm respectively, the depth of irrigation water (in mm) required for irrigating the crop is ________


[Ans. B] Area under crop = 2 Volume

But the moisture content should not fall below 50%. ∴ Depth of water required 0.5 0.331 0.165 m Actual depth stored, 0.165 60 0.099m 99mm 100 Frequency of irrigation, Actual depth stored 99 Consumptive use 15 6.63 d ≈ 6 d

10

0.8 10 m Area ∆ 0.8 10 14 10 1.2 10 m .

Discharge, Q

4.63 m ⁄sec 2.

[Ans. B] Depth of maximum available moisture, γ d F γ Field capacity, weight of water retained by soil weight of same area of soil γ V γ V But

γ γ γ ⇒ γ ∴ d

[Ans. B] Consumptive use of water = 2.0 mm/day Maximum depth of available water = 60 mm If amount of available water is reduced by 50%, then irrigation is required for a depth 50 60 30 mm 100 30 ∴ Frequency of irrigation 2 15 days

4.

[Ans. C] Area to be irrigated = Culturable command area × intensity of irrigation 50 20.00 100

n

∴ Field capacity ⇒

3.

γ γ

n

n Field capacity 0.45 0.38 0.45 1 0.38 0.38 0.331 m

0.1

th

th

th



10,000 ha Duty, D

.

7.

[Ans. D] Moisture holding capacity of soil for crop Y= 18 × 0.8 = 14.4 cm Allowable depletion of moisture, 50 14.4 7.2cm 100 Consumptive use of water for crop Y = 0.4 cm/day . ∴ Frequency of irrigation = 18 days . Thus 7.2 cm depth of water is to be applied to the crop Y in the next 18 days. . Irrigation requirement = 9.6 cm . Quantity of water required = Discharge through pump 9.6 ⇒ A 10 100 40 10 18 10 60 60 ⇒ A 27 hectares

8.

[Ans. A] Given, Base period, B = 120 days 55 10 45cm 0.45m B 8.64 120 8.64 , ∴ D 2304 D 0.45 Area, A 2304 0.01 23.04 As efficiency of Irrigation is 60% Then required area for wheat crop 60 23.04 13.82 ha. 100

9.

[Ans. B]

.

=2160 ha/cumec Area to be irrigated ∴ Outlet discharge Duty of crop = 5.

,

4.63 m s

[Ans. D] Total depth of water required by the crop = 105 cm Contribution of rainfall during the base period =15 cm ∴ Delta, 105 15 90 cm Duty of crop

864

ha/cumec 6.

[Ans. B] Moisture holding capacity of soil = 18 cm/m Root zone depth of crop X = 1.0 m Moisture holding capacity of soil for crop X = 18 × 1 = 18 cm Allowable depletion of moisture, 50 18 9 cm 100 Consumptive use of water for crop, X 0.5 cm⁄day ∴ Frequency of irrigation

.

18 days

Thus 9 cm depth of water is to be applied to the crop X in the next 18 days. Irrigation requirement, Net water depth Water application efficiency 9 12cm 0.75 Quantity of water required, 12 36 10 43200 m 100 ∴ Discharge through pump 43200 10 litres sec 18 10 60 60 litres 66.67 ≈ 67litres sec sec


Duty = 8.64 Duty = 8.64

hec cumec

Duty = 216 hec/cumec. 10.

[Ans. 91] Given, Root zone depth d = 70 cm Field capacity F 28 Existing moisture content w = 18% Density of soil γ = 1.3 gm/cm Density of water γ = 1.0 gm/cm Depth of irrigation water required, γ d d F w γ . .

70

1.3

70

10 28

18

10

91mm th

th

th




Sediment, Transport and Design of Irrigation Channels CE – 2005 1. Which one of the following equations represents the downstream profile of Ogee spillway with vertical upstream face? (x, y) are the coordinates of the point on the downstream profile with origin at the crest of the spillway and H is the design head. (A)

0.5 ( )

(B)

0.5 ( )

(C)

2.0 ( )

(D)

2.0 ( )

CE – 2008 3. A stable channel is to be designed for a discharge of Q m s with silt factor f as per Lacey’s method. The mean flow velocity (m/s) in the channel is obtained by

.

(A) *

+

(C) *

(B) *

+

(D) 0.48

+ * +

. . .

CE – 2007 2. As per the Lacey’s method for design of alluvial channels, identify the TRUE statement from the following: (A) Wetted perimeter increases with an increase in design discharge. (B) Hydraulic radius increases with an increase in silt factor. (C) Wetted perimeter decreases with an increase in design discharge. (D) Wetted perimeter increases with an increase in silt factor.

CE – 2009 4. The depth of flow in a alluvial channel is 1.5m. If critical velocity ratio is 1.1 and Manning’s n is 0.018, the critical velocity of the channel as per ennedy’s method is (A) 0.713 m/s (C) 0.879 m/s (B) 0.784 m/s (D) 1.108 m/s


[Ans. A] For a spillway having a vertical u/s face, the d/s crest is given by the equation, x . 2H . y H ⇒x . 2H . y H . H y ⇒ x . 2 H . x y ⇒ . H 2H y x . ⇒ 0.5 ( ) H H

2.

[Ans. A] As per Lacey’s method of design of alluvial channels, P

4.75√Q

3.

[Ans. A]

4.

[Ans. B] The critical velocity as per ennedy’s method is given by V 0.55 my . 0.55 1.1 1.5 . 0.784 m s

th

th

th




Hydrology CE– 2005 1. The intensity of rain fall and time interval of a typical storm are Time interval Intensity of rainfall (minutes ) (mm/minute ) 0-10 0.7 10-20 1.1 20-30 2.2 30-40 1.5 40-50 1.2 50-60 1.3 60-70 0.9 70-80 0.4 The maximum intensity of rainfall for 20 minutes duration of the storm is (A) 1.5 mm/minute (B) 1.85 mm/minute (C) 2.2 mm/minute (D) 3.7 mm/minute

2.

3.

4.

Statement for Linked Answer Questions 2&3 A four hour unit hydrograph of a catchment is triangular in shape with base of 80 hours. The area of the catchment is 720 km . The base flow and – index are 30 m /s and 1mm/h, respectively. A storm of a 4 cm occurs uniformly in 4 hours over the catchment. The peak discharge of four hour unit hydrograph is (A) 40 m /s (C) 60 m /s (B) 50m /s (D) 70 m /s The peak flood discharge due to the storm is (A) 210 m /s (C) 260 m /s (B) 230 m /s (D) 720 m /s When the outflow from a storage reservoir is uncontrolled as in a freely operating spillway, the peak of outflow hydrograph occurs at

(A) At point of inter section of the inflow and outflow hydrographs (B) A point, after the inter section of the inflow and outflow hydrographs (C) The tail of inflow hydrographs (D) A point, before the inter section of the inflow and outflow hydrographs 5.

Two observation wells penetrated into a confined aquifer and located 1.5 km apart in the direction of flow, indicate head of 45 m and 20 m. If the coefficient of permeability of the aquifer is 30 m/day and porosity is 0.25, the time of travel of an inert tracer from one well to another is (A) 416.7 days (C) 750 days (B) 500 days (D) 3000 days

CE– 2006 6. During a 3 hour storm event, it was observed that all abstractions other than infiltration are negligible. The rainfall was idealized as 3 one hour storms of intensity 10 mm/hr, 20 mm/hr and 10 mm/hr respectively and the infiltration was idealized as a Horton curve, f = 6.8 + 8.7 exp( t) (f in mm/hr and t in hr). What is the effective rainfall? (A) 10.00 mm (C) 12.43 mm (B) 11.33 mm (D) 13.63 mm Common Data for Questions 7and 8 For catchment, the S – curve (or S – hydrograph) due to a rainfall of intensity 1 cm/hr is given by Q = 1 (1+t)exp( t)(t in hr and Q in m /s). What is the area of the catchment? (A) 0.01 km (C) 1.00 km (B) 0.36 km (D) 1.28 km

7.

8.

What will be the ordinate of a 2 – hour unit hydrograph for this catchment at t = 3 hour ? (A) 0.13 m /s (C) 0.27 m /s (B) 0.20 m /s (D) 0.54 m /s th

th

th



9.

For steady flow to a fully penetrating well in a confined aquifer, the drawdowns at radial distances of r1 and r2 from the well have been measured as s1 and s2 respectively , for a pumping rate of Q. The transmissivity of the aquifer is equal to (A)

(C)

(B)

(D) 2 Q

11.

12.

(C) Increased peak with increased time – base (D) Increased peak with reduced time – base 14.

A volume of 3 10 m of ground water was pumped out from an unconfined aquifer uniformly from an area of 5 . The pumping lowered the water table from initial level of 102 m to 99m. the specific yield of the aquifer is (A) 0.20 (C) 0.40 (B) 0.30 (D) 0.50

15.

An outlet irrigates an area of 20 ha. The discharge (l/s) required at this outlet to meet the evapo-transpiration requirement of 20 mm occurring uniformly in 20 days neglecting other field losses is (A) 2.52 (C) 2.01 (B) 2.31 (D) 1.52

l ( ) ( )

CE– 2007 Common Data for Questions 10 and 11 Ordinates of a 1 – hour unit hydrograph at 1 hour intervals, starting from time t = 0, are 0, 2, 6, 4, 2, 1 and 0 m /s. 10. Catchment area represented by this unit hydrograph is (A) 1.0 km (C) 3.2 km (B) 2.0 km (D) 5.4 km Ordinate of a 3 – hour unit hydrograph for the catchment at t = 3 hour is (A) 2.0 m /s (C) 4.0 m /s (B) 3.0 m /s (D) 5.0 m /s An isolated 4-hour storm occurred over a catchment as follows: Time 1st hr 2nd hr 3rd hr 4th hr Rainfall 9 28 12 7 (mm) The index for the catchment is 10 mm/h. The estimated runoff depth from the catchment due to the above storm is (A) 10 mm (C) 20 mm (B) 16 mm (D) 23 mm

CE– 2008 13. A flood wave with a known inflow hydrograph is routed through a large reservoir. The outflow hydrograph will have. (A) Attenuated peak with reduced time – base (B) Attenuated peak with increased time – base


CE– 2009 Common Data for Questions 16 and 17 One hour triangular unit hydrograph of a watershed has the peak discharge of 60 m /sec.cm at 10 hour and time base of 30 hours. The index is 0.4 cm per hour and base follow is 15 m /sec. 16. The catchment area of the watershed is (A) 3.24 km (C) 324 km (B) 32.4 km (D) 3240 km 17.

If these is rainfall of 5.4 cm in 1 hour, the ordinate of the flood hydrograph at 15th hour is (A) 225 m /sec (C) 249 m /sec (B) 240 m /sec (D) 258 m /sec

18.

The relationship among specific yield (S ), specific retention (S ) and porosity ( ) of an aquifer is (A) S S (C) S S (B) S S (D) S S 2

th

th

th



CE– 2010 19. A well of diameter 20 cm fully penetrates a confined aquifer. After a long period pumping at a rate of 2720 liters per minute, the observations of drawdown taken at 10 m and 100 m distances from the center of the wall are found to be 3 m and 0.5 m respectively. The transmissivity of the aquifer is (A) 676 m /day (C) 526 m /day (B) 576 m /day (D) 249 m /day 20.

Match List-I with List-II and select the correct answer using the codes given below the lists: List-I A. Evapotranspiration B. Infiltration C. Synthetic Unit Hydrograph D. Channel Routing List-II 1. Penman method 2. Snyder’s method 3. Muskingum method 4. Horton’s method Codes: A B C D (A) 1 3 4 2 (B) 1 4 2 3 (C) 3 4 1 2 (D) 4 2 1 3

CE– 2011 21. In an aquifer extending over 150 hectare, the water table was 20 m below ground level. Over a period of time the water table dropped to 23 m below the ground level. If the porosity of aquifer is 0.40 and the specific retention is 0.15, what is the change in ground water storage of aquifer? (A) 67.5 ha.m (C) 180 ha.m (B) 112.5 ha.m (D) 450 ha.m

22.

23.

24.


A watershed got transformed from rural to urban over a period of time. The effect of urbanization on storm runoff hydrograph from the watershed is to (A) Decrease the volume of runoff (B) Increase the time to peak discharge (C) Decrease the time base (D) Decrease the peak discharge Common Data Questions for 23 and 24 The ordinates of 2- h unit hydrograph at 1 hour intervals starting from time t = 0, are 0, 3, 8, 6, 3, 2 and 0 m /s. Use trapezoidal rule for numerical integration, is required. What is the catchment area represented by the unit hydrograph? (A) 1.00 km (C) 7.92 km (B) 2.00 km (D) 8.64 km A storm of 6.6 cm occurs uniformly over the catchment in 3 hours. If – index is equal to 2mm/h and base flow is 5 m /s , what is the peak flow due to the storm? (A) 41.0 m /s (C) 53.0 m /s (B) 43.4 m /s (D) 56.2 m /s

CE– 2012 25. The ratio of actual evapo – transpiration to potential evapo – transpiration is in the range (A) 0.0 to 0.4 (C) 0.0 to 1.0 (B) 0.6 to 0.9 (D) 1.0 to 2.0

26.

th

Statement for Linked Answer Questions 26 & 27 The drainage area of a watershed is 50 km . The index is 0.5 cm/ hour and the base flow at the outlet is 10 m s. One hour unit hydrograph (unit depth = 1 cm) of the watershed is triangular in shape with a time base of 15 hours. The peak ordinate occurs at 5 hours. The peak ordinate (in m /s/cm) of the unit hydrograph is (A) 10.00 (C) 37.03 (B) 18.52 (D) 185.20 th

th



27.

For a storm of depth of 5.5 cm and duration of 1 hour, the peak ordinate (in m /s) of the hydrograph is (A) 55.00 (C) 92.60 (B) 82.60 (D) 102.60

28.

Group I contains parameters and Group II lists methods/ instruments. Group I Group II P. Stream flow 1. Anemometer velocity Q. Evapo – 2. Penman’s transpiration method rate R. Infiltration 3. Hortons’s rate method S. Wind 4. Current velocity meter The correct match of Group I & Group II is (A) P – 1, Q – 2, R – 3, S – 4 (B) P – 4, Q – 3, R – 2, S – 1 (C) P – 4, Q – 2, R – 3, S – 1 (D) P – 1, Q – 3, R – 2, S – 4

32.

30.

31.

A 1 hour rainfall of 10 cm magnitude at a station has return period of 50 year. The probability that 1 hour of rainfall of magnitude 10 cm or more will occur in each of two successive year is (A) 0.04 (C) 0.02 (B) 0.2 (D) 0.0004

CE– 2014 33. A conventional flow duration curve is a plot between (A) Flow and percentage time flow is exceeded (B) Duration of flooding and ground level elevation (C) Duration of water supply in a city and proportion of area receiving supply exceeding this duration (D) Flow rate and duration of time taken to empty a reservoir at that flow rate 34.

In reservoirs with an uncontrolled spillway, the peak of the plotted outflow hydrograph (A) lies outside the plotted inflow hydrograph (B) lies on the recession limb of the plotted inflow hydrograph (C) lies on the peak of the inflow hydrograph (D) is higher than the peak of the plotted inflow hydrograph

35.

An isolated 3-h rainfall event on a small catchment produces a hydrograph peak and point of inflection on the falling limb of the hydrograph at 7 hours and 8.5 hours respectively, after the start of the rainfall. Assuming, no losses and no base flow contribution, the time of concentration (in hours) for this catchment is approximately (A) 8.5 (C) 6.5 (B) 7.0 (D) 5.5

CE– 2013 29. An isohyet is a line joining points of (A) Equal temperature (B) Equal humidity (C) Equal rainfall depth (D) Equal evaporation Statement for Linked Answer Questions 30 & 31 At a station, storm I of 5 hour duration with intensity 2 cm/ h resulted in a runoff of 4 cm and Storm II of 8 hour duration resulted in a runoff of 8.4 cm. assume that the - index is the same of both the storms. The - index ( in cm/h) is : (A) 1.2 (C) 1.6 (B) 1.0 (D) 1.4


The intensity of storm II ( in cm/h) is : (A) 2.00 (C) 1.50 (B) 1.75 (D) 2.25 th

th

th



36.


The Muskingum model of routing a flood through a stream reach is expressed as O I I O , where k , k and are the routing coefficients for the concerned reach, I and I are the inflows to the reach, and O and O are the outflows from the reach corresponding to time steps 1 and 2 respectively. The sum of , and of the model is (A) 1 (C) 0.5 (B) 0.5 (D) 1


[Ans. B] Maximum intensities of rainfall for two consecutive 10min, are 2.2 mm/min and 1.5 mm/min Maximum intensity for 20 min is, .

.

Coefficient of permeability, K 30 m/day ∴ Discharge velocity V = K.i 30 0.01667 0.5 m/day Actual velocity or seepage velocity, .

V

1.85 mm/min

.

2m/day

n = porosity = 0.25

3.

[Ans. B] Area of UH = Area of catchment unit depth 1 80 60 60 Q 2 720 10 0.01 Q 50 m /sec

time of travel, t 1.5

6.

[Ans. A]

5.

[Ans. C] Hydraulic gradient, I

750

[Ans. D]

[Ans. A] Rainfall excess, R = P – t = 4 – 0.1 4 3.6 cm Peak of DRH = Peak of UH R = 50 3.6 180 m /s Peak of flood hydrograph = Peak of DRH + Base flow 180 30 210 m /s

4.

10 2

Intensity (m/hr)

2.

20 15.5 Horton’s infiltration curve

0

1 2 3 Time (hr)

Horton’s equation, f 6.8 8.7 e At t =0 hr f = 6.8 +8.7 e =15.5 mm/hr At t = 1 hr f = 6.8 +8.7 e =10 mm/hr t = 2 hr f = 6.8 +8.7 e =7.98 mm/hr t = 3 hr f = 6.8 +8.7 e =7.23 mm/hr

.

0.01667 th

th

th



The Horton’s curve is super imposed on the hyetograph. The hetched portion indicates run off. Area below the Horton’s curve from t = 1 hr to 3 hr is

9.

[Ans. A] In a confined aquifer, 2 kb s s Q log Q

∫ 6.8

8.7e


2 T s

s

log ( )

dt

Transmissibility, T = kb *6.8t 6.8

8.7e + 1 3

8.7e 6.8

T 1

8.7e

19.97 3.60 16.37mm(equal to infiltration loss from t = 1 hr to 3 hr) Total rain fall depth = 20 1 10 1 (in 2ndhr and 3rdhr) =30mm Rain fall excess = 30 16.37 =13.63 mm 7.

8.

[Ans. B] The saturation discharge four S – curve Q Lim Q Q 1m s Rainfall intensity cm 1 i 1 m s hr 360000 ∴ catchment are i Q 1 ⇒ catchment are 1 360000 ⇒ catchment are 360000 m 0.36 km [Ans. C] The duration of S - curve =1hr The ordinate of 2 - h OH is obtained by the procedure. Step 1: The ordinate of S - curve at t = 3h S 1 1 3 e 0.8 m s Step 2: The ordinate of S- curve at t = 3 – 2 = 1h S 1 1 1 e 0.26 m s Step 3: S S 0.8 0.26 0.54 m s

10.

Q. log ( ) 2

s

s

[Ans. D] Volume of UH = ∆t. ∑ 0 1 60 60 0 2 6 4 2 1 0 54000 m Area of UH = catchment area 0.01 m 54000 A 0.01 A 5.4 10 m 5.4 km

11.

[Ans. C] Method of superposition is used to derive 3hr UH from the given 1hr UH as follows. Time Ordinates DRH Ordinate of (h) of 1 h – of 3 3h. UH cm UH( ) ( ) in 0 hr 1 2 3h hr hr ( ) 0 1 2 3 4 5 6 7 8

12.

0 2 6 4 2 1 0

0 0 2 0 2 2 3 0.67 2 8 8 3 2.67 6 12 12 3 4 4 12 12 3 4 2 7 7 3 2.33 1 3 3 3 1 0 1 1 3 0.33 0 0 0 Thus at time t = 3 h, the ordinate of 3h – UH is 4 m s The ordinate of 3hr UH at t = 3 hr is 4m /sec 0 2 6 4 2 1 0

[Ans. C] -index = ⇒ 10

th

th

28 12 Runoff Time of rainfall excess th



But precipitation in the 1st and 4th hours is less than the -index i.e. the rainfall during these two hours is ineffective which means time of rainfall excess will be 2 hrs. 40 Runoff ∴ 10 2 ⇒ Runoff 20mm

17.


[Ans. B]

60 m S

Q 0 10

15

30

20

By interpolation, the ordinate of UH at 13.

[Ans. B]

15 hrs is :

14.

[Ans. A] Area of the aquifer = 5 km = 5 10 m Depth by which W.T is lowered = 102 99 =3m ∴ Volume of soil involved = 5 10 3 15 10 m Volume of water pumped out = 3 10 m

Rainfall excess (R)=P – t Ordinate of DRH at 15 hrs is : 45 255 m /s Ordinate of flood hydrograph = ordinate of DRH + Base flow 225 15 240m /s

Specific yield = 3 10 15 10 15.

16.

5

18.

[Ans. C]

19.

[Ans. D] The transmissivity of a confined aquifer is given by

0.2

[Ans. B] Volume of water required for evapotranspiration, 20 10 20 10 10 4 10 litres ∴ Dishcarge required at outlet, 4 10 20 24 60 60 2.31 liters sec

45 m /sec

T=

log

=

( )

.

log

(

)

= 249.4 m day 20.

[Ans. B]

21.

[Ans. B] S S m S 0.40 0.15 S 0.25 Specific yield,

[Ans. C] Volume of UH (m = Area of catchment m unit depth m 1 30 60 3600 A 0.01 2 A 324 10 m 324 m

S change in ground water storage = volume of water extracted, S total volume of aquifer 0.25 150 23 20 112.5 ha. m 22.

th

[Ans. C]

th

th



23.

[Ans. C] Volume of UH = ∆t. ∑ O 1 60 60 0 3 8 6 3 2 0 79200 m Area of UH (m Area of catchment (m 0.01 m Area of catchment =

.

.

7.92 10 m 7.92 km 24.

[Ans. A] (Required confirmation) Ordinates of 2-h unit hydrograph are given in the question. Therefore, the ordinates of 3 – hour unit hydrograph are to be derived by method of S – curve as shown below. Time 2S -Curve S - Curve, SA hr. Addition (3) (2) + (3) (1) UH (2) 0 0 0 1 3 3 2 8 0 8 3 6 3 9 4 3 8 11 5 2 9 11 6 0 11 11 7 11 11 8 11 9 11 SSA 3hr.U.H S - Curve, SB Curve, SB Lagged SB by 3hr . Lagged by 3hr 0 0 3 2 8 5.34 0 9 6 0 3 8 5.34 3 8 3 2 8 9 2 1.33 9 11 0 0 11


Peak of 3hr. U.H = 6 m /sec Peak of D.R.H = Peak of U.H R

P

t

R cm

6.6

R

6 cm Peak of D.R.H = 6 6 36 m /sec Peak of storm hydrograph = 36 + Base flow 36 5 41 m /sec 25.

[Ans. C]

26.

[Ans. B]

Discharge m s QP 0

5hr

15hr Time (hr) Volume of UH = Area of catchment 0.01 m 1 15 60 60 Q 50 10 0.01 2 Q 18.52 m /sec

27.

28.

[Ans. D] Rainfall excess, R = P – t 5.5 0.5 1 5 cm Peak of DRH = Peak of UH R 18.52 5 92.6 m /sec Peak of flood hydrograph = Peak of DRH + BF 92.6 10 102.60 m /sec [Ans. C]

29.

[Ans. C] Isohyet joins points of same rainfall depth.

30.

[Ans. A] Storm I D = 5h i = 2 cm/hr Runoff = 4cm Storm II D = 8h i =?

th

th

th




Run off = 8.4cm /hr 6 5 31.

Inflow hydrograph Recession limb Outflow hydrograph

1.2

[Ans. D] Let intensity of storm – II be P cm/hr P 8 8.4 1.2 8 1.2 8 8.4 P 8 P 2.25 cm /hr ∴ intensity 2.25 cm/hr [Ans. D] Return period of rainfall T=50 years ∴Probabilbity of occurrence once in 50 year, 1 p 0.02 5 Probability of occurrence in each of 2 Successive year p 0.02 0.004

33.

[Ans. A] A typical flow duration curve loops like

[Ans. D] In a small catchment Time of concentration = Lag time of peak flow = 7.0 – 1.5 = 5.5 hr

36.

[Ans. D]

Mean daily flow

32.

35.

Flow exceedence percentile ∴ A is best choice 34.

[Ans. B] Outflow hydrograph is similar to inflows hydrograph but with time lag ∴Peak of outflow hydrograph must lie on recession limits of plotted inflow hydrograph for uncontrolled spillway since time lag will be very less For controlled spillway, peak of outflow hydrograph may lie on recession limits or outside of plotted inflow hydrograph.

th

th

th



Environmental Engineering

Quality Standards of Water CE – 2005 1. If tomato juice is having a pH of 4.1 the hydrogen ion concentration will be (A) (B) (C) (D) CE – 2007 2. The alkalinity and the hardness of a water sample are 250mg/L and 350 mg/L as CaCO3 respectively. The water has (A) 350 mg/L carbonate hardness and zero non-carbonate hardness (B) 250mg/L carbonate hardness and zero non-carbonate hardness. (C) 250mg/L carbonate hardness and 350 mg/L non-carbonate hardness (D) 250 mg/L carbonate hardness and 100 mg/L non-carbonate hardness CE – 2009 Common Data for Q.No. 3 and 4 Following chemical species were reported for water sample from a well: Concentration Species (milli equivalent/L) Chloride ( ) 15 15 Sulphate ( ) 05 Carbonate ( ) Bicarbonate ( ) 30 12 Calcium ( ) 18 Magnesium ( ) pH 8.5 3. Total hardness in mg/L as CaCO3 is (A) 1500 (C) 3000 (B) 2000 (D) 5000 4.

Alkalinity present in the water in mg/L as CaCO3 is (A) 250 (C) 1750 (B) 1500 (D) 5000

CE – 2010 Common Data for Q.No. 5 and 6 Ion concentration obtained for a groundwater sample (having pH = 8.1) are given below. Ion Ion Atomic Weight concentration (mg/L) Ca2+ 100 Ca = 40 Mg2+ 6 Mg = 24 + Na 15 Na = 23 250 H = 1, C = 12 O =16 45 S = 12 O = 16 39 CI = 35.5 5. Total hardiness (mg/L as CaCO3) present in the above water sample is (A) 205 (C) 275 (B) 250 (D) 308 6.

Carbonate hardness (mg/L as CaCO3) present in the above water sample is (A) 205 (B) 250 (C) 275 (D) 289

CE – 2011 7. Anaerobically treated effluent has MPN of total coliform as 106/100mL. After chlorination, the MPN value declines to /100mL. The percent removal (%R) and log removal (log R) of total coliform MPN is (A) (%R) = 99.90; log R = 4 (B) (%R) = 99.90; log R = 2 (C) (%R) = 99.99; log R = 4 (D) (%R) = 99.99; log R = 2 CE – 2014 8. Some of the nontoxic metals normally found in natural water are (A) arsenic, lead and mercury (B) calcium, sodium and silver (C) cadmium, chromium and copper (D) iron, manganese and magnesium th

th

th





2.

3.

4.

5.

6.

[Ans. D] pH = log10 [ ] 4.1= log10 [ ] = [ ] = [ ] = 7.94×

[Ans. A] Total alkalinity in water consists of alkalinity caused by C ,H and A tittle negative alkalinity is also caused by H+ t ty , wt [ wt

.

[Ans. D] Carbonate hardness is equal to the total hardness or alkalinity whichever is lesser. In this case alkalinity (250 mg/L) is less than total hardness (350 mg/L) Carbonate hardness = 250 mg/L as CaC Non-carbonate hardness is total hardness in excess of the alkalinity i.e. Non-carbonate hardness, = Total hardness – alkalinity = 350 – 250 = 100 mg/L as CaC [Ans. A] Total hardness is mg/l as CaC =[ w t of CaC [ ]× combining weight of CaC = 12 × 50 + 18 × 50 = 600 + 900 = 1500 [Ans. C] Total alkalinity in mg/L as CaC =[ ] × combining weight of CaC +[ ] × combining weight of CaCO3+ {[ ]× combining weight of CaCO3 [H+] × combining weight of CaCO3} ([ ] and [ ] can be neglected for lower values) = 5 × 50 + 30 × 50 = 250 + 1500 = 1750

= 250 ×

= 205 mg /L

Carbonate hardness is equal to the total hardness or alkalinity whichever is less. The carbonate hardness of sample will be 205 mg/L N r t rd ss, = total hardness carbonate hardness = 275 205 = 70 mg/L

7.

[Ans. C] (

)

= 99.99%

8.

[Ans. D] Silver, lead, mercury, cadmium are most commonly present in industrial waste – water (effluent) and are toxic in nature Also calcium, sodium are not metals. Option D is most appropriate.

[Ans. C] Total hardness [

wt wt

[ wt wt

=6×

th

th

th




Water Supply and its Treatment CE – 2005 Statement for Linked Answer Q.No. 1 & 2 A city is going to install a rapid sand filter after the sedimentation tanks. Use the following data. Design loading rate to the filter 200m3/m2d Design flow rate 0.5m3/s Surface area per filter box 50m2 1. Surface area required for the rapid sand filter will be (A) 210 (C) 216 (B) 215 (D) 218 2.

The no. of filter required (A) 3 (C) 6 (B) 4 (D) 8

3.

Match List-I with List-II and select the correct answer using the code given below the lists List I List II A. Release 1. Reduce high valve inlet pressure to lower outlet pressure B. Check valve 2. Limit the flow of water to single direction C. Gate valve 3. Remove air from the pipeline D. Pilot valve 4. Stopping the flow of water in the pipeline Codes A B C D (A) 3 2 4 1 (B) 4 2 1 3 (C) 3 4 2 1 (D) 1 2 4 3

4.

List-I contains some properties of water/waste water and List-II contains list of some tests on water /waste water. Match List-I with List-II and select the correct answer using the codes given below the lists List 1 (properties) List 2 (Test) (A) Suspended 1) BOD solid concentration (B) Metabolism of 2) MPN Biodegradable organics (C) Bacterial 3) Jar test concentration (D) Coagulant 4) Turbidity dose Codes A B C D (A) 2 1 4 3 (B) 4 1 2 3 (C) 2 4 1 3 (D) 4 2 1 3

5.

1 TCU is equivalent to the color produced (A) 1 mg/L of chloroplatinate ion (B) I mg/L of platinum ion (C) 1 mg/L platinum in form of chloroplatinate ion (D) 1 mg/L of organo – chloroplatinate ion

CE – 2007 Common Data Questions Q No 6 & Q No 7 A plain sedimentation tank with a length of 20m, width of 10m and a depth of 3m is used in a water treatment plant to treat 4 million litres of water per day(4 MLD). The average temperature of water is C. The dynamic viscosity of water is 1.002 × 10-3 N.s/m2 at C. Density of 3 water is 998.2 kg/m . Average specific gravity of particle is 2.65

th

th

th



6.

What is the surface overflow rate in the sedimentation tank? (A) 20 m3/m2/day (B) 40 m3/m2/day (C) 67 m3/m2/day (D) 133 m3/m2/day

7.

What is the minimum diameter of the particle which can be removed with 100% efficiency in the above sedimentation tank? (A) 11.8 × 10-3 mm (C) 50 × 10-3 mm -3 (B) 16.0 × 10 mm (D) 160 × 10-3 mm

list of titrants Group – I Group – II P. Alkalinity 1. N/35.5 AgNO3 Q. Hardness 2. N/40 Na2S2O3 R. Chloride 3. N/50 H2SO4 S. Dissolved 4. N/50 EDTA oxygen The correct match of water quality parameters in Group –I with titrants in Group – II is: (A) P – 1, Q – 2, R – 3, S – 4 (B) P – 3, Q – 4, R – 1, S – 2 (C) P – 2, Q – 1, R – 4, S – 3 (D) P – 4, Q – 3, R – 2, S – 1

.

CE – 2011 8. Consider the following unit processes commonly used in water treatment; rapid mixing (RM), flocculation (F), primary sedimentation (PS), secondary sedimentation (SS), chlorination (C) and rapid sand filtration (RSF). The order of these unit processes (first to last) in a conventional water treatment plant is (A) P → F → F → → → (B) P → F → → F→ → (C) P → F → → F → → (D) P → →F→ → F→ CE – 2012 9. A town is required to treat 4.2 min of raw water for daily domestic supply. Flocculating particles are to be produced by chemical coagulation. A column analysis indicated that an overflow rate of 0.2 mm/s will produce satisfactory particle removal in a settling basin at a depth of 3.5 m. The required surface area (in m2) for settling is (A) 210 (C) 1728 (B) 350 (D) 21000


11.

A water treatment plant, having discharge 1 s has 14 filter to treat the water. Each filter is having area, but due to backwashing activity 2 filters are non operational. Calculate hydraulic loading rate in

CE – 2014 12. 16 MLD of water is flowing through a 2.5 km long pipe of diameter 45 cm. The chlorine at the rate of 32 kg/d is applied at the entry of this pipe so that disinfected water is obtained at the exit. There is a proposal to increase the flow through this pipe to 22 MLD from 16 MLD. Assume the dilution coefficient, n = 1. The minimum amount of chlorine (in kg per day) to be applied to achieve the same degree of disinfection for the enhanced flow is (A) 60.50 (C) 38.00 (B) 44.00 (D) 23.27 13.

CE – 2013 10. Some of the water quality parameters are measured by titrating a water sample with a titrant. Group-I gives a list of parameters and Group-II gives the th

The potable water is prepared from turbid surface water by adopting the following treatment sequence. (A) ur d sur w t r→ u t → F u t → d t t → F tr t →Ds t → t r & Supply

th

th



(B)

ur d sur w t r→Ds t → F u t → d t t → F tr t → u t → t r & Supply (C) ur d sur w t r → F tr t → d t t → Ds t → F u t → u t → Storage & Supply (D) ur d sur w t r → d t t → F u t → Coagulation → D s t → F tr t → t r & upp y 14.

16.

A suspension of sand like particles in water with particles of diameter 0.10 mm and below is flowing into a settling tank at s . Assume g = 9.81 s , specific gravity of particles = 2.65, and kinematic viscosity of water = /s. The minimum surface area (in ) required for this settling tank to remove particles of size 0.06 mm and above with 100% efficiency is _______________

17.

A surface water treatment plant operates round the clock with a flow rate of 35 /min. The water temperature is and jar testing indicated an alum dosage of 25 mg/l with flocculation at a Gt value of producing optimal results. The alum quantity required for 30 days (in kg) of operation of the plant is ____________

18.

A conventional flow duration curve is a plot between (A) Flow and percentage time flow is exceeded (B) Duration of flooding and ground level elevation (C) Duration of water supply in a city and proportion of area receiving supply exceeding this duration (D) Flow rate and duration of time taken to empty a reservoir at that flow rate

For a sample of water with the ionic composition shown in the figure below, the carbonate and non - carbonate hardness concentrations (in mg/ as CaCO3), respectively are: 7 N

0

(A) 200 and 50 (B) 175 and 75 15.

3.5

(C) 75 and 175 (D) 50 and 200

7


A straight 100 m long raw water gravity main is to carry water from an intake structure to the jack well of a water treatment plant. The required flow through this water main is 0.21 /s. Allowable velocity through the main is 0.75 m/s. Assume s . The minimum gradient (in cm/100 m length) to be given to this gravity main so that the required amount of water flows without any difficulty is ___________

th

th

th





[Ans. C]

7.

t sur

r

D s D s

[Ans. B]

wr t d r t

Settling velocity = v u r t

ut t

= t w

= 12960 seconds s 2.

[Ans. C]

Let the velocity of particle which can be removed with 100% efficiency be then

No. of filter =

100 = 100 ( -1)

As per the options select 6 filters. ∵ t rs > 3.

4.

5.

6.

[Ans. B] Properties Suspended solid concentration Metabolism of biodegradable organics Bacterial concentration Coagulant dose

Test Turbidity

MPN Jar test

[Ans. D]

9.

[Ans. B] Q= /min /s=

/min

10.

[Ans. B] Alkalinity is measured by acid base titration & hardness by EDTA. P So, } pt s rr t

11.

[Ans. *]Range 143 to 145 ydr u

d

r t s

= /

= 0.2315

mm

8.

[Ans. A] Surface overflow rate = 20

1) d2

d = 2.58 d = 1.6 = 16

BOD

[Ans. C] 1 TCU is the colour produced by 1 mg of platinum – cobalt in the form of choroplatinate ions dissolved in 1 litre of distilled water

=

(2.65



[Ans. A] All of these are pipe appurtenances which are required for the proper functioning of the pipeline.

= 0.2315

d y

/day d y

th

th

th



12.

[Ans. A] For disinfection, we have t Where t = time required to kill all organisms c = concentration of disinfectant n = dilution coefficient k = constant t t r t t t

t v

ty

w

t

⁄s

d d d u 16.

w

ds

[Ans. *] Range 4.7 to 4.9 s ⁄s

r d

t

pp

ds

w

15.


r

t

p rd y

[Ans. *] Range 31.0 to 32.0 d

s

tp rd y BL

(

)

(

= 31.21

)

d y 13.

[Ans. A]

14.

[Ans. B] Carbonate hardness CH ⁄ s N r t rd ss N t rd s r t s

rd

ss

17.

[Ans. 37800] Given Q = 35 Gt = Alum dosage = 25 mg/ltr Alum quantity for 30 days

18.

[Ans. A]

⁄ s

th

th

th




Waste Water Treatment CE – 2005 1. In a certain situation waste water discharges into a river mixes with the river water instantaneously and completely. Following is the data available. Waste water DO = 2.00 mg/L Discharge rate = 1.10 m3/s River water DO = 8.3 mg/L Flow rate = 8.70 m3/s Temperature = 20oC Initial amount of DO in the mixture of waste and river shall be (A) 5.3 mg/L (C) 7.6 mg/L (B) 6.5 mg/L (D) 8.4 mg/L 2.

Total Kjeldahl nitrogen is a measure of (A) Total organic nitrogen (B) Total organic and ammonia nitrogen (C) Total ammonia nitrogen (D) Total inorganic and ammonia nitrogen

3.

In aerobic environment, nitrosomonas convert (A) N t N (C) N t N (B) N t N (D) N t N

CE – 2006 4. A synthetic sample of water is prepared by adding 100 mg Kaolinite (a clay mineral), 200 mg glucose, 168 mg NaCl, 120 mg MgSO4, and 111 mg CaCl2 to 1 liter of pure water. The concentration of total solids (TS) and fixed dissolved solids (FDS) respectively in the solution in mg/L are equal to (A) 699 and 599 (C) 699 and 199 (B) 599 and 399 (D) 699 and 399

[ ⁄ [ . Water pH is 7 Atomic weights: Ca:40;Mg: 24; Al: 27; H: 1; C: 12; O: 16; Na: 23; Cl: 35.5 The total hardness of the sample in mg/L as Ca is (A) 484 (C) 242 (B) 450 (D) 225

5.

6.

The non - carbonate hardness of the sample in mg/L as is (A) 225 (C) 86 (B) 156 (D) 0 Common Data for Question for 7 & 8 In a rapid sand filter, the time for reaching particle break through is defined as the time elapsed from start of filter run to the time at which the turbidity of the effluent from the filter is greater than 2.5 NTU. The time for reaching terminal head loss is defined as the time elapsed from the start of the filter run to the time when head loss across the filter is greater than 3m.

7.

The effect of increasing the filter depth (while keeping all other condition same) on and s (A) increases and decreases (B) Both and increases (C) decreases and increases (D) t and decreases

8.

The effect of increasing the filer loading rate (while keeping all other condition same) on and is (A) increases and decreases (B) t and increase (C) decreases and increases (D) Both and decreases

Linked Data for Q.No. 5 and Q.No. 6 A water contains the following dissolved ions [N ⁄ [ ⁄ [ ⁄ [ ⁄ th

th

th



9.

10.

To determine the BOD5 of a waste water sample, 5, 10 and 50 mL aliquots of the waste water were diluted to 300 mL and incubated at in BOD bottles for 5 days. The results were as follows. S.No. Waste- Initial DO water DO, after volume, mg/L 5 mL days, mg/L 1. 5 9.2 6.9 2. 10 9.1 4.4 3. 50 8.4 0.0 Based on the data, the average D of the waste water is equal to (A) 139.5 mg/L (C) 109.8 mg/L (B) 126.5 mg/L (D) 72.2 mg/L The composition of a certain MSW sample and specific weights of its various components are given below: Component Percent Specific by weight weight (kg/ ) Food 50 300 Dirt and 30 500 Ash Plastics 10 65 Wood and 10 125 Yard waste Specific weight (kg/ ) of the MSW sample is (A) 319 (C) 209 (B) 217 (D) 199

CE – 2007 Common Data Question Q No 11 & Q No 12 A completely mixed activated sludge process is used to treat a wastewater flow of 1 million litres per day (1 MLD) having a BOD5 of 200 mg/L. The biomass concentration in the aeration tank is 2000 mg/L and the concentration of the net biomass leaving the system is 50 mg/L. the aeration tank has a volume of 200 m3


11.

What is the hydraulic retention time of the wastewater in aeration tank? (A) 0.2h (C) 10h (B) 4.8h (D) 24h

12.

What is the average time for which the biomass stays in the system? (A) 5h (C) 2 days (B) 8h (D) 8 days

13.

The presence of hardness in excess of permissible limit causes (A) Cardio – vascular problem (B) Skin discolouration (C) Calcium deficiency (D) Increased laundry expenses

14.

50 g of d 25 g of are produced from the decomposition of municipal solid waste (MSW) with a formula weight of 120 g. What is the average per capita greenhouse gas production in a city of 1 million people with a MSW production rate of 500 ton/day? (A) 104 g/day (C) 208 g/day (B) 120 g/day (D) 313 g/day

CE – 2008 15. Match group 1 (Terminology) with Group 2 (Definition / Brief Description) for wastewater treatment system Group 1 Group 2 P. Primary 1. Contaminant removal by treatment physical forces Q. Secondary 2. Involving biological treatment and/or chemical reaction R. Unit 3. Conversion of soluble operation organic matter to biomass S. Unit 4. Removal of solid process materials from incoming wastewater (A) P-4,Q-3, R-1, S-2 (B) P-4,Q-3, R-2, S-1 (C) P-3,Q-4, R-2, S-1 (D) P-1,Q-2, R-3, S-4 th

th

th



16.

The 5 – day BOD of a wastewater sample is obtained as 190mg/L (with k = 0.01 ). The ultimate oxygen demand (mg/L) of the sample will be (A) 3800 (C) 272 (B) 475 (D) 190

17.

Group 1 lists estimation method of some of the water and wastewater quality parameter. Group 2 lists the indicators used in the estimation methods. Match the estimation method (Group 1) with the corresponding indicator (group 2) Group 1 Group 2 P. Azide modified 1. Eriochrome Winkler method Black T for dissolved oxygen Q. Dichromate 2. Ferrion method for Chemical oxygen demand R. EDTA titrimetric 3. Potassium method for chromate hardness S. Mohr or 4. Starch Argentometric method for chlorides (A) P-3,Q-2, R-1, S-4 (B) P-4,Q-2, R-1, S-3 (C) P-4,Q-1, R-2, S-3 (D) P-4,Q-2, R-3, S-1

18.

A wastewater sample contains 10–5.6 mmol/l of O ions at 250C. The pH of this sample is (A) 8.6 (C) 5.6 (B) 8.4 (D) 5.4

19.

A water treatment plant is required to process 28800 d of raw water (density = 1000 kg/ , kinematic viscosity = s). The rapid mixing tank imparts a velocity gradient of to blend 35 mg/L of alum with the flow for a detention time of 2 minutes. The power input (W) required for rapid mixing is (A) 32.4 (C) 324 (B) 36 (D) 32400


CE – 2009 20. Match the following Column 1 Column 2 P. Grit chamber 1. Zone settling Q. Secondary 2. t ’s w settling tank R. Activated 3. Aerobic sludge process S. Trickling 4. Contact filter stabilization The correct match of the column 1 with column 2 is (A) P-1, Q-2, R-3, S-4 (B) P-2, Q-1, R-3, S-4 (C) P-1, Q-2, R-4, S-3 (D) P-2, Q-1, R-4, S-3 21.

An aerobic reactor receives wastewater at a flow rate of 500 m3/d having a COD of 2000 mg/L. The effluent COD is 400 mg/L. Assuming that wastewater contains 80% biodegradable waste, the daily volume of methane produced by the reactor is (A) 0.224m3 (C) 224m3 (B) 0.2804m3 (D) 280m3

CE – 2010 22. If the BOD3 of a wastewater sample is 75 mg/L and reaction rate constant k (base e) is 0.345 per day, the amount of BOD remaining in the given sample after 10 days is (A) 3.12 mg/L (C) 3.69 mg/L (B) 3.45 mg/L (D) 3.92 mg/L CE – 2011 23. Chlorine gas (8mg/L as Cl2) was added to a drinking water sample. If the free chlorine residual and pH was measured to be 2 mg/L (as Cl2) and 7.5, respectively, what is the concentration of residual OCI – ions in the water? Assume that the chlorine gas added to the water is

th

th

th



completely converted to HOCI and OCI–. Atomic Weight of Cl: 35.5 Given OCI + H+ ⃗⃗⃗ HOCl, K = 107.5 (A) 1.408 × 10–5 moles/L (B) 2.817 × 10–5 moles/L (C) 5.634 × 10–5 moles/L (D) 1.127 × 10–4 moles/L 24.

Total suspended particulate matter (TSP) concentration in ambient air is to be measured using a high volume sampler. The filter used for this purpose had an initial dry weight of 9.787 g. The filter was mounted in the sampler and the initial air flow rate through the filter was set at 1.5 m3/min. Sampling continued for 24 hours. The airflow after 24 hours was measured to be 1.4 m3/min. The dry, weight of the filter paper after 24 hour sampling was 10.283 g. Assuming a linear decline in the air flow rate during sampling, what is the 24 hour average TSP concentration in the ambient. Air? 3. 3. (A) μ (C) μ 3 3. (B) μ . (D) μ

CE – 2012 27. Assertion [a]: At a manhole, the crown of the outgoing sewer should not be higher than the crown of the incoming sewer. Reason [r]: Transition from a larger diameter incoming sewer to a smaller diameter outgoing sewer at a manhole should not be made. The CORRECT option evaluating the above statements is (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 true but [r] is false Common Data Question 28 and 29 An activated sludge system (sketched below) is operating at equilibrium with the following information. Wastewater related data: flow rate = 500 m3/hour, influent BOD = 150 mg/L, effluent BOD = 10 mg/L. Aeration tank related data: hydraulic retention time = 8 hours, mean-cell-residence time = 240 hours, volume = 4000 mixed liquor suspended solids, = 2000 mg/L.

Linked Data Question for 25 & 26 The sludge from the aeration tank of the activated sludge process (ASP) has solids content (by weight) of 2%. This sludge is put in a sludge thickener, where sludge volume is reduced to half. Assume that the amount of solids in the supernatant from the thickener is negligible, the specific gravity of sludge solids is 2.2 and the density is 1000 kg/ 25.

What is the density of the sludge removed from the aeration tank? (A) 990 kg/ (C) 1011 kg/ (B) 1000 kg/ (D) 1022 kg/

26.

What is the solids content (by weight) of the thickened sludge? (A) 3.96% (C) 4.04% (B) 4.00% (D) 4.10%


Influent

Aeration Tank

Secondary Effluent Clarifier Sludge Recycle Solids Wasted

28.

th

The food-to-biomass (F/M) ratio (in kg BOD per kg biomass per day) for the aeration tank is (A) 0.015 (C) 0.225 (B) 0.210 (D) 0.240

th

th



29.

The mass (in kg/day) of solids wasted from the system is (A) 24000 (C) 800 (B) 1000 (D) 33

33.

CE – 2013 30. A student began an experiment for determination of 5day, BOD on th Monday. Since the 5 day fell on Saturday,the final DO readings was taken on next Monday. On calculation, BOD (i.e. 7 day, )was found to be 150 mg/L. What would be the 5 day, BOD (in mg/L)? Assume value of BOD rate constant (k) at standard temperature of as 0.23/day (base e)._____________


A waste water stream w s ut t D is joining a small river (flow=12 /s, ultimateBOD=5mg/l). Both water streams get mixed up instantaneously. Cross- sectional area of the river is 50 . Assuming the de-oxygenation rate constant, K = 0.25/day, the BOD (in mg/l) of the river water, 10 km downstream of the mixing point is (A) 1.68 (C) 15.46 (B) 12.63 (D) 1.37

CE – 2014 31. The dominating microorganisms in an activated sludge process reactor are (A) aerobic heterotrophs (B) anaerobic heterotrophs (C) autotrophs (D) phototrophs 32.

An effluent at a flow rate of 2670 /d from a sewage treatment plant is to be disinfected. The laboratory data of disinfection studies with a chlorine dosage of 15 mg/l yield the model N N where N = number of micro-organisms surviving at time t (in min.) and N = number of microorganisms present initially (at t = 0). The volume of disinfection unit (in ) required to achieve a 98% kill of microorganisms is ______________

th

th

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[Ans. C] D

D

2.

[Ans. A]

4.

[Ans. A] Total solids

[Ans. C] If filter depth is increased, the static head of the filter increases thus it takes more time to reach a head loss of 3 m. also, the time required to reach a turbidity of 2.5 NTU will decrease as there will be a large area for suspended particles to be entrapped in it.

8.

[Ans. D] If the filter loading increases, time for reaching terminal head will decrease and same will happen with the time of reaching particle break through.

9.

[Ans. C] BOD in mg/L = [initial DO – Final DO] × dilution factor

D

[Ans. B] The sum total of organic nitrogen and ammonia nitrogen is called Kjeldahl

3.

7.

Fixed dissolved solids = Total solid – 100 = 699 – 100 = 599 mg/L 5.

[Ans. C] Total hardness wt wt

[

[

D

[

D

[

D

[ wt wt

v r

D = 109.8 mg/L

[ (

) ⁄

6.

(

)

10.

s

[Ans. A] Specific weight of the MSW sample (

)

[Ans. C] Carbonate hardness wt [ wt

⁄ s Non carbonate hardness = Total hardness – Carbonate hardness ⁄ s

( ( (

11.

) ) )

[Ans. B] The hydraulic retention time is given by t = × 24 Where V = vol of the aeration tank

th

th

th



Q = quality of wastewater flow into the aeration tank excluding the quality of recycled sludge

where, p[ p[ ut P p p

t= 12.

[Ans. D] The Average time for which the biomass stays in the system is known as sludge age ( ) and it is given by

19.

= = = 8 days 13.

[Ans. D]

14.

[Ans. D] We know that both d are green house gases. Total green house gases produced by 120 gm of MSW


is in mol/l.

P –

[Ans. D] The power input (W) required for rapid mixing is expressed in terms of temporal mean velocity gradient, G, expressed by the equation P [ ] P μ μ r μ dy v s s ty r w water = kinematic viscosity × density N s s t = 2 minutes = 120 seconds Discharge of raw water s Volume of water in tank,

500 ton of MSW will produce green house gases

P = 32400 watts

d y Per capita average production of green house gases d y d y

20.

[Ans. B]

21.

[Ans. D] Q = 500 m3/day = 0.5 MLD Influent COD: 2000 mg/L Effluent COD: 400 mg/L COD removed = 2000 – 400 = 1600 mg/L Biodegrades in COD = 1600 0.8 = 1280 mg/L

d y

15.

[Ans. A]

16.

[Ans. C] BOD of t days may be given as Lt = L (1 ) [ –

d r d s ssu d t s t v t s w st w t r w

L= 17.

[Ans. B]

18.

[Ans. D] p[ p[

Gas

produced

@

Methane gas

0.9m3/kg t

t t

d y of VSS s

[

th

th

th



22.


[Ans. C] The BOD at any instant is given by

Secondary Effluent clarifies

Aeration Tank

Solid wasted

F D

t r

= 0.225 kg Biomass per day

d ys 29.

u t

[Ans. C] Sludge

age,

or

sludge

residence

Dr since

is not given

is in days 23.

[Ans. B]

(

24.

[Ans. C]

Mass of solids,

v r

r

mg/day = 800 kg/day

wr t

(

30.

)

[Ans. *]Range 127 to 132 BOD5= ( ) D

Total airflow through the sampler during 24 hours

P

tr t

(

)

t

(

)

(

)

D D

t r

)

D

μ 25.

[Ans. C]

31.

[Ans. A]

26.

[Ans. A]

32.

27.

[Ans. D]

28.

[Ans. C] Given, Flow rate ur Influent BOD t Effluent BOD, = 10 mg/lit

[Ans. *] Range 49.0 to 51.0 N N w r Nt = No.of micro org @ time t No = No.of micro org @ time t = 0 For 98% kill means, 2% of bacteria will remain

Hydraulic retention time

ur

t

d y

t sd s

Log (

Mean - cell resistance time, = 240 hour = 10 day M.L.S.S, (X) = 2000 mg/lit.

)

t = 26.97 min Discharge,Q = th

th

th




Volume = Discharge = Volume = 50 33.

[Ans. C] Du

⁄s t s

D

ty

v r

w s

t

t tr v s

d

th

th

th




Sludge Disposal CE – 2005 1. Match the following List I List II A Thickening 1. Decrease in of sludge volume of sludge by chemical oxidation B Stabilization 2. Separate of water of sludge by heat and chemical treatment C Conditioning 3. Digestion of of sludge Sludge D Reduction of 4. Separation of sludge water by floatation or gravity Codes: A B C D (A) 4 3 1 2 (B) 3 2 4 1 (C) 4 3 2 1 (D) 2 1 3 4 2.

4.

Direction: determine the correctness or otherwise of the following Assertion [A] and the Reason [R]: Assertion: The crown of the outgoing larger diameter sewer is always matched with the crown of incoming smaller diameter sewer Reason: It eliminates backing up of sewage in the incoming smaller diameter sewer. (A) Both [a] & [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 true but [r] is false

CE – 2013 5. A settling tank in a water treatment plant is designed for a surface overflow rate of 30

sediment particles = 2.65, density of water , dynamic viscosity of water μ Ns , d st s’ w s v d ppr x t minimum size of particles that would be completely removed is : (A) 0.01mm (C) 0.03 mm (B) 0.02 mm (D) 0.04 mm

Bulking sludge refers to having (A)

. Assume specific gravity of

< 0.3/d

(B) 0.3/d < F/M < 0.6/d (C) F/M = zero (D) F/M > 0.6/d CE – 2008 3. Two biodegradable components of municipal solid waste are (A) Plastic and wood (B) Cardboard and glass (C) Leather and tin cans (D) Food wastes and garden trimmings

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[Ans. C]

2.

[Ans. A]

3.

[Ans. D] Biologically active or biodegradable components of wastes oxidise by bacteria. Food wastes and garden trimmings, both are organic wastes which are biodegradable.

4.

[Ans. A]

5.

[Ans. B] d

μ

Surface overflow rate = s

→d

th

th

th




Domestic Waste Water Treatment CE – 2005 1. A Circular primary clarifier processes an average flow of 5005 /d of municipal waste water. The over flow rate is 35 / /d. The diameter of clarifier shall be (A) 10.5 m (C) 12.5 m (B) 11.5 m (D) 13.5 m CE – 2009 2. A horizontal flow primary clarifier treats wastewater in which 10%, 60% and 30% of particles have settling velocities of 0.1 mm/s, 0.2 mm/s and 0.1mm/s respectively. What would be the total percentage of particles removed if clarifier operates at a Surface Overflow Rate (SOR) of 43.2 -d? (A) 43% (C) 86% (B) 56% (D) 100% CE – 2010 3. A Coastal city produces municipal solid waste (MSW) with high moisture content, high organic materials. Low calorific value and low inorganic materials. The most effective and sustainable option for MSW management in that city is (A) Composting (B) Dumping in sea (C) Incineration (D) Landfill

CE – 2012 4. A sample of domestic sewage is digested with silver sulphate, sulphuric acid, potassium dichromate and mercuric sulphate in chemical oxygen demand (COD) test. The digested sample is then titrated with standard ferrous ammonium sulphate (FAS) to determine the un – reacted amount of (A) Mercuric sulphate (B) Potassium dichromate (C) Silver sulphate (D) Sulphuric acid 5.

A water sample has a pH of 9.25. The concentration of hydroxyl ions in the water sample is (A) 10 9.25 moles/L (B) 10 4.75 moles/L (C) 0.302 mg/L (D) 3.020 mg/L

CE – 2013 6. Elevation and temperature data for places are tabulated below: Elevation, m Temperature, 4 21.25 444 15.70 Based on the above data, lapse rate can be referred as: (A) Super – adiabatic (B) Neutral (C) Sub – adiabatic (D) Inversion

th

th

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[Ans. D] Area of clarifier = =

= 146

D = 143 D = 2.

= D = 13.5m

[Ans. B] Surface overflow rate d = 43.2 m/d s It means that particles which have setting velocities more that surface overflow rate will be 100% removed. t p r t p rt s r v d

3.

[Ans. A] Incineration can be adopted when the calorific value of the MSW is high. Landfill can be adopted when the density of MSW is high. Since in the given MSW the quantity of inorganic material is low. Its density is less compositing can be adopted when the MSW contains high organic content. Barging the MSW into sea is now a days generally not used and has becomes obsolete

4.

[Ans. B]

5.

[Ans. C] pH = 9.25 p0H = 14 – 9.25 =4.75 = mol/l

/l 6.

[Ans. A] th

th

th




Air Pollution is the overall efficiency of the system for the same d ? (A) 100% (C) 80% (B) 93% (D) 65%

Percent of time greater than stated value

CE – 2006 1. The cumulative noise power distribution curve at a certain location is given below. 100

0

50

The value of (A) 90 dBA (B) 80 dBA 2.

Noise Level, dBA 100

is equal to (C) 70 dBA (D) 60 dBA

The mean indoor airborne Chloroform concentration in a room was determined to be 0.4 μ Use the following data T = 293K, P = 1 Atmosphere, R = 82.05 × atm/mol-K. Atomic weights: C = 12, H=1, Cl = 35.5. This concentration expressed in parts per billion (volume basic, ppbv) is equal to (A) 1.00 ppbv (C) 0.10 ppbv (B) 0.20 ppbv (D) 0.08 ppbv

CE – 2007 3. The dispersion of pollutants in atmosphere is maximum when (A) Environment lapse rate is greater than adiabatic lapse rate (B) Environment lapse rate is less than adiabatic lapse rate (C) Environment lapse rate is equal to adiabatic lapse rate (D) Maximum mixing depth is equal to zero 4.

CE – 2008 5. Two primary air pollutants are (A) Sulphur oxide and ozone (B) Nitrogen oxide and peroxyacetylnitrate (C) Sulphur oxide and hydrocarbon (D) Ozone and peroxyacetylinitrate CE – 2009 6. The reference pressure used in the determination of sound pressure level is (A) 20 μP (C) 10 μP (B) 20 d (D) 10 d 7.

Particulate matter (fly ash) carried in effluent gases from the furnaces burning fossil fuels are better removed by (A) Cotton bag house filter (B) Electrostatic precipitator (ESP) (C) Cyclone (D) Wet scrubber

8.

Match List – I with List – II and select the correct answer by using the codes given below the lists: List – I List – II A. Coriolis effect 1. Rotation of earth B. Fumigation 2. Lapse rate and vertical temp. profile C. Ozone layer 3. Inversion D. Max. mixing 4. Dobson depth (mixing height) Codes: A B C D (A) 2 1 4 3 (B) 2 1 3 4 (C) 1 3 2 4 (D) 1 3 4 2

Two electrostatic precipitators (ESPs) are in series. The fractional efficiencies of the upstream and downstream ESPs for size d are 80% and 65%, respectively. What

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CE – 2010 9. According to the Noise pollution (Regulation and Control) Rules, 2000, of the Ministry of Environment and Forests, India, the day time and night time noise level limits in ambient air for residential area expressed in dB(A) are (A) 50 and 40 (C) 65 and 55 (B) 55 and 45 (D) 75 and 70 10.

An air parcel having temperature moves from ground level to 500 m elevation in dry air following the “ d t ps r t ” r su t temperature of air parcel at 500 m elevation will be (A) (C) (B) (D)


CE – 2011 11. Consider four common air pollutants found in urban environment, N Soot and . Among these which one is the secondary air pollutant? (A) (C) (B) N (D) t CE – 2014 12. The two air pollution control devices that are usually used to remove very fine particles from the flue gas are (A) Cyclone and Venturi Scrubber (B) Cyclone and Packed Scrubber (C) Electrostatic Precipitator and Fabric Filter (D) Settling Chamber and Tray Scrubber 13.

The amount of generated (in kg) while completely oxidizing one kg of to the end products is ____________


[Ans. C] is the sound pressure level in dB which is exceeded for 40% of the gauging time.

v u t = 22.4 L/mol Given P t

Now, slope of the given

∵

d

y v

r

s p r

μ ⁄

dr ’s w w

wt

in ppm = pp v pp v 3.

[Ans. A] When the environment lapse rate is more than the adiabatic lapse rate, a rising parcel of warm lighter air (pollutants) will continue to lift up; where as parcel of

t

P r P

sp

pp

[Ans. D] The relation between μ and ppm is 1 ppm × molecular mass of ⁄ p ut t μ ⁄ p ut t t v

P

t

P P

∵ Curve is a straight line, hence slope is constant

2.

t

atmosphere; th

th

th



heavier cooler air will continue to come down. In such circumstances the environment is unstable and the dispersion of pollutants will be rapid due to marked vertical mixing of the air 4.

[Ans. B] Given data: Since the efficiency of upstream ESP is 80%, only 20% of the particulate are not removed. These remaining particulates will face the downstream ESP whose efficiency is 65%. Particulates removed by downstream

the Ministry of Environment & Forest, India are tabulated below: Category of Day Night Area/Zone time time Industrial 75 70 Area Commercial 65 55 Area Residential 55 45 Area Silence 50 40 Zone 10.

[Ans. A] Dry air cools at the rate of per km and it is called dry adiabatic lapse rate. In saturated (wet)air, this rate is calculated to be per km and is known as wet adiabatic lapse rate. Resulting temperature of air

11.

[Ans. A] Soot is a general term that refers to impure carbon particles resulting from the incomplete combustion of hydrocarbons. Soot, as an air borne contaminant in the environment has many different sources but they are all result of some form of pyrolysis.

12.

[Ans. C]

13.

[Ans. *] Range 2.7 to 2.8 Oxidization reaction of CH4 → i.e 1 mole of → → → → 2.75 is correct response

P The two ESPs are connected in series, therefore the overall efficiency =80 + 13 = 93%. 5.

6.

[Ans. C] Sulphur dioxide carbon monoxide, nitrogen oxides, lead, hydrocarbons, allergic agents like pollens and spores and radioactive substances are primary pollutants. Sulphuric acid, ozone, formaldehydes & peroxyacylnitrates (PAN) are secondary pollutants. [Ans. A] The sound pressure of the faintest sound that can be heard by a normal healthy individual is about 20 μP . Hence, this pressure is used as reference pressure in determination of sound pressure level.

7.

[Ans. B]

8.

[Ans. D]

9.

[Ans. B] Ambient air quality standards in respect of Noise as per Noise pollution (Regulation and Control) Rules, 2000, of


th

th

th



Structural Analysis

Trusses and Arches CE – 2005 Common Data for Q.No.1 & Q.No.2 A truss is shown in the figure. Members are of equal cross section A and same modulus of elasticity E. A vertical force P is applied at point C

4.

The members EJ and IJ of a steel truss shown in the figure below are subjected to a temperature rise . The coefficient of thermal expansion of steel is 0.000012 pe per unit length. The displacement (mm) of joint E relative to joint H along the direction HE of the truss, is

P C

J

E

A

3000 mm

L B

H

G

I

2L 3000 mm

1.

Force in the member AB of the truss is (A) (B)

2.

(C) (D)

√ √

Deflection of the point C is (A) (

√

)

(B) √ (C) ( √ (D) (√

(A) 8 (B) 7

(A) 0.255 (B) 0.589

(C) 0.764 (D) 1.026

CE – 2009 5. The degree of static indeterminacy of a rigidly jointed frame in a horizontal plane and subjected to vertical load only, as shown in figure below , is Ends clamped to rigid wall

) )

CE – 2008 3. The degree of static indeterminacy of the rigid frame having two internal hinges as shown in the figure below, is I H J

G

3000 mm

F

E

(C) 6 (D) 5

(A) 6 (B) 4

(C) 3 (D) 1

CE – 2010 6. A three hinged parabolic arch having a span of 20 m and rise of 5 m carries a point load of 10 kN at quarter span from the left end as shown in the figure. The resultant reaction at the left support and its inclination with the horizontal respectively

th

th

th


. GATE QUESTION BANK 10 kN

CE – 2014 9. A box of weight 100 kN shown in the figure is to be lifted without swinging. If all forces are coplanar, the magnitude and di ection (θ) of force (F) with respect to x – axis should be y

5m 5m

(A) (B) (C) (D) 7.

5m

10 m

9.01 kN and 9.01 kN and 7.50 kN and 2.50 kN and

40kN

90 kN

F θ

(A) k nd θ (B) F= 56.389kN and θ (C) k nd θ (D) k nd θ

L R

Q

10.

L

S

T

P

(A) Zero (B)

(C) P (D) √

√

x

100kN

For the turss shown in the figure, the force in the member QR is

Mathematical idealization of a crane has three bars with their vertices arranged as shown in the figure with a load of 80 kN hanging vertically. The coordinates of the vertices are given in parentheses. The force in the member QR, will be P (0, 4)

80 kN

22.8

CE – 2013 8. The pin-jointed 2-D truss is loaded with a horizontal force of 15 kN at joint S and another 15 kN vertical force at joint U, as shown. Find the force in member RS (in kN) and report your answer taking tension as positive and compression as negative. __________ 4m

y

104.0 Q (1, 0)

4m

4m R

S

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

15kN 4m

Q

Structural Analysis

V

U

15kN

T 4m

11.

P

th

53.1 R (3, 0)

30 kN Compressive 30 kN Tensile 50 kN Compressive 50 kN Tensile

For the truss shown below, the member PQ is short by 3 mm. the magnitude of vertical displacement of joint R (in mm) is _______________

th

th



12.

R 3M

Structural Analysis

The tension (in kN) in a 10 m long cable, shown in the figure neglecting its self – weight is

Q

P

3m

3m S

P 4M

4M

Cable

Q

y

Cable R

120 kN

(A) 120 (B) 75

(C) 60 (D) 45


[Ans. C] Since the truss is loaded symmetrically, the reaction will be equal Thus reaction at B = Now considering joint equilibrium at B

√

√

k

in θ

(i)

co θ From (i), we have in co θ ( 2.

√

)

√

(ii)

√

√ Assuming tensile forces as positive and compressive forces as negative. Member P K L

co

[Ans. A] It is evident from the diagram that all the interior members will carry zero force

A

√

BC

√

√

√

√

√ √

A

2L

√

Deflection at C, Where K is the force in member when a unit load is applied at C. th

th

th



√ √ ( 3.

√ √ √

√ √

)

√

[Ans. D] The degree of static indeterminacy for a rigid hybrid frame is given by m (j j ) Where, m = total number of members = 9 tot l number of external reactions =2+1+1=4 tot l number of released reactions at hybrid joint ) ( ) (m ) (

√

√ Considering joint I in

j = total number of rigid joints = 6 j = total number of hybrid joints = 2 ( ) ( ) –

√

√

Considering joint E

[Ans. *] 1

√

√ E

√

√

J

1

H

G

3000 mm

√

1

3000 mm

4.

Structural Analysis

√

I

The deflection at E in the direction HE is given by k But only EJ and IJ subjected to temperature change k

3000 mm

√ √

(

√

)

√

√ √

√ mm Note: None of the option is correct.

√ th

th

th



5.

6.

[Ans. A] Seeing the load pattern, it is to be treated as space frame. The number of reactions at each fixed support of a rigid jointed space frame = 6 Hence total no. of reactions = No. of equilibrium equations =6 Static indeterminacy = 12 – 6 =6

7.

Structural Analysis

[Ans. C] Using method of joints and considering joint S, we get

nd (ten ile)

[Ans. A] 10 kN

5m 5m

5m

Considering joint R, we get

10 m

in √ (comp )

Let the vertical reactions at left and right support be nd upwards respectively. Taking moment about right support, we get

co √

√

k

(ten ion)

k Let the horizontal reaction at left support be H from left to right. Taking moment about the crown from left, we get k e lt nt e ction

√

8.

[Ans. 0] S

R

( ) √ k Let the resultant reaction at the left support makes and angle θ with the horizontal

15 kN

15 kN Q

V

U

T

15 kN

θ P

H=0

t nθ θ θ

t n (

)

th

th

th



Structural Analysis

X

Take moment about V

R

Z

40 o x sin 53.13 = 40 x = 50 kN Z = 30 kN ie. Compressive

So, Force in RS = 0

9.

[Ans. A] ∑ ∑ From (1) 90 sin ( ) From (2) 90 cos ( ) From (3) in θ From (1) co θ (5) / (6) θ

( ) ( ) in

o θ

in θ

11.

co

( )

1 R

( )

n

(

)

From (5) F = 56.389 kN 10.

[Ans. A] At point

[Ans. *] Range 1.0 to 2.5 is short by 3 mm We have to find out vertical displacement of joint R in mm ( ) Let us apply unit load at R as shown below

P

∑

θ

Q

⁄

⁄

in θ

⁄

co θ co θ in θ

cot θ

Q

∑

∑

co θ

⁄

⁄

∑

Let force in PR be x and PQ be y and force in QR be Z (all assumed to be tensile)

( mm p

) d

k At point R:

th

th

th



12.

Structural Analysis

[Ans. B] 3m P 5m

3m S 4m θ θ

θ 5m

R 120 kN

Free body diagram S T 4m θ θ 5m 5m

T

120 kN

θ o θ

= 75 kN

th

th

th



Structural Analysis

Influence Line Diagram and Rolling Loads CE – 2005 1. Match List – I with List –II and select the correct answer using the codes given below the lists List – I List – II A. Slope deflection 1. Force method method B. Moment 2. Displacement distribution method C. Method of three moments D. tigli no’ second theorem Codes A B C D (A) 1 2 1 2 (B) 1 1 2 2 (C) 2 2 1 1 (D) 2 1 2 1 CE – 2006 2. Consider the beam ABCD and the influence line as shown below. The influence line pertains to A

B

C

Internal hinge L L

D

6m R

P

4.5m

(A) 2.47 mm (B) 10.25 mm

4.

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

B

Q

R

S

Tension

(A) PS (B) RS

(C) PQ (D) QS

CE – 2008 5. The span(s) to be loaded uniformly for maximum positive (upward) reaction at support P, as shown in the figure below, is (are) Q

P

R

S

T

D

(A) PQ only (B) PQ and QR

Reaction at A, Shear force at B, Shear force on the left of C, Shear force on the right of C,

CE – 2007 3. The right triangular truss is made members having equal sectional area 1550 mm nd Yo ng’ mod l MPa. The horizontal deflection the joint Q is

P

Compression

2L

C

(C) 14.31 mm (D) 15.68 mm

The influence line diagram (lLD) shown is for the member

1 A

135 kN

Q

of of of of

(C) QR and RS (D) PQ and RS

CE – 2013 6. Beam PQRS has internal hinges in spans PQ and RS as shown. The beam may be subjected to a moving distributed vertical load of maximum intensity 4kN/m of any length any where on the beam. The maximum absolute value of the shear force (in kN) that can occur due to this

th

th

th



loading just to the right of support Q shall be: P

Q

R

Structural Analysis 0.75

P.

0.6

Q.

S 0.25

5m

5m

5m

20 m

(A) 30 (B) 40

5m

(C) 45 (D) 55

1.0

0.5

R.

CE – 2014 7. In a beam of length L, four possible influence line diagrams for shear force at a section located at a distance of

0.6

S.

0.5

from

(A) P (B) Q

the left end support (marked as P, Q, R and S) are shown below. The correct influence line diagram is

(C) R (D) S


[Ans. C]

2.

[Ans. B]

3.

[Ans. D]

P 135 kN Q

135 kN

1 kN

135 kN

180 kN

180 kN

A+R

6m

P

θ

A+ P co θ

θ 4.5 m

P

R

R

= 225 kN k Tension (+), compression ( ) and k for each member

k t nθ

in θ

k

and

co θ

Taking k k Considering joint R and joint P

th

Member

P

K

L

PQ

225

1:67

7.5

QR

6

RP

4.5

th

th



When a cut is made just to right of Q and displacement are given such that is parallel to As is very close to Q, displacement of to the left will be zero and that to right will be 1

mm 4.

[Ans. A]

5.

[Ans. D] With the help of Muller Breslau principle, we can draw the ILD for reaction at P P

+

Structural Analysis

Q

R

Hence rope of Slope of or dinate at Or dinate at

S

If UDL is loading span PR, we get max SF just to the right of Q

T

(

)

k ILD for vertical reaction at P

The reaction is positive between P and Q and R and S respectively. Hence the spans PQ and RS should be loaded uniformly for maximum positive reaction at P. 6.

7.

[Ans. A] ⁄

⁄

⁄

[Ans. C] P

Q

5m 5m

R 20m

⁄

S 5m 5m

1

⁄

θ

⁄ ⁄

l o ⁄

0.25

( ) 0.25

⁄

ILD for SF to right of Q

th

th

th



Structural Analysis

Slope and Deflection Method CE – 2005 1. All members of the frame shown below have the same flexural rigidity El and length L. if a moment M is applied at joint B, the rotation of the joints is M

A

C

B El, L

El, L

CE – 2007 Common Data Questions 4 and 5 A two span continuous beam having equal spans each of length L is subjected to a uniformly distributed load w per unit length. The beam has constant flexural rigidity. 4.

El, L

The reaction at the middle support is (A) wL (C)

D

(B)

(A)

(C)

(B)

(D)

5.

CE – 2006 2. Carry – over factor for the beam shown in the figure below is C B A Internal hinge L

(A) ¼ (B) ½ 3.

(C) ¾ (D) 1

Consider the beam AB shown in the figure below. Part AC of the beam is rigid while Part CB has the flexural rigidity El. Identify the correct combination of deflection at end B and bending moment at end A, respectively

(D)

The bending moment at the middle support is (A)

(C)

(B)

(D)

CE – 2008 Statement for linked Answer Questions 6&7 Beam GHI is supported by three pontoons as shown in the figure below. The horizontal cross sectional area of each pontoon is 8m2, the flexural rigidity of the beam is 10000 kN-m2 and the unit weight of water is 10kN/m3. P = 48 kN G

H

I

P Pontoons

A

B

C L

5m

5m

L

(A)

(C)

(B)

(D)

6.

When the middle pontoon is removed, the deflection at H will be (A) 0.2 m (C) 0.6m (B) 0.4m (D) 0.8m

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th

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7.

When the middle pontoon is brought back to its position as shown in the figure above, the reaction at H will be (A) 8.6 kN (C) 19.2 kN (B) 15.7 kN (D) 24.2 kN

Structural Analysis 3m

4m

S 2m T

P

CE – 2009 Statement for linked Answer Questions 8&9 In the cantilever beam PQR shown in the figure below, the segment PQ has flexural rigidity EI and the segment QR has infinite flexural rigidity W EI Rigid Q R P L

8.

(B) 9.

100kNm 2m Q

12.

and and

A uniform beam (EI = constant ) PQ in the form of a quarter circle of radius R is fixed at end P and free at the end Q, where a load W is applied as shown. The vertical downward displacement, at the loaded point Q is given by:

L

(C)

and

(D)

and

(C)

(B)

(D)

CE – 2012 10. A simply supported beam is subjected to a uniformly distributed load of intensity w per until length, on half of the span from one end. The length of the span and the flexural stiffness are denoted as l and EI, respectively. The deflection at mid – span of the beam is (A)

(C)

(B)

(D)

)

Q

The deflection of the beam at R is (A)

(

ind the value of (correct to 4 – decimal places).______________________ W

The deflection and slope of the beam at Q are respectively (A)

R

R P

13.

A uniform beam weighing 1800 N is supported at E and F by cable ABCD. Determine the tension (in N ) in segment AB of this cable (correct to 1 – decimal place). Assume the cables ABCD, BE and CF to be weightless,----------------------A

D 1m

C B E

CE – 2013 11. All members in the rigid – jointed frame shown are prismatic and have the same flexural stiffness EI. Find the magnitude of the bending moment at Q (in kNm) due to the given loading.______________________

1m

F 2m

0.5 m

th

th

0.5 m 0.5 m

th



CE – 2014 14. For the cantilever beam of span 3 m (shown below), a concentrated load of 20 kN applied at the free end causes a vertical displacement of 2 mm at a section located at a distance of 1 m from the fixed end. If a concentrated vertically downward load of 10 kN is applied at the section located at a distance of 1 m from the fixed end (with no other load on the beam), the maximum vertical displacement in the same beam (in mm) is __________ 2 mm

1m

15.

16.

Structural Analysis

The beam of an overall depth 250 mm (shown below) is used in a building subjected to two different thermal environments. The temperatures at the top and bottom surfaces of the beam are 36°C and 72°C respectively. Considering coefficient of thermal expansion ( ) per °C, the vertical deflection of the beam (in mm) at its mid-span due to temperature gradient is ________

20kN

2m

Considering the symmetry of a rigid frame as shown below, the magnitude of the bending moment (in kNm) at P (preferably using the moment distribution method) is

(A) 170 (B) 172

(C) 176 (D) 178

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th

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Structural Analysis


[Ans. B]

3.

[Ans. A] Part AC of the beam is rigid. Hence C will act as a fixed end. Thus the deflection at B will be given as

For AB, θ tiffne fo

θ

l

tiffne fo

l

t

t

θ

l

tiffne

But the bending moment does not depend on the rigidity or flexibility of the beam. t 2L = 2PL

l

4.

l

t

[Ans. C]

W per limit length B

A

Total stiffness of the joint l l l l

C L

L

Rotation of joint B = 2.

0.5 0.5 Initial F.E.M

[Ans. D] Carry over factor

Balancing Carry over

M

A

B

C

Final F.E.M

0

0

A B L

0

L

et For ∑

L

ppl moment ‘ ’ t take moment @ C = 0 M/L (upward) (do n

g in ∑

C

B L

0

d)

Take moment at A: Take moment at C,

f om ight ide

∑

∑

Carry over factor =

ot l

th

th

th



Structural Analysis

Hint: The given continuous beam can be treated as two symmetrical propped cantilevers. Each span

H

( It is a standard case,

which can be remembered) 5.

[Ans. B]

6.

[Ans. B] The reaction at the ends are zero as there are hinges to left of G and right of I. Hence when the middle pontoon is removed, the beam GHI acts as a simply supported beam

Also, we have Q+Q+R=P 2Q + R = 48 l o ( ) e of c o pontoon x =R ( ) [from (ii)]

48 kN G

[from (iii)] I

H

24kN

24kN

(

The deflection at H will be due to the load at H as well as due to the displacement of pontoons at G and I in water. Since the loading is symmetrical, both the pontoons will be immersed to same height Let it be x. area of cross-section of pontoon unit weight of water = 24 x 8 10 = 24 x = 0.3m Also, deflection at H due to load P = (

8.

)

= k

[Ans. C] et the el tic deflection t (

[from (i)]

[Ans. A] The given cantilever beam can be modified into beam as shown below w EI L

P

Q

Deflection at Q =

=

)

m

WL

+

=

= 0.1m in l deflection t 7.

(iii) – section of

=

lope t

be

)

(i) The reactions at G and I will be same, as the beam is symmetrically loaded, Let the reaction at each G and I be Q Using principle of buoyancy, we get x area of cross-section of pontoon x =Q

9.

[Ans. C] Since the portion QR of the beam is rigid, QR will remain straight. Deflection of R = Deflection at Q + Slope at Q × L =

+

=

×L=

=

(ii)

th

th

th



10.

Structural Analysis

[Ans. B]

k m l l ( )( )

(l) l

(

A

k m

B l⁄

)

l⁄

12.

[Ans. *] Range 0.785 to 0.786 W Q

d d

(

( )

d d ( )

)(

) Q

(

)

(

)

R P Applying method of strain

To find energy ∫

d

he e

@ x = l, y = 0 ( )

l l )

( l

l

in θ in pol

(l)

nd d

dθ ∫ in θ

l l l ( ) ( )

[

in θ

l

l

11.

fo m

l l )( )

(

in θ

∫ in θdθ

l

] l

13.

[Ans. 25] 3m

[Ans. *]Range 1310 to 1313 R

4m

s

D H

2m P

dθ

A

R

T

H 1m

100kNm 2m

C

B

Q

1m

1350N

450N

Weight of beam = 1800 N. Taking moment about F:-

For equilibrium,

th

th

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Structural Analysis

Fixed End Moment fo c ble k m

∑

√( 14.

)

(

k m k m m k m Distribution table

)

[Ans. 1]

20 kN 0

2mm

d

16

ecip oc l theo em d 16.

32 32

48

0

0

16

3

0

[Ans. *] Range 2.38 to 2.45

m

24 kN/m C

P

h

6m

(

A

8m

Distribution factor Joint Member

0

0

0

[Ans. C] B

0

128

6 of bending 2 The magnitude moment at P = 176 kNm

x = 1 mm 1 is answer 15.

128

32 16

ell’

0

E

) (

D

8m

( RS

TRS

) )

mm

DF

BA B BP PB P

PE PC CP

C CD

th

th

th



Structural Analysis

Degree of Static Indeterminacy CE – 2005 1. Considering beam as axially rigid, the degree of freedom of a plane frame shown below is

ble

F

(A) 9 (B) 8

(C) 7 (D) 6

CE – 2014 2. The degree of static indeterminacy of a rigid joined frame PQR supported as shown in figure is.

(A) Zero (B) One 3.

(C) Two (D) Unstable

The static indeterminacy of the two-span continuous beam with an internal hinge shown below, is ___________


[Ans. B] Degree of kinematic indeterminacy or degree of freedom, j m

2.

[Ans. A] ( (

3.

[Ans. 0] Number of member, m = 4 Number of external reaction, Number of joint, j = 5 Number of reactions released, Degree of static indeterminacy, m j

) )

th

th

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Structural Analysis

Force Energy Method CE – 2006 1. Vertical reaction developed at B in the frame below due to the applied load of 100 kN (with 150,000 mm cross-sectional area and mm moment of inertia for both members) is A

B Internal hinge 1m

1m

(A) 5.9 kN (B) 30.2 kN

(C) 66.3 kN (D) 94.1 kN


[Ans. A] 100 kN A

A

B

R R

C R

Deflection at A in beam AB = Compression in column AC ( ) l ( ) ( )

k k

th

th

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Structural Analysis

Matrix Method of Structural Analysis CE – 2005 1. For a linear elastic frame, if stiffness matrix is doubled with respect to the existing stiffness matrix, the deflection of the resulting frame will be (A) Twice the existing value (B) Half the existing value (C) The same as existing value (D) Indeterminate value

CE – 2013 Common Data Questions 4 and 5 A propped cantilever made of a prismatic steel beam is subjected to a concentrated load P at mid span as shown. P

R

CE – 2007 2. The stiffness coefficient k indicates (A) Force at i due to a unit deformation at j (B) Deformation at j due to a unit force at i (C) Deformation at i due to a unit force at j (D) Force at j due to a unit deformation at i

1.5m

4.

If load P = 80 kN, find the reaction R (in kN) (correct to 1 – decimal place) using elastic analysis.__________

5.

If the magnitude of load P is increased till collapse and the plastic moment carrying capacity of steel beam section is 90 kNm, determine reaction R(in kN) (correct to 1decimal place) using plastic analysis._______

CE – 2012 3. A symmetric frame PQR consists of two inclined members PQ and QR, connected at ‘ ’ ith igid joint nd hinged t ‘ ’ nd ‘ ’ he ho i ont l length i l f eight i pended t ‘ ’ the bending moment t ‘ ’ i (A)

(C)

(B)

(D) Zero

1.5m

th

th

th



Structural Analysis


[Ans. B] Stiffness matrix and deflection are related as Thus, when stiffness matrix is doubled, then deflection will reduced to half of the existing value.

2.

[Ans. A]

3.

[Ans. D] The given frame can be treated as a linear inverted arch. For a lined arch, No S.F & B.M, only axial forces. Hence B.M at every point incl ding t ‘ ’ i e o

4.

5.

[Ans. 60] P

m

m

θ

θ

θ

k m

MP = 90kN m θ

m R

θ

[Ans. 25]

k B

A

l

C

l

(

R

)

k

From compatibility equation Net deflection at c = 0 l l ( )

( *

(

) )

(

) (

)

+

k

th

th

th



Mechanics

Simple Stress and Strain Relationship CE – 2005 1. The symmetry of stress tensor at a point in the body under equilibrium is obtained from (A) conservation of mass (B) force equilibrium equations (C) moment equilibrium equations (D) conservation of energy 2.

3.

6.

For an isotropic material, the relationship between the Young's modulus (E), shear modulus (G) and Poisson's ratio (μ) is given by (A) G = (B) E =

(

)

(C) (D)

(

) (

)

A metal bar of length 100 mm is inserted between two rigid supports and its temperature is increased by 10°C. If the coefficient of thermal expansion is per °C and the Young's modulus is 2 × 105 MPa, the stress in the bar is (A) Zero (C) 24MPa (B) 12 MPa (D) 2400 MPa

8.

A rigid bar is suspended by three rods made of the same material as shown in the figure. The area and length of the central rod are 3 A and L, respectively while that of the two outer rods are 2A and 2L, respectively. If a downward force of 50kN is applied to the rigid bar, the forces in the central and each of the outer rods will be

If principal stresses in a two-dimensional case are MPa and 20 MPa respectively, then maximum shear stress at the point is (A) 10 MPa (C) 20 MPa (B) 15 MPa (D) 30 MPa

CE – 2006 4. Mohr's circle of the state of stress defined (A) (B) (C) (D)

)

7. The components of strain tensor at a point in the plane strain case can be obtained by measuring longitudinal strain in following directions (A) Along any two arbitrary direction (B) Along any three arbitrary directions (C) Along two mutually orthogonal directions (D) Along any arbitrary direction

by *

(

+ MPa is a circle with

center at (0,0) and radius 30 MPa center at (0,0) and radius 60 MPa center at (30,0) and radius 30 MPa center at (30, 0) and zero radius

CE – 2007 5. An axially loaded bar is subjected to a normal stress of 173 MPa. The shear stress in the bar is (A) 75MPa (C) 100MPa (B) 86.5MPa (D) 122.3MPa

50 kN

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

th

16.67kN each 30kN and 15kN 30kN and 10kN 21.4kN and 14.3kN

th

th



9.

U1 and U2 are the strain energies stored in a prismatic bar due to axial tensile forces P1 and P2, respectively. The strain energy U stored in the same bar due to combined action of P1 and P2 will be (A) U = U1 + U2 (C) U < U1 + U2 (B) U = U1 U2 (D) U > U1 + U2

CE – 2008 10. A mild steel specimen is under uniaxial tensile stress. Young's modulus and yield stress for mild steel are 2 × 105 MPa and 250 MPa respectively. The maximum amount of strain energy per unit volume that can be stored in this specimen without permanent set is (A) l56Nmm/mm3 (B) 15.6Nmm/mm3 (C) 1.56Nmm/mm3 (D) 0.156Nmm/mm3 11.

A vertical rod PQ of length L is fixed at its top end P and has a flange fixed to the bottom end Q. A weight W is dropped vertically from a height h (
CE – 2009 12. Consider the following statements : 1. On a principal plane, only normal stress acts. 2. On a principal plane, both normal and shear stresses act. 3. On a principal plane, only shear stress acts. 4. Isotropic state of stress is independent of frame of reference. Which of the above statements is /are correct? (A) 1 and 4 (C) 2 and 4 (B) 2 only (D) 2 and 3

Mechanics

CE – 2010 13. The major and minor principal stresses at a point are 3 MPa and MPa respectively. The maximum shear stress at the point is (A) Zero (C) 6 MPa (B) 3 MPa (D) 9 MPa 14.

The number of independent elastic constants for a linear elastic isotropic and homogeneous material is (A) 4 (C) 2 (B) 3 (D) 1

CE – 2011 15. Consider a simply supported beam with a uniformly distributed load having a neutral axis (NA) as shown. For points P (on the neutral axis) and Q (at the bottom of the beam) the state of stress is best represented by which of the following pairs? P

NA

Q L (A)

(B)

(C)

(D)

L

P

Q

P

Q

P

Q

P

Q

CE – 2012 16. The poisson’s ratio is defined as

th

(A) |

|

(C) |

|

(B) |

|

(D) |

|

th

th



17.

If a small concrete cube is submerged deep in still water in such a way that the pressure exerted on all faces of the cube is p , then the maximum shear stress developed inside the cube is (A) 0 (C) p (D) 2p (B)

22.

Mechanics

For the state of stresses (in MPa) shown in the figure below, the maximum shear stress (in MPa) is ________ 4 4 2

CE – 2013 18. The ‘plane section remains plane’ assumption in bending theory implies: (A) Strain profile is linear (B) Stress profile is linear (C) Both strain and stress profiles are linear (D) Shear deformations are neglected 19.

20.

2

4

23.

The creep strains are (A) Caused due to dead loads only (B) Caused due to live loads only (C) Caused due to cyclic loads only (D) Independent of loads

A box of weight 100kN shown in the figure is to be lifted without swinging. If all forces are coplanar, the magnitude and direction (θ) of the force (F) with respect to x-axis should be y 40 kN 90 kN F θ 100 kN

2D stress at a point is given by a matrix *

+

+ Pa

*

x

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

The maximum shear stress in MPa is (A) 50 (C) 100 (B) 75 (D) 110

F F F F

kN and θ kN and θ kN and θ kN and θ

CE – 2014 21. The values of axial stress ( ) in kN/m , bending moment (M) in kNm, and shear force (V) in kN acting at point P for the arrangement shown in the figure are respectively Cable P

(

Beam m 3m

Frictionless Pulley m)

Q

kN

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

1000, 75 and25 1250, 150 and50 1500, 225 and75 1750, 300 and100 th

th

th



Mechanics


[Ans. C] Taking moment equilibrium about the Centre, we get

8.

[Ans. C] If the force in each of outer rods is P0, and force in the central rod is Pc then 2P0 + Pc … (i) Also, the elongation of central rod and outer rods is same.

d/2 d/2

yx

∴

d/2 d/2

yx

xy

∴ P P … (ii) Solving (i) and (ii), we get PC = 30kN and PO = 10kN

xy

xy

[Ans. B] When strain is measured along any three arbitrary directions, the strain diagram is called rosette.

3.

[Ans. B] Maximum

4.

5.

shear

)

stress

[Ans. D] The maximum and minimum principal stresses are same, So, radius of circle becomes zero and centre is at (30, 0). The circle is represented by a point.

10.

[Ans. A] (

(

=

11.

)

N

mm/mm

[Ans. A] The kinetic energy of the weight (W) is stored in the form of strain energy in the rod. We know that U=

Pa

area of cross section of rod ×

length of rod Since, energy remains constant, hence to reduce the axial stress in the rod, the length of the rod should be increased or area of cross-section of the rod should be increased or modulus of elasticity should be decreased.

μ) μ)

[Ans. C] Temperature stress

[Ans. D] The strain energy per unit volume may be given as U=

[Ans. B] Shear stress =

(

7.

=

It is obvious from the above equation that strain energy is proportional to the square of load applied. We know that sum of squares of 2 number is less than square of their sum. Thus U >

Pa

∴Shear stress =

6.

[Ans. D] We know that strain energy, U =

2.

(

9.

yx

αT

= 24MPa th

th

th



12.

[Ans. A] On a principal plane, only normal stresses act. No shear stresses act on the principal plane.

13.

[Ans. B] Maximum shear stress at the point is given by max

=

=

(

)

15.

[Ans. *] No answer is correct. Correct stress element will be P

20.

[Ans. A] √(

)

√(

(

)

)

Q

There will be no stress at P because of zero shear force at mid span and location of P at neutral axis.

18.

[Ans. A] Creep strains are caused due to prolonged loading (static loading ) for long duration of time. Thus creep is caused by dead loads whereas fatigue is caused by moving / cyclic loading.

⁄

[Ans. C]

17.

19.

Pa

14.

16.

Mechanics

∴

21.

[Ans. B] lateral strain μ linear strain

Pa

[Ans. B] Free body diagram 50 kN

50 kN

(0.2m × 0.2m)

[Ans. A] A member subjected to hydrostatic pressure has mohr circle as a point. The radius of mohr circle, i.e, maximum shear stress is zero. Pressure is uniform in all directions in absence of shear stress [By pascal’s law]

3m

50 kN

Axial stress =

kN/m

Bending moment = shear force = 50 × 3 = 150 kNm

22.

distance

[Ans. 5.0] 4

[Ans. A] Recalling discussion from LSM of concrete design, the limitations/ assumptions of simple bending theory hold true with stress and strain distribution 0.446fcp

4 2

2

4

ax shear stress

fy

here

s 0.87fy (Nonlinear above N. A.) Strain distribution (linear)

√(

)

Hence, (A) is most appropriate option th

th

th



√(

Mechanics

( )

)

Pa √(

o

23.

(

) Pa )

Pa

[Ans. A] For no swinging F 40 kN

90 kN

F θ

x

100 kN cos cos 49.658 F cos θ F cos θ cos kN From options ) cos( So, F = 56.389 kN θ

F cos θ

kN

th

th

th



Mechanics

Bending Moment and Shear Force Diagram CE – 2005 1. The bending moment diagram for a beam is given Below.

A

b'

0.5 m 0.5 m

1m

1m

The shear force at sections aa' and bb' respectively are of the magnitude (A) 100 kN, 150 kN (B) Zero, 100 kN (C) Zero, 50 kN (D) 100 kN,100 kN CE – 2006 2. A simply supported beam AB has the bending moment diagram as shown in the following figure: M C

A

D

4.

(A)

(upwards)

(B)

(downwards)

(C)

(upwards)

(D)

(downwards)

The rotation at B is (A)

(clockwise)

(B)

(anticlockwise)

(C)

(clockwise)

(D)

(anticlockwise)

CE – 2008 5. The stepped cantilever is subjected to moments, M as shown in the figure below. The vertical deflection at the free and (neglecting the self weight) is M

M

M L

B

L

The reaction at C is

100 kN-m

a'

L

C

a

L

3.

a

a

P

200 kN-m

b

P

B

L

(A) Couples of M at C and 2M at D (B) Couples of 2M at C and M at D (C) Concentrated loads of M/L at C and 2M/L at D (D) Concentrated load of M/L at C and couple of 2M at D Common data questions 3 and 4 Consider a propped cantilever beam ABC under two loads of magnitude P each as shown in the figure below, Flexural rigidity of the beam is EI

EI

2EI

The beam is possibly under the action of following loads

L/2

L/2

(A)

(C)

(B)

(D) Zero

M

CE – 2009 6. Match List – I (Shear force diagrams) beam with List – II (Diagrams of beams with supports and loading) and select the correct answer by using the codes given below the lists:

th

th

th



q

CE – 2010 7. Two people weighting W each are sitting on a plank of length L floating on water at L/4 from either end. Neglecting the weight of the plank, the bending moment at the centre of the plank is

q

+

+

q

q q +

B

q q

q

+

q

+ q

q

q

Mechanics

8.

(A)

(C)

(B)

(D) Zero

For the simply supported beam of length L, subjected to a uniformly distributed moment M kN-m per unit length as shown in the figure, the bending moment (in kNm) at the mid – span of the beam is

q

M kN – m per unit length

q

q/unit length

q/unit length

L

q

q

q

/

/

/

/

q

B 1 4 1 4

(C) ML (D) M/L

CE – 2011 Linked Answer Questions 9 and 10: A rigid beam is hinged at one end and supported on linear elastic springs (both having a stiffness of ‘k’) at points ‘ ’ and ‘ ’ and an inclined load acts at ‘ ’ as shown. √ P

inge

q

/

/

Codes: A (A) 3 (B) 3 (C) 2 (D) 2

(A) Zero (B) M

/

/

C 2 2 4 3

D 4 1 3 1

Fixed

9.

Which of the following options represents the deflections and at points ‘ ’ and ‘ ’?

th

(A)

( ) and

(B)

( ) and

(C)

(

(D)

( th

√ √

( ) ( )

) and

(

) and

(

th

√ √

) )



10.

If the load P equals 100 kN, which of the following options represents forces and in the springs at points ‘ ’ and ‘ ’? (A) kN and kN (B) kN and kN (C) kN and kN (D) kN and kN

Mechanics Q

h/2 P M h/2 R

CE – 2012 11. The sketch shows a column with a pin at the base and rollers at the top. It is subjected to an axial force P and a moment M at mid – height. The reactions (s) at R is /are

(A) A vertical force equal to P (B) A vertical force equal to P/2 (C) A vertical force equal to P and horizontal force equal to M/h (D) A vertical force equal to P/2 and a horizontal equal to M/h


[Ans. C] The bending moment to the left as well as right of section aa’ is constant which means shear force is zero at aa’ hear force at bb’

2.

=

(

= Thus,

)

(downwards)

will be upwards

=

(upwards)

=

=

[Ans. A] The Shear force diagram is

(upwards)

D

A

B

4.

[Ans. A] The rotation at B (i) Due to moment

SFD

M

)=

∴ The reaction at

kN

C

(

θ

=

(clockwise)

(ii) Due to reaction R

2M

θ

=

+

=

(anticlockwise) θ RA

=(

RB Loading diagram

3.

θ

=

θ ) (clockwise)

[Ans. C] The moment at point B = 2Pa In the cantilever beam ABC, the deflection at C due to moment 2Pa will be given as th

th

th



5.

=M Thus, the reaction at the left support will be M downwards

[Ans. C] Using moment area method M M

2EI

A

EI

L/2

Mechanics

∴

B

oment at the mid – span, = M× +M× =0

L/2

Infact the bending moment throughout the beam is zero

9.

2M

[Ans. B]

M

√ P

P

BMD

P k

k

k

k

diagram

Deflection at B w.r.t A = moment of area of

diagram between A and B about B

6.

[Ans. A]

7.

[Ans. D] The plank will be balanced by the buoyant force acting under its bottom. Let the intensity of buoyant force be w, w

The free diagram of the beam is show below From similar triangles, we get

…( ) Taking moments about hinge, we get P k k ) P k( [∴ from (i)] ) P k( P k From (i). we get P P k k

w

/

/

/

For equilibrium, w × L = W + W w=

upwards

Thus, the bending moment at the centre of the plank will be M=

× ×

W×

M=

10.

[Ans. D] k

M=0

8.

k

P k kN

[Ans. A] Let the reaction at the right hand support be upwards. Taking moments about left hand support, we get × L ML = 0

k

k

P k kN

th

th

th



11.

Mechanics

[Ans. C] Q P M

P

R

H V

Taking moment about θ h h And axial force =P

th

th

th



Mechanics

Thin Walled Pressure Vessel CE – 2006 1. A thin-walled long cylindrical tank of inside radius r is subjected simultaneously to internal gas pressure p and axial compressive force F at its ends. In order to produce 'pure shear' state of stress in the wall of the cylinder, F should be equal to (A) pπr2 (C) pπr2 (B) pπr2 (D) pπr2

CE – 2009 2. A thin walled cylindrical pressure vessel having a radius of 0.5 m and wall thickness of 25 mm is subjected to an internal pressure of 700 kPa. The hoop stress developed is (A) 14 MPa (C) 0.14MPa (B) 1.4MPa (D) 0.014MPa


[Ans. C]

pr t

oop stress

pr t Now, for pure shear state, compressive and is equal to ∴ pr pr F t t πrt pr F t πrt F πpr ongitudinal stress

2.

F πrt should be

[Ans. A] oop stress

pd t

Pa

th

th

th



Mechanics

Simple Bending Theory shear force of 3000 N, the glue at any of the four joints will be subjected to a shear force (in kN per meter length) of

CE – 2006 1. A beam with the cross-section given below is subjected to a positive bending moment (causing compression at the top) of 16 kN-m acting around the horizontal axis. The tensile force acting on the hatched area of the cross-section is

50 mm

200 mm

75 mm

50 mm

25 mm

50 mm 75 200 mm mm

50 mm 50 mm

50 mm

(A) Zero (B) 5.9 kN 2.

(C) 8.9 kN (D) 17.8 kN

If a beam of rectangular cross-section to subjected to a vertical shear force V, the sheer force earned by the upper one-third of the cross-section is (A) Zero (C) (B)

3.

(D)

For the section shown below, second moment of the area about an axis d/4 distance above the bottom of the area is. b

(A) 3.0

(B) 4.0

(C) 8.0

(D) 10.7

CE – 2007 5. The shear stress at the neutral axis in a beam of triangular section with a base of 40 mm and height 20 mm, subjected to a shear force of 3 kN is (A) 3 MPa (C) 10 MPa (B) 6 MPa (D) 20 MPa CE – 2008 6. The maximum tensile stress at the section X-X shown in the figure below is L/3

L/3 X

L/3 d/2

b

d/2 X

L/2

(A)

L/2

(B)

(C)

(D)

d

4.

(A)

(C)

(B)

(D)

I – section of a beam is formed by gluing wooden planks as shown in the figure. If this beam transmits a constant vertical

CE – 2009 7. The point within the cross sectional of a beam through which the resultant of the external loading on the beam has to pass through to ensure pure bending without twisting of the cross section of the beam is called (A) Moment centre (B) Centroid (C) Shear centre (D) Elastic centre th

th

th


d


CE – 2010 8. A disc of radius r has a hole of radius r/2 cut – out as shown. The centroid of the remaining disc (shaded portion) at a radial distance from the “O” is r/2 O O

(A)

(C)

(B)

(D)

CE – 2011 9. For the cantilever bracket, PQRS, loaded as shown in the adjoining figure(PQ=RS=L, and, QR =2L) which of the following statements is FALSE? S

Fixed

R

Mechanics

CE – 2012 10. The following statements are related to bending of beams: I. The slope of the bending moment diagram is equal to the shear force. II. The slope of the shear force diagram is equal to the load intensity III. The slope of the curvature is equal to the flexural rotation IV. The second derivative of the deflection is equal to the curvature. The only FALSE statements is (A) I (C) III (B) II (D) IV CE – 2013 11. A symmetric l-section (with width of each flange = 50 mm, thickness of web = 10mm depth of web = 100mm) of steel is subjected to a shear force of 100 kN. Find the magnitude of the shear stress (in N⁄mm ) in the web at its junction with the top flange. 50mm 10mm

10mm

Q W L

(A) The portion RS has a constant twisting moment with a value of 2WL. (B) The portion QR has a varying twisting moment with a maximum value of WL. (C) The portion PQ has a varying bending moment with a maximum value of WL. (D) The portion PQ has no twisting moment.

100mm

2L

P

10mm 50mm

th

th

th



Mechanics


[Ans. C]

b d * mm

b

d

d

d

+

d

25 mm mm

b bd

x f

mm

Bending stress distribution

mm

3.

[Ans. C] Using parallel axis theorem, we get the second moment of inertia as

y N/mm From similar triangles, we have x

d

I= 4.

+ bd (

+

=

[Ans. B] shear flow q =

x n/mm ∴ Tensile force

I =

+ 2 ×* += 3.5 ×

kN 2.

) =

mm

For any of the four joints, Q = 50 × 75 × 125 = 468750 mm

[Ans. B]

∴q = 4.0 N/mm = 4.0 kN/m d/

5.

y

[Ans. C] ̅

shear stress

(

̅ (

dF

( / )

)

(

Where S = shear force A = Area above the level where shear stress is desired y̅ = distance of CG of area A from neautral axis I = Moment of inertia about neutral axis b = width of the section at the level where shear stress is desired

)

)

b dy (

mm

y )

b dy

20 mm

Integrating both sides, we get b / d F ∫ ( y ) dy /

b d * y

y

40 mm

/

Width at a distance of

+ /

= th

mm from the top

mm th

th



)

9.

[Ans. B] Statement B is false it should be the portion QR has a constant twisting moment of WL.

10.

[Ans. C] d d d y dx dx dx But slope of curvature is not flexural rotation

11.

[Ans. *] Range 70 to 72

) Pa (hy

y ) [

( ) ]

=10 MPa

50mm 10mm

60mm

[Ans. A] The section at X – X may be shown as in the figure below b P d

d

60mm

6.

The maximum tensile stress at the section X – X is

10mm 50mm

+ = =

mm

Alternative q=

y̅

(

(

100mm

)

10mm

(

Mechanics

( / )

+

( / ) (

l

/ )

mm

+ q

7.

[Ans. C]

8.

[Ans. C] The centroid of the shaded portion of the disc is given by

y̅ lb

N⁄mm

X= Where x is the radial distance from O. πr ; x = 0;

=

π

( ) =

= πr πr r

πr

( ve sign indicates centroid lies in left of the origin)

th

th

th



Mechanics

Torsion CE – 2005 1. A circular shaft shown in the figure is subjected to torsion T at two points A and B. The torsional rigidity of portions CA and BD is GJ1, and that of portion AB is GJ2. The rotations of shaft at points A and B are θ and θ . The rotation θ is C A B D

L

T

T

L

(A)

(C)

(B)

(D)

L

CE – 2006 2. A long shaft of diameter d is subjected to twisting moment T at its ends. The maximum normal stress acting at its cross-section is equal to (A) Zero (C) (B)

CE – 2008 4. The maximum shear stress in a solid shaft of circular cross – section having diameter d subjected to a torque T is . If the torque is increased by four times and the diameter of the shaft is increased by two times, the maximum shear stress in the shaft will be (A) (C) / (B) (D) /

(D)

CE – 2007 3. The maximum and minimum shear stresses in a hollow circular shaft of outer diameter 20 mm and thickness 2 mm, subjected to a torque of 92.7 N-m will be (A) 59 MPa and 47.2 MPa (B) 100 MPa and 80 MPa (C) 118 MPa and 160 MPa (D) 200 MPa and 160 MPa

CE – 2009 5. A hollow circular shaft has an outer diameter of 100 mm and a wall thickness of 25 mm. The allowable shear stress in the shaft is 125 MPa. The maximum torque the shaft can transmit is. (A) 46 kN-m (C) 23 kN-m (B) 24.5 kN-m (D) 11.5kN-m CE – 2010 6. A solid circular shaft of diameter d and length L is fixed at one end and free at the other end. A torque T is applied at the free end. The shear modulus of the material is G. The angle of twist at the free end is. (A)

(C)

(B)

(D)

CE – 2014 7. Polar moment of inertia ( ) in cm of a rectangular section having width, b=2cm and depth, d = 6 cm is ______________

th

th

th



Mechanics


[Ans. B] The symmetry of the shaft shows that there is no torsion on section AB. T ∴ otation θ

5.

[Ans. C] …… ( ) T T [ ][ ] Td π(d

2.

[Ans. A] π(

Maximum shear stress Normal stress = 0 3.

π

(

6.

)mm

N

m

mm

( )

(

and f

Pa

7.

T ( )

(

[Ans. 40] Polar moment of inertia, bd db

)

Pa

Pa

bd 4.

]

θ

)

Pa

[

πd

But

T

f

]

[Ans. B] Angle of twist is given by T θ

mm T

) T

π[ T kNm So correct option is (C)

[Ans. B] f T ere

putting in ( )

d ) Td

[Ans. C] We know that T

(b

d ) (

)

cm

T T T T

(

)

th

th

th



Mechanics

Column and Struts CE – 2006 1. The buckling load P = Pcr for the column AB in figure, as KT approaches infinity,

4.

A rigid bar GH of length L is supported by a hinge and a spring of stiffness K as shown in the figure below, the bucking load, for the bar will be

becomes

P

P A Flexural rigidity, EI L Torsional spring of stiffness B

Where α is equal to (A) 0.25 (B) 1.00

(C) 2.05 (D) 4.00

CE – 2007 2. A steel column pinned at both ends, has a buckling load of 200 kN. If the column is restrained against lateral movement at its mid-height, its buckling load will be (A) 200 kN (C) 400 kN (B) 283 kN (D) 800 kN CE – 2008 3. Cross-section of a column consisting of two steel strips, each of thickness t and width b is shown in the figure below. The critical loads of the column with perfect bond and without bond between the strips are P and respectively. The ratio P/P is.

(A) 2 (B) 4

(C) 6 (D) 8

(A) 0.5 KL (B) 0.8 KL

(C) 1.0KL (D) 1.2KL

CE – 2010 5. The effective length of a column of length L fixed against rotation and translation at one end and free at the other end is (A) 0.5 L (C) 1.414 L (B) 0.7 L (D) 2 L CE – 2012 6. The ratio of the theoretical critical buckling load for a column with fixed ends to that of another column with the same dimensions and material, but with pinned ends, is equal to (A) 0.5 (C) 2.0 (B) 1.0 (D) 4.0 CE – 2013 7. Two steel column P(length L and yield strength f Pa) and Q(length 2L and yield strength f MPa) have the same cross-section and end condition the ratio of bucking load of column P to that of column Q is (A) 0.5 (C) 2.0 (B) 1.0 (D) 4.0

th

th

th



Mechanics

CE – 2014 8. The possible location of shear centre of the channel section, shown below is,

P

Q

R

S

(A) P (B) Q

(C) R (D) S


[Ans. C] Support A behaves like a hinge and support B like a fixed end

When the steel strips are perfectly bonded then

l=

When the steel strips are not bonded, then bt bt

P

( )

√

π

π

( ) √

Nearest approximate answer is (C) 2.

[Ans. D] When both ends are hinged, the bucking load is given by π P

/

P P 4.

[Ans. C] Let the deflection in the spring be and force in the spring be F. Taking moments about G, we get (But F ) P F

π When the lateral movement at the mid height is not available, than buckling load. kN π π where ( ) 3.

/

∴

P P P

P

F

[Ans. B] We know that critical load for a column is proportional to moment of inertia irrespective of end conditions of the column i.e. , P th

th

th



5.

Mechanics

[Ans. D] Force end Left =2L

Fixed end

6.

[Ans. D] Eulers critical load, π P P P

7.

8.

(l ) * + (l )

[Ans. D] Buckling load, π P (l ) π ∴P P ( ) P ∴ P

[

l ] l⁄

π ( )

[Ans. A]

th

th

th



Mechanics

Analysis of Statically Determinate Structures CE – 2014 1. The axial load (in kN) in the member PQ for the arrangement/assembly shown in the figure given below is ____________ P

2m

160 kN

Q 2m

Beam S

R 2m


[Ans. 50] Free body diagram

160 kN

S

Q 2m

2m

We can neglect the axial deformation as deflection due to axial forces will be less compared to bending forces.

kN

th

th

th



RCC

Concrete Technology CE – 2009 1. Match List I (list of test methods for evaluating properties of concrete) with List II (List of properties) and select the correct answer using the codes given below the list. List I List II (A) Resonant 1. Tensile Frequency test strength (B) Rebound 2. Dynamic hammer test Modulus of Elasticity (C) Split cylinder 3. Workability test (D)Compacting 4. Compressive factor test strength Codes A B C D (A) 2 4 1 3 (B) 2 1 4 3 (C) 2 4 3 1 (D) 4 3 1 2 CE – 2011 2. The cross-section of a thermo-mechanically treated (TMT) reinforcing bar has (A) Soft ferrite-pearlite throughout. (B) Hard martensite throughout (C) A soft ferrite-pearlite core with a hard martensitic rim. (D) A hard martensitic core with a soft pearlite-bainitic rim. CE – 2013 3. Maximum possible value of Compacting Factor for fresh (green) concrete is (A) 0.5 (C) 1.5 (B) 1.0 (D) 2.0

CE – 2014 4. Group I contains representative stress – strain curve as shown in the figure, while Group II gives the list of materials. Match the stress – stress curve with corresponding materials. Stress J K

L

Strain

Group I P. Curve J Q. Curve K R. Curve L (A) (B) (C) (D)

Group II 1. Cement paste 2. Coarse aggregate 3. Concrete

–

5.

The modulus of elasticity, E = 5000√ where is the characteristic compressive strength of concrete, specified in IS: 45-2000 is based on (A) Tangent modulus (B) Initial tangent modulus (C) Secant modulus (D) Chord modulus

6.

The first moment of area about the axis of bending for a beam cross-section is (A) Moment of inertia (B) Section modulus (C) Shape factor (D) Polar moment of inertia

th

th

th



RCC


[Ans. A]

2.

[Ans. C]

3.

[Ans. B] Compacting factor is ratio of actual density after compaction/theoretical density (maximum) and for fresh concrete it approaches unity semi compact sections.

4.

[Ans. B]

5.

[Ans. B] Modulus of electricity of concrete is based on initial tangent modulus

6.

[Ans. B] i

th

th

th



RCC

Basic of Mix Design CE – 2005 1. The flexural strength of M-30 concrete as per IS : 456-2000 is (A) 3.83MPa (C) 21.23MPa (B) 5.47MPa (D) 30.0MPa

2.

In a random sampling procedure for cube strength of concrete, one sample consists of X number of specimens. These specimens are tested at 28 days and average strength of these X specimens is considered as test result of the sample, provided the individual variation in the strength of specimens is not more than Y percent of average strength the value of X and Y as per IS :456-2000 are (A) 4 and 10 respectively (B) 3 and 10 respectively (C) 4 and 15 respectively (D) 3 and 15 respectively

CE – 2006 3. If the characteristic strength of concrete fck is defined as the strength below which not more than 50% of the results are expected to fall, the expression for fck in terms of mean strength fm and standard deviation would be (A) fm 0.1645 (C) fm (B) fm 1.645 (D) fm+ 1.645 CE – 2007 4. Consider the following statements 1. The compressive strength of concrete decreases with increase in water cement ratio of concrete mix 2. Water is added to concrete mix for hydration of cement and workability 3. Creep and shrinkage of concrete are independent of the water cement ratio in concrete mix. The TRUE statements are (A) 1 and 2 (C) 2 and 3 (B) 1, 2 and 3 (D) 2 only

5.

Consider the following statements 1. Modulus of elasticity of concrete increases with increase in compressive strength of concrete 2. Brittleness of concrete increases with decrease in compressive strength of concrete. 3. Shear strength of concrete increases with increase in compressive strength of concrete. The TRUE statements are (A) 2 and 3 (C) 1 and 2 (B) 1, 2 and 3 (D) 1 and 3

CE – 2008 6. A reinforced concrete structure has to be constructed along a sea coast. The minimum grade of concrete to be used as per IS : 456-2000 is (A) M 15 (C) M 25 (B) M 20 (D) M 30 CE – 2009 7. The modulus of Rupture of concrete in terms of its characteristic cube compressive strength (fck)in MPa according to IS : 456 -2000 is (A) 5000 fck (C) 5000√ (B) 0.7 fck (D) 0.7√ CE – 2011 8. A 16mm thick plate measuring 650mm × 420 mm is used as a base plate for an ISHB 300 column subjected to a factored axial compressive load of 2000 kN. As per IS : 456:2000, the minimum grade of concrete that should be used below the base plate for safely carrying load is (A) M 15 (C) M 30 (B) M 20 (D) M 40

th

th

th



9.

Consider a reinforcing bar embedded in concrete. In a marine environment this bar undergoes uniform corrosion, which leads to deposition of corrosion products on its surface and an increase in apparent volume of the bar. This subjects the surrounding concrete to expansive pressure. As a result, corrosion cracks appear at surface of concrete. Which of the following statement is true? (A) Corrosion causes circumferential tensile stress in concrete and the cracks will be parallel to corroded reinforcing bar. (B) Corrosion causes radial tensile stresses in concrete and cracks will be parallel to corroded reinforcing bar. (C) Corrosion causes circumferential tensile stresses in concrete and cracks will be perpendicular to the direction corroded reinforcing bar. (D) Corrosion causes radial tensile stresses in concrete and cracks will be perpendicular to direction of the corroded reinforcing bar.

CE – 2012 Statement for Linked Answer Questions 10 & 11 The cross – section at mid – span of a beam at the edge of a slab is shown in the sketch. A portion of the slab is considered as the effective flange width for the beam. The grades of concrete and reinforcing steel are M25 and Fe415, respectively. The total area of reinforcing bars (A), is 4000 . At the ultimate limit state, denotes the depth of the neutral axis from the top fiber. Treat the section as under – reinforce and flanged ( .

RCC 1000 100

650

570

325 All dimensions are in mm.

10.

The value of (in mm) computed as per the Limit State Method of IS 456:2000 is (A) 200.0 (C) 236.3 (B) 223.3 (D) 273.6

11.

The ultimate moment capacity (in kNm) of section, as per the Limit State Method is IS 456:2000 is (A) 475.2 (C) 756.4 (B) 717.0 (D) 762.5

CE – 2014 12. In a reinforced concrete section, the stress at the extreme fibre in compression is 5.80 MPa. The depth of neutral axis in the section is 58 mm and the grade of concrete is M25. Assuming linear elastic behavior of the concrete, the effective curvature of the section (in per mm) is (A) (C) (B) (D)

13.

th

The target mean strength for concrete mix design obtained from the characteristic strength and standard vi i σ , i i : 6-2000, is (A) σ (C) σ (B) σ (D) 6 σ

th

th



RCC


[Ans. A] Flexural strength of concrete is given as

7.

[Ans. D] The modulus of rupture of concrete

= 0.7√ = 0.7 × √ 2.

3.

is = 3.83 N/mm2 = 3.83 MPa 8.

[Ans. B]

[Ans. D] As per IS : 456 – 2000, the variation should not be more than 15% for 3 sample listed for a cube for 28 days compressive strength.

Working axial load =

The allowable bearing concrete may be given as i σ

[Ans. C] As per clause 8.2.8 of IS:456-2000 concrete in sea water or exposed directly along the sea coast shall be atleast M 20 grade in the case of plain concrete and M 30 in case of reinforced concrete.

4.

[Ans. C] For 50% fck=fm Where fm = average or mean compressive strength.

5.

[Ans. B] 1. Compressive strength of concrete increase with decrease of water – cement ratio 2. Main function of water in concrete is hydration of cement and make concrete workable. 3. Creep and shrinkage are independent of water-cement ratio.

6.

[Ans. B] 1.

2.

3.

= 0.7√

Ec = 5000√ . Therefore modulus of elasticity of concrete increases with compressive strength of concrete. Brittleness is opposite to elasticity of the concrete as brittleness increases, elasticity decreases so the strength of concrete. More compressive strength means more compacted material and hence, more shear strength of concrete.

pressure

on

6

⁄ The permissible stress in direct compression in various grades of concrete as per IS: 456-2000 are tabulated below: Grades Stress M10 2.5 M15 4.0 M20 5.0 M25 6.0 M30 8.0 M35 9.0 M40 10.0 M45 11.0 M50 12.0 The permissible stress in concrete should be more than the allowable bearing pressure. Thus the minimum grade of concrete which should be used is M20. 9.

[Ans. C]

Sectional view of reinforcement bar

Corrosion on the surface of steel bar will cause apparent increase in volume of concrete as it is obvious if stress acts in X th

th

th



– direction thus cracks will develop in direction perpendicular to X ie. Y. By some reason as circumferential tensile stress develops in bar, hence crack will develop parallel to surface of bar. 10.

12.

RCC

[Ans. C] From flexure formula

Here E = 5000 √

[clause 6.2.1.1 IS 456]

= 5000 √ Y = 58 mm σ

[Ans. C] 1000 mm

570 mm

100 mm

30 mm

C is correct answer 13.

[Ans. D] 6 σ

570

325

Given, And section is under reinforced ( (6 6 6

Take Check

11.

6

6 ( (

6

[Ans. B] 6

(

( (6

6 6 ( 6 6

(

, take M = 717.00 kNm

th

th

th



RCC

Design of RCC structures CE – 2013 1. As per IS 456:2000 for M20 grade concrete and plain barsin tension the design bond stress . Further, IS 456:2000 permits this design bond stress value to be increased by 60% for HSD bars. The stress in the HSD reinforcing steel barsin tension, σ = 6 . Find the required development length, , for HSD barsin terms of the bar diameter, .____

CE – 2014 2. Match the information given in Group – I with those in Group – II. Group – I A. Factor to decrease ultimate strength to design strength B. Factor to increase working load to ultimate load for design C. Statical method of ultimate load analysis D. Kinematical mechanism method of ultimate load analysis Group – II 1. Upper bound on ultimate load 2. Lower bound on ultimate load 3. Material partial safety factor 4. Load factor A B C D (A) 1 2 3 4 (B) 2 1 4 3 (C) 3 4 2 1 (D) 4 3 2 1


[Ans. *] Range 46 to 47 σ ( 6 6 6 6

2.

[Ans. C]

th

th

th



RCC

Analysis of Ultimate Load Capacity CE – 2005 1. A rectangular column section of 250mm × 400mm is reinforced with five steel bars of grade Fe-500, each of 20mm diameter. Concrete mix is M-30. The axial load on the column section with minimum eccentricity as per IS:456-2006 using limit state method can be applied upto. (A) 1707.37 (C) 1806.40 (B) 1805.30 (D) 1903.7

Stress

Common Data for Questions 2 and 3 Assume straight line instead of parabola for stress-strain curve of concrete as given below and partial factor of safety as 1.0

0.67

0.002

CE – 2006 Statement for Linked Answer Questions 4 and 5 In the design of beams for the limit state of collapse in flexure as per IS:456-2000, let the maximum strain in concrete be limited to 0.0025 (in place of 0.0035). For this situation, consider a rectangular beam section with breadth as 250mm, effective depth as 350mm, area of tension steel as 1500 and characteristic strengths of concrete and steel as 30 and 250 MPa respectively 4. The depth of neutral axis for the balanced failure is (A) 140mm (C) 168mm (B) 156mm (D) 185mm 5.

At the limiting state of collapse in flexure, the force acting on the compression zone of section is (A) 326 kN (C) 424 kN (B) 389 kN (D) 542 kN

6.

Consider the following statements: 1. The width-to-thickness ratio limitations on the plate elements under compression in steel members are imposed by IS: 800-1984 in order to avoid fabrication difficulties. 2. In a doubly reinforce concrete beam, the strain in compressive reinforcement is higher than the strain in adjoining concrete. 3. If a cantilever I-section supports slab construction all along its length with sufficient friction between them, the permissible bending stress in compression will be the same as that in tension. The true statements are (A) 1 and 2 (C) 1 and 3 (B) 2 and 3 (D) 1, 2 and 3

0.0035 Strain

A rectangular under-reinforced concrete section of 300 mm width and 500mm effective depth is reinforced with 3 bars of grade Fe-415, each of 16mm diameter. Concrete mix is M-20. 2.

The depth of neutral axis from the compressive fibre is (A) 76mm (C) 87mm (B) 81mm (D) 100mm

3.

The depth of the neutral axis abtained as per IS : 456-2000 differs from the depth of neutral laxis obtained in Q.2 by (A) 15mm (C) 25mm (B) 20mm (D) 32mm

th

th

th



7.

8.

As per IS: 456-2000, consider the following statements 1. The modular ratio considered in the Working stress methods depends on the type of steel used. 2. There is an upper limit on the nominal shear stress in beams (even with shear reinforcement) due to the possibility of crushing of concrete in diagonal compression. 3. A rectangular slab whose length is equal to its width may not be a two – way slab for some support conditions. The true statement are (A) 1 and 2 (C) 1 and 3 (B) 2 and 3 (D) 1, 2 and 3 Assuming concrete below the neutral axis to be cracked, the shear stress across the depth of a singly-reinforced rectangular beam section. (A) Increases parabolically to the neutral axis and then drops suddenly to zero value. (B) Increases parabolically to the neutral axis and then remains constant over the remaining depth. (C) Increases linearly to the neutral axis and then remains constant up to the tension steel (D) Increases parabolically to the neutral axis and then remains constant up to the tension steel.

9.

The limiting value of the moment of resistance of the beam in kN-m is (A) 0.14 (C) 45.08 (B) 0.45 (D) 156.82

10.

The limiting area of tension steal in mm2 is (A) 473.9 (C) 373.9 (B) 412.3 (D) 312.3

CE – 2008 11. In the design of a reinforced concrete beam the requirement for bond is not getting satisfied. The economical option to satisfy the requirement for bond is by (A) Bundling of bars (B) Providing smaller diameter bars more in number (C) Providing larger diameter bars less in number (D) Providing same diameter bars more in number.

12. CE – 2007 Common Data for Questions 9 and 10 A single reinforced rectangular concrete beam has a width of 150mm and an effective depth of 330mm. The characteristic compressive strength of concrete is 20 MPa and the characteristic tensile strength of steel is 415 MPa. Adopt the stress block for concrete as given is IS:456-2000 and take limiting value of depth of neutral axis as a 0.48 times the effective depth of the beam.

RCC

13.

th

Common Data for Questions 12 & 13 A reinforced concrete beam of rectangular cross-section of breadth 230 mm and effective depth 400mm is subjected to a maximum factored shear force of 120kN. The grades of concrete, main steel and stirrup steel are M 20, Fe-415 and Fe-250 respectively. For the area of main steel provided, the design h g h c as per IS: 456-2000 is 2 0.48N/mm . The beam is designed for collapse limit state. The spacing (mm) of 2-legged 8mm stirrups to be provided is (A) 40 (C) 250 (B) 115 (D) 400 In addition, the beam is subjected to a torque whose factored value is 10.90 kN-m. The strirrups have to be provided to carry a shear (kN) equal to (A) 50.42 (C) 151.67 (B) 130.56 (D) 200.23

th

th



14.

A reinforce concrete column contains longitudinal steel equal to 1% of net cross-sectional area of column. Assume modular ratio as 10. The loads carried (using elastic theory) by the longitudinal steel and net area of concrete are Ps and Pc respectively. The ratio Ps/Pc expressed as percent is (A) 0.1 (C) 1.1 (B) 1 (D) 10

15.

Un-factored maximum bending moments at a section of a reinforced concrete beam resulting from a frame analysis are 50, 80, 120 and 180 kN-m under dead, live, wind and earthquake loads respectively. The design moment (kNm) as per IS 456: 2000 for the limit state of collapse (flexure) is (A) 195 (C) 345 (B) 250 (D) 372

CE – 2010 Statement for Linked Answer Questions 16 & 17 A doubly Reinforced rectangular concrete beam has a width of 300 mm and an effective depth of 500 mm. the beam is reinforced with 2200 mm2 of steel in tension and 628 mm2 of steal in compression. The effective cover for compressive steel is 50mm. Assume that both tension and compressive steel yield. The grades of concerete and steel used are M 20 and Fe 250 respectively. The stress block parameters (rounded off to first two decimal places) for concrete shall be as per IS: 456-2000) 16.

The depth of neutral axis is (A) 205.30mm (C) 160.91mm (B) 184.56mm (D) 145.30mm

17.

The moment of resistance of section is (A) 206.00 KN-m (C) 237-80 kN-m (B) 209.20KN-m (D) 251.90Kn-M

RCC

CE – 2011 18. Consider two RCC beams, P and Q, each having the section 400 mm × 750mm (effective depth, d = 750mm) Made with 2 h vi g cmax = 2.1N/mm . For the reinforcement provided and the grade of concrete used , it may be assumed that h c = 0.75 N/mm2. The design shear in beam P is 400kN and in beam Q is 750kN. Considering the provisions of IS:4562000, which of the following statements is true? (A) Shear reinforcement should be designed for 175kN for beam P and the section for beam Q should be revised. (B) Nominal shear reinforcement is required for beam P & shear reinforcement should be designed for 120kN for beam Q (C) Shear reinforcement should be designed for 175kN for beam P and shear reinforcement should be designed for 525 kN for beam Q (D) The sections for both beams P & Q need to be revised. 19.

th

Consider a bar of diameter D embedded in a large concrete block as shown in the adjoining figure, with a pull out force P i g i σb σst be the bond strength (between the bar and concrete) and the tensile strength of the bar, respectively. If the block is held in position and it is assumed that the material of block does not fail, which of the following options represents the maximum value of P?

th

th


. GATE QUESTION BANK Concrete block

22.

A rectangular beam of width (b) 230 mm and effective depth (d) 450 mm is reinforced with four bars of 12 mm diameter. The grade of concrete is M20 and grade of steel is Fe500. Given that for M20 grade of concrete the ultimate shear strength, 6 for steel percentage, p = 0.25, and for p = 0.50. For a factored shear force of 45 kN, the diameter (in mm) of Fe500 steel two legged stirrups to be used at spacing of 375 mm, should be (A) 8 (C) 12 (B) 10 (D) 16

23.

For a beam of cross-section, width = 230 mm and effective depth = 500 mm, the number of rebars of 12 mm diameter required to satisfy minimum tension reinforcement requirement specified by IS:456-2000 (assuming grade of steel reinforcement as Fe500) is _____________

24.

The flexural tensile strength of M25 grade of concrete, in , as per IS:456-2000 is __________

Embedded Steel bar L

P

(A) Maximum of (

σ )

(π

σst)

(B) Maximum of (

σ )

(π

σb)

(C) Minimum of (

σ )

(π

σ

(D) Minimum of (

σ )

(π

σ

CE – 2012 20. As per IS 456:2000, in the Limit State Design of a flexural member, the strain in reinforcing bars under tension at ultimate state should not be less than (A)

(C)

(B)

(D)

RCC

CE – 2014 21. While designing, for a steel column of Fe250 grade, a base plate resting on a concrete pedestal of M20 grade, the bearing strength of concrete (in N/ ) in limit state method of design as per IS : 456 – 2000 is _______

th

th

th



RCC


6

[Ans. A] The axial Load on the column can be given as 6 (

*

π

(

=3

( 6

= 100.38 So, difference = 100.38 76 = 24.38 mm = 25mm.

+

4.

(

6

[Ans. B] B

(

0.0035

( 6

0.002 ,

d

2.

[Ans. A] Variation of stress is taken as straight line instead of parabola for the strain upto 0.002 and rest rectangular. Here d, = 500mm, let x be depth of neutral axis so force of tension = Force of compression 3× (16)2 × 0.87 × fy = 0.67 fck × b × xu + 0.67

similar triangle, we have ,

,

× b × xu ×

Calculation of depth of rectangular section from strain diagram.

, ,

X=

, ⁄

,

(

(

= 75.84.38 = 76mm 3.

,

[Ans. C] Difference between above result when variation of stress for strain upto 0.002 is taken as straight line instead of parabolically IS : 456 – 2000 provided parabolic variation of stress upto strain 0.002 partial factor of safety as 1.5 for concrete equating, force of compression = Force in tension

,

,

[

,

66

,

th

]

th

6

th



5.

[Ans. B]

Parabolic

B

A 0.0035 B

E d

0.002 F

O

D

,

C

9.

[Ans. C] B = 150mm, d = 330 mm, fck = 20 MPa fy = 4.15, , = 0.48d, =? Limiting value of moment of Resistance = Force of compression x lever arm 6 ( ) , , ( 6 ( = 45.07 kN-m

10.

[Ans. A] Limiting area of tension in steel, Force of tension = Force in compression 6 ( 6

In strain diagram

T

6.

7.

iv

,

For limiting state of collapse in flexure 6 , T iv , 6 6

11.

[Ans. B] To avoid buckling not for fabrication difficulty

[Ans. B] Provide small diameter bars in more number

12.

[Ans. B] Here, B = 230 mm, d = 400 mm, V = 120 kN 2 c = 0.48 N/mm Total shear carried by beam = 0.48 × 230 × 400 = 44.16 kN< 120 kN Net shear force to be used for design = 120 – 44.16 = 75.84KN For vertical stirrups, spacing is given as

= 473.82mm2

[Ans. B] Modular ratio in wsm method is defined as m =

where σ

permissible comp.

stress due to bending in concrete Therefore, it does not depend upon type of steel used, but type of grade of concrete. 8.

RCC

[Ans. D] Shear stress distribution along the depth for a singly reinforced rectangular beam section is given as

π

th

th

th



13.

[Ans. C] Now, T = 10.90 kN-m Equivalent Shear force due to concrete 6T 6 = 75.82 Total Shear Force = 75.82 + 75.84 = 151.66 kN

14.

15.

16.

(

=

= 158.29 mm 6 (Discrepancy in deduction of area of concrete taken over by steel in compression) 17.

[Ans. B] Moment of resistance (MR) calculated by using compressive force ( 6 ( 6 6 ( 6 (( ( 6 ( ) = 209.21 kNm

18.

[Ans. A] 𝛕cmax = 2.1 N/mm2, c = 0.75 N/mm2 Pr = 400 KN, Qr = 750kN For P,

[Ans. D] If A is the net cross sectional area of the column, is the area of steel, is the area of concrete, m is modular ratio and is stress in concrete, then

[Ans. D] Wind and earthquake effects are not considered simultaneously. (i) Design moment when wind effect is considered = 1.2 (DL + LL + WL) = 1.2 (50 + 80 + 120) = 300 kN-m (ii) Design moment when earthquake effect is considered = 1.2 (DL + LL + EL) = 1.2 (50 + 80 + 180) = 372 kN-m

RCC

v

=

as h

, , h

i

i

( v c) bd = (1.33 0.75) = 175 kN ,

V

400

vi 750 N

=

v> max

hence, sections should be

revised. 19.

[Ans. C] σ will be working for peripheral area embedded in concrete B g h σb (π σst will act on x –section are of bar

[Ans. C] Both compressive steal and tensioned steal yield now, let x be the depth of neautral axis, Total force of compression = 6 and force intension = 0.87 fy Equating both, = 6

T

i

gh

σst(

) D

P 10 L

(

th

th

th



Maximum value of P will be minimum of bond strength and tensile strength. P=

σ ( i [ σ (π

)

23.

[Ans. 2] Minimum tension reinforcement as per 26.5.11 is

]

20.

[Ans. D]

21.

[Ans. 9] As per clause 344 of IS 456: 2000 Permissible bearing stress =

RCC

1

(

=9 22.

12 rebar =

are required

[Ans. A] 230m m

24.

[Ans. 3.5] Flexural strength =

= 0.7 √ = 3.5 N/

450 mm

i

√

(

i ( (

6

, i i i i

h h

i

i

i

i

π

,

th

th

th



RCC

Basic Elements of Pre-stressed Concrete CE – 2005 1. IS 1343: 1980 limits the minimum characteristic strength of pre-stressed concrete for post tensioned work and pretension work as (A) 25 MPa, 30 MPa respectively (B) 25 MPa, 35 MPa respectively (C) 30 MPa, 35 MPa respectively (D) 30 MPa, 40 MPa respectively

2.

CE – 2010 5. As per Indian standard code of practice for pre-stressed concrete (IS: 1343-1980) the minimum grades of concrete to be used for post-tensioned and pretensioned structural elements are respectively (A) M 20 for both (C) M 15 and M 20 (B) M 40 and M 30 (D) M 30 and M 40

A concrete beam of rectangular cross section of 200 mm × 400 mm is prestressed with a force 400 kN at eccentricity 100 mm. The maximum compressive stress in the concrete is (A) (C) 5.0 (B) 7.5 (D) 2.5

CE – 2007 3. The percentage loss of pre-stress due to anchorage slip of 3 mm in a concrete beam of length 30m, which is post tensioned by a tendon with an initial stress of 1200 N/mm2 and modulus of elasticity equal to 2.1 × 105 N/mm2 is (A) 0.0175 (C) 1.75 (B) 0.175 (D) 17.5 CE – 2009 4. A rectangular concrete beam of width 120 mm & depth 200 mm is pre-stressed by pre-tensioning to a force of 150 KN at an eccentricity of 20mm. The cross sectional area of the pre-stressing steel is 187.5 mm2. Take modulus of elasticity of steel and concrete as 2.1 × 105 MPa and 3 × 104 MPa respectively. The percentage loss of stress in pre-stressing steel due to elastic deformation of concrete is. (A) 8.75 (C) 4.81 (B) 6.125 (D) 2.19

th

th

th



RCC


[Ans. D]

2.

[Ans. A]

So, strain in steel = strain in concrete ⁄

σ

⁄

ii

6 (

,

g

6 6

3.

[Ans. C] Slip 5 =3mm Length = 30m Ec = 2.1×105 N/mm2 ,

5.

[Ans. D] A high strength concrete is always required for pre-stressed concrete work the pre-tension losses are more than the post tension losses therefore minimum M 30 is used for post tension work while minimum M 40 is used for pre-tension work.

i

So, stress loss = strain × ⁄ , 4.

[Ans. B] Here, B

, ,

, ,

So Percentage loss of pre-stresses due to elastic shortening = ? Initial direct stresses 6

⁄

And stress due to eccentricity

( ⁄ Maximum compression in concrete, = 6.25+0.75= 7N/mm2 This stress will cause compression in steel strain concrete

th

th

th



RCC

Design of Pre-Stressed Concrete Beams The stress (in N/ at mid – span is

CE – 2007 1. A concrete beam of rectangular crosssection of size 120 mm (width) and 200 mm (depth) is prestressed by a straight tendon to an effective force of 150 kN at an eccentricity of 20 mm (below the centroidal axis in the depth direction). The stresses at the top and bottom fibres of section are (A) 2.5 N/mm2 (compression), 2 10 N/mm (compression) (B) 10 N/mm2 (tension), 2.5 N/mm2 (compression) (C) 3.75 N/mm2 (tension), 3.75 N/mm2 (compression) (D) 2.75 N/mm2 (compression), 3.75 N/mm2 (compression) CE – 2008 2. A pre-tensioned concrete member of section 200 mm 250 mm contains tendons of area 500 mm2 at centre of gravity of section. The pre-stress in the tendons is 1000 N/mm2. Assuming modular ratio as 10, the stress (N/mm2) in concrete is (A) 11 (B) 9 (C) 7 (D) 5 CE – 2012 3. Which one of the following is categorized as a long –term loss of prestress in a prestressed concrete member? (A) Loss due to elastic shortening (B) Loss due to friction (C) Loss due to relaxation of strands (D) Loss due to anchorage slip 4.

)in the bottom fiber

145 7300 Sectional elevation All dimensions are in mm 500 750 Cross – section (tendon not shown)

(A) Tensile 2.90 (B) Compressive 2.90 (C) Tensile 4.32 (D) Compressive 4.32 CE – 2013 5. A rectangular concrete beam 250 mm wide and 600 mm deep is per – stressed by means of 16 high tensile wires, each of 7 mm diameter, located at 200 mm from the bottom face of the beam at a given section. If the effective pre - stress in the wires is 700 MPa, what is the maximum sagging bending moment (in kNm) (correct to 1 – decimal place) due to live load that this section of the beam can withstand without causing tensile stress at the bottom face of the beam? Neglect the effect of dead load of beam._______________

A concrete beam prestressed with a parabolic tendon is shown in the sketch. The eccentricity of the tendon is measured from the centroid of the cross – section. The applied prestressing force at service is 1620 kN. The uniformly distributed load of 45 kN/m includes the self – weight.

th

th

th



RCC


[Ans. A] B=120 mm, D=200 mm P=150 kN, e=20 mm Stress due to direct stresses

5.

[Ans. *] Range 85.5 to 86.5

6

6 Stress due to moment produced due to eccentricity of load M(P.e) ( So, stress at top, 6 Stress at bottom, 6 2.

Since the tensile stress at bottom face of the beam is zero

(compression) (

i

π

6

i

4.

6

6 Since the prestressing force is located at 200 mm from the bottom face of the beam i i

[Ans. B] Compressive force in steel = Area of section in terms of concrete =A+( ( (

3.

(

6 66 6

[Ans. C] The only time dependent loss relaxation of pre stress in strands.

6 6 6

6

is

[Ans. B] σ 6

6 ( (

[

) )

] (

th

th

th



RCC

Concrete Design CE – 2005 1. The partial factor of safety for concrete as per IS: 456-2000 is (A) 1.50 (C) 0.87 (B) 1.15 (D) 0.446

CE – 2009 2. For limit state of collapse, the partial safety factors recommended cube by IS : 456-2000 for estimating the design strength of concrete and reinforcing steel are respectively (A) 1.15 and 1.5 (C) 1.5 and 1.15 (B) 1.0 and 1.0 (D) 1.5 and 1.0


[Ans. A] Partial factor of safety = 1.50 The High FOS is because of larger variation in strength of concrete in comparison to steel.

2.

[Ans. C] Partial safety factor for concrete is 1.5 and for steel is 1.15 per IS: 456 – 2000

th

th

th



Steel structure

Introduction CE – 2013 1. As per IS 800:2007, the cross – section in which the extreme fiber can reach the yield stress, but cannot develop the plastic moment of resistance due to failure by local buckling is classified as (A) Plastic section (B) Compact section (C) Semi – compact section (D) Slender section CE – 2014 2. Match the information given in Group – I with those in Group - II. Group – I Group – II P. Factor to decrease 1. Upper bound ultimate strength on ultimate to design strength load Q. Factor to increase 2. Lower bound working load to on ultimate ultimate load for load design R. Statical method of 3. Material partial ultimate load safety factor analysis S. Kinematical 4. Load factor mechanism method of ultimate load analysis (A) (B) (C) (D) – 3.

(A) 30.33 and 20.00 (B) 30.33 and 25.00 (C) 33.33 and 20.00 (D) 33.33 and 25.00 4.

The ultimate collapse load (P) in terms of plastic moment by kinematic approach for a propped cantilever of length L with P acting at its mid – span as shown in the figure, would be

(A)

(C)

(B)

(D)

5.

The first moment of area about the axis of bending for a beam cross-section is (A) moment of inertia (B) section modulus (C) shape factor (D) polar moment of inertia

6.

A prismatic beam (as shown below) has plastic moment capacity of then the collapse load P of the beam is

The tension and shear force (both in kN) in each bolt of the joint, as shown below, respectively are

th

(A)

(C)

(B)

(D) th

th



Steel structure


[Ans. C ] (fail by local lusting )

Stress Compact sections (full elastic capacity only )

Internal work done: Only point C and A will form plastic hinge (since B is free to rotate roller joint). = By principle of virtual work

Plastic section (full plastic moment capacity)

Strain

2.

3.

[Ans. C] Steel structure: Plastic analysis P and Q are definitions of partial factor of safety for material and load respectively R: Statical method of ultimate load analysis is based on lower bound theorem which states that actual collapse load cannot be less than collapse load obtained from static method of load analysis. S: kinematic mechanism method of ultimate load analysis follows upper bound theorem.

P= 5.

[Ans. B]

6.

[Ans. C]

Degree of static indeterminacy = 0 No.of plastic winges required

[Ans. D] Horizontal force = = 200 kN

p

Vertical force = 250 = 150 kN Horizontal force will exert tensite pull of magnitude

= 33.33 kN on each bolt.

Vertical force will exert shear force of magnitude

= 25 kN on each bolts

Mp Mp

D is correct choice 4.

From principal of virtual work

[Ans. C]

(

)

External work done th

th

th



Steel structure

Plastic Analysis CE – 2005 1. A cantilever beam of length L, width b and depth d is loaded with a concentrated vertical load at the tip. If yielding starts at a load P, the collapse load shall be (A) 2.0 P (C) 1.2 P (B) 1.5 P (D) P CE – 2006 2. When the triangular section of a beam as shown below becomes a plastic hinge, the compressive force acting on the section (with denoting the yield stress) becomes

h

b

(A)

(C)

(B)

(D)

R

(C)

(B)

(D)

Continuous beam is loaded as shown in the figure below. Assuming a plastic moment capacity equal to , the minimum load at which the beam would collapse is P P J I G H L

(A)

(C)

(B)

(D)

CE – 2009 6. The square root of the ratio of moment of inertia of the cross-section to its crosssectional area is called (A) second moment of area (B) slenderness ration (C) section modulus (D) radius of gyration 7.

CE – 2007 3. The plastic collapse load for the propped cantilever supporting two point loads as shown in figure in terms of plastic moment capacity, is given by w w L/3 L/3 L/3 (A)

5.

In the theory of plastic bending of beams, the ratio of plastic moment to yield moment is called (A) Shape factor (B) Plastic section modulus (C) Modulus of resilience (D) Rigidity modulus

CE – 2011 8. The value of W that results in the collapse of the beam shown in the adjoining figure and having a plastic moment capacity of is W

Fixed

CE – 2008 4. The shape of the cross-section, which has the largest shape factor, is (A) Rectangular (C) Diamond (B) I-section (D) Solid circular

7m

3m

(A) (4/21) (B) (3/10)

th

th

Hinge

(C) (7/21) (D) (13/21)

th



CE – 2013 Common data Questions (9 & 10) A propped cantilever made of a prismatic steel beam is subjected to a concentrated load P at mid span as shown.

9.

If load P= 80 kN, find the reaction R(in kN)(correct to 1-decimal place) using elastic analysis. _________

10.

If the magnitude of load P is increased till collapse and the plastic moment carrying capacity of steel beam section is 90 kNm, determine reaction R(in kN)(correct to 1decimal place) using plastic analysis. _________________

P

R

1.5 m

1.5 m

Steel structure


[Ans. B] Yielding starts at P, therefore yielding moment at the fixed end is given by … P

3.

[Ans. B] w A L/3

L/3

L/3

=1 No. of plastic hinges formed,

L

1st case: Plastic hinge formed at A and B w w B C D A

PL

Now, if is the collapse load, then collapse moment at the fixed end is give by … But, for rectangular beam [

w

L/3

L/3

L/3

] By virtual work method External work = Internal work (

2.

[Ans. A] We know that the neutral axis of the plastified section is the equal area axis. Compressive force,

)

(

)

Put … 2nd case: Plastic hinge at (A) and (C) w w B C A

th

th

th

D



Steel structure

6.

[Ans. D]

7.

[Ans. A]

8.

[Ans. D] Number of possible hinges, n = 2 Statical indeterminacy, r = 1 Number of independent mechanisms,

External work = Internal work

Put

Hence, collapse load

W

4.

[Ans. C] Shape factors of some cross-section are as follows: (i) Rectangle – 1.5 (ii) I-section 1.14 (iii) Diamond – 2 (iv) Triangle 2.34 (v) Circle 1.7

5.

[Ans. B]

7m

3m

By principle of virtual work, we get P

G

H

P I

J

L

For collapse in IJ

9.

[Ans. 25]

From compatibility equation net deflection at C = 0

For collapse in HI [

]

Minimum load for collapse th

th

th



10.

Steel structure

[Ans. 60]

th

th

th



Steel structure

Welded Connections CE – 2005 1. A fillet-welded joint of 6 mm size is shown in the figure. The welded surfaces meet at 60-90 degree and permissible stress in the fillet weld is 108 MPa. The safe load that can be transmitted by the joint is

P = 10 kN

e =100

30 30

100 mm 50 mm

F

(A) 162.7 kN (B) 151.6 kN 2.

40

F

(C) 113.4 kN (D) 109.5 kN

An unstiffened web I-section is fabricated from a 10 mm thick plate by fillet welding as shown in the figure. If yield stress of steel is 250 MPa, the maximum shear load that section can take is

(all distance are in mm)

(A) 5 kN (B) 6.5 kN

(C) 6.8 kN (D) 7.16 kN

CE – 2008 4. Rivets and bolts subjected to both shear stress and axial tensile stress shall be so proportioned that the stresses do not exceed the respective allowable stresses and and the value of ( (A) 1.0

) does not exceed (B) 1.2

(C) 1.4

(D) 1.8

300 mm

CE – 2009 5. A 12 mm thick plate is connected to two 8 mm thick plates, on either side through a 16 mm diameter power driven field rivet as shown in the figure below. Assuming permissible shear stress as 90 MPa and permissible bearing stress as 270 MPa in the rivet, the rivet value of the joint is

200 mm

(A) 750 kN (B) 350 kN

40

(C) 337.5 kN (D) 300 kN

P/2

8mm 12mm

CE – 2007 3. A bracket connection is made with four bolts of 10 mm diameter and supports a load of 10 kN at an eccentricity of 100 mm. The maximum force to be resisted by any bolt will be.

P/2

P

8mm

(A) 56.70 kN (B) 43.29 kN

(C) 36.19 kN (D) 21.65 kN

CE – 2010 6. A double cover butt riveted joint is used to connect two flat plates of 200 mm width and 14 mm thickness as shown in the figure. There are twelve power driven rivets of 20 mm diameter at a pitch of th

th

th



50 mm in both directions on either side of the plate. Two cover plates of 10 mm thickness are used. The capacity of the joint in tension considering bearing and shear ONLY, with permissible bearing and shear stresses as 300 MPa and 100 MPa respectively is

7.

Steel structure

Two plates, subjected to direct tension, each of 10 mm thickness and having widths of 100 mm and 175 mm, respectively are to be fillet welded with an overlap of 200 mm. Given that the permissible weld stress is 110 MPa and the permissible stress in steel is 150 MPa, the length of the weld required using the maximum permissible weld size as per IS 800: 1984 is

50 mm 50 mm 50 mm 50 mm 50 mm

100 mm

50 mm

(A) 1083.6 kN (B) 871.32 kN

175 mm

50 mm

(C) 541.8 kN (D) 433.7 kN 200 mm

(A) 245.3 mm (B) 229.2 mm

(C) 205.5 mm (D) 194.8 mm

Answer Keys & Explanations

2.

[Ans. C] Total weld length = (100×2)+50 =250mm Strength of weld per mm length, = 0.70 × 6 × 108 = 453.6 N Maximum load taken by joint,

[Ans. D]

3.

[Ans. D] e =100

10 kN

30 mm 30 mm

1.

The web depth to thickness ratio 40 mm

40 mm

Bolt will be subjected to direct shear and shear force due to twisting moment. Force on bolt due to direct shear = Force on bolt due to twisting moment

√

th

th

th



Steel structure

Thus the strength of joint will be governed by shearing and it will be equal to 871.32 kN Resultant force,

7.

[Ans. B] The maximum size of a fillet weld is obtained by subtracting 1.5 mm from the thickness of the thinner member to be jointed.

√ √

4.

[Ans. C] As per clause 8.9.4.5 of IS:800-1984 rivets and bolts subject to both shear and axial tension shall be so proportioned that the shear and axial stresses calculated do not exceed the respective allowable stresses and and the expression (

5.

Strength of the thinner plate

Strength of the weld √

) does not exceed 1.4.

√

[Ans. B] Rivet value is the lesser of shearing strength or bearing strength of rivet. Shearing strength of rivet,

Bearing strength of rivet

Rivet value of the joint 6.

[Ans. B] Strength of one rivet in double shear,

Strength of the riveted joint in double shear Strength of one rivet in bearing , = 3 kN Strength of the riveted joint in bearing

th

th

th



Steel structure

Design of Tension Member CE – 2005 1. The permissible stress in axial tension in steel member on the net effective area of the section shall not exceed is the yield stress) (A) 0.80 (C) 0.60 (B) 0.75 (D) 0.50 CE – 2006 2. In the design of welded tension members, consider the following statements: 1. The entire cross-sectional area of the connected leg is assumed to contribute to the effective area in case of angles. 2. Two angles back-to-back and tackwelded as per the codal requirements may be assumed to behave as a tee section. 3. A check on slenderness ratio may be necessary in some cases. The TRUE statements are (A) Only 1 and 2 (C) Only 1 and 3 (B) Only 2 and 3 (D) 1, 2 and 3

CE – 2007 3. A steel flat of rectangular section of size 70 × 6 mm is connected to a gusset plate by three bolts each having a shear capacity of 15 kN in holes having diameter 11.5 mm. If the allowable tensile stress in the flat is 150 MPa, the maximum tension that can be applied to the flat is

15 20 20 15

T

35

(A) 42.3 kN (B) 52.65 kN

(C) 59.5 kN (D) 63.0 kN


[Ans. C]

2.

[Ans. D]

3.

[Ans. B]

Along (1)-(1) ( ) =52.65 kN Along (2)-(2) 2 1

Force to shear 1 bolt at position (1) – (1)

15 20 20 15

T

Hence, the maximum tension that we can applied = 52.65 kN

35 mm 1

2

th

th

th



Steel structure

Compression member CE – 2005 1. Which one of the following is NOT correct for steel sections as per IS 800: 1984? (A) The maximum bending stress in tension or in compression in extreme fiber calculated on the effective section of a beam shall not exceed 0.66 (B) The bearing stress in any part of a beam when calculated on the net area shall not exceed 0.75 (C) The direct stress in compression on the cross sectional area of axially loaded compression member shall not exceed 0.6 (D) None of the above CE – 2006 2. Consider the following statements: 1. Effective length of a battened column is usually increased to account for the additional load on battens due to the lateral expansion of columns. 2. As per IS 800: 1984, permissible stress in bending compression depends on both Euler bucking stress and the yield stress of steel.

3. As per IS 800: 1984, the effective length of a column effectively held in position at both ends but not restrained against rotation, is taken to be greater than that in the ideal end conditions. The TRUE statements are (A) 1 and 2 (C) 1 and 3 (B) 2 and 3 (D) 1,2 and 3 CE – 2009 3. Consider the following statements for a compression member: 1. The elastic critical stress in compression increases with decrease in slenderness ratio. 2. The effective length depends on the boundary conditions at its ends. 3. The elastic critical stress in compression is independent of the slenderness ratio 4. The ratio of the effective length to its radius of gyration is called as slenderness ratio. Which of the above statements is/are correct? (A) 2 and 3 (C) 2, 3 and 4 (B) 3 and 4 (D) 1, 2 and 4


[Ans. D]

2.

[Ans. A] The ideal condition is that column is effectively held in position at both ends but not restrained against rotation. IS 800: 1984, prescribes the same value of effective length as taken for ideal end condition. Hence 3 is false.

3.

[Ans. D] The elastic critical stress in compression depends on the slenderness ratio.

Where is slenderness ratio of the compression member

th

th

th



Steel structure

Beams CE – 2011 1. ‘ ’ adjoining figure the effective throat thickness is

CE – 2012 3. In a steel plate with bolted connections, the rupture of the net section is a mode of failure under (A) Tension (C) Flexure (B) Compression (D) Shear 4.

(A) 0.61s (B) 0.65s 2.

Two plates are connected by fillet welds of size 10 m and subjected to tension, as shown in the sketch. The thickness of each plate is 12 mm. The yield stress and the ultimate tensile stress of steel are 250 MPa and 410 MPa, respectively. The welding is done in the workshop ( . As per the Limit State Method of IS 800.2007. The minimum length (rounded off to the nearest higher multiple of 5mm) of each weld to transmit force P equal to 270 kN is

(C) 0.70s (D) 0.75s

The adjoining figure shows a schematic representation of a steel plate girder to be used as a simply supported beam with a concentrated load. For stiffeners, PQ (running along the beam axis) and RS (running between the top and bottom flanges) which of the following pairs of statements will be TRUE?

P 100mm

R P

Q S L

(A) (i) RS should be provided under the concentrated load only (ii) PQ should be placed in the tension side of the flange. (B) (i) RS helps to prevent local buckling of the web. (ii) PQ should be placed in the compression side of the flange. (C) (i) RS should be provided at supports. (ii) PQ should be placed along the neutral axis (D) (i) RS should be provided away from points of action of concentrated loads. (ii) PQ should be provided on the compression side of the flange.

150mm P

(A) 100 mm (B) 105 mm

(C) 110 mm (D) 115 mm

CE – 2014 5. A steel section is subjected to a combination of shear and bending actions. The applied shear force is V and the shear capacity of the section is . For such a section, high shear force (as per IS: 800 – 2007) is defined as (A) (C) (B) (D) th

th

th



Steel structure


[Ans. B] Throat thickness T = K fillet size K is depends upon angle between fusion forces

T

RS is a vertical stiffener in the given plate girder. Vertical stiffeners are also called transverse stiffeners. It is assumed that the vertical stiffener is not subjected to any load and is selected to provide necessary lateral stiffness only and can therefore, be crimped or joggled for tight fitting. Such stiffeners increase the buckling resistance of the web caused by shear.

49.5

S

3.

[Ans. A]

4.

[Ans. B] Design strength of fillet weld,

S

√ 49.5 T

+ s

√

990

Rounded off to 105 mm 5.

[Ans. A] As per clause 9.2.1 of IS 800: 2007, V > 0.6

s

T

x = 0.65s

2.

[Ans. B] PQ is a horizontal stiffener in the given plate girder. Horizontal stiffeners are also called longitudinal stiffeners. The horizontal stiffener are provided in the compression zone of the web. The first horizontal stiffener is provided at onefifth of the distance from the compression flange to the tension flange. If required another stiffener is provided at the neutral axis. Horizontal stiffeners are not continuous and are provided between vertical stiffeners.

th

th

th



Geotechnical Engineering

Three Phase System, Fundamental Definitions and Relationship CE – 2005 1. A saturated soil mass has a total density 22kN/m3 and a water content of 10%. The bulk density and dry density of this soil are (A) 12kN/m3 and 20 kN/m3 respectively (B) 22kN/m3 and 20 kN/m3 respectively (C) 19.8kN/m3 and 19.8 kN/m3 respectively (D) 23.2kN/m3 and 19.8 kN/m3 respectively CE – 2007 2. The water content of a saturated soil and the specific gravity of soil solids were found to be 30% and 2.70, respectively. Assuming the unit weight of water to be 10 kN/m3, the saturated unit weight (kN/m3) and the void ratio of the soil are (A) 19.4,0.81 (C) 19.4,0.45 (B) 18.5,0.30 (D) 18.5,0.45

CE – 2009 3. Deposit with flocculated structure is formed when (A) clay particles settle on sea bed (B) clay particles settle on fresh water lake bed (C) sand particles settle on river bed (D) sand particles settle on sea bed CE – 2012 4. As per the Indian Standard soil classification system, a sample of silty clay with liquid limit of 40% and plasticity index of 28% is classified as (A) CH (C) CL (B) CI (D) CL - ML CE – 2014 5. The clay mineral primarily governing the swelling behavior of Black Cotton soil is (A) Halloysite (C) Kaolinite (B) Illite (D) Montmorillonite


[Ans. B] Total density is bulk density,

3.

[Ans. B]

4.

[Ans. B] Since liquid limit lies between 35 and 50 it is intermediate compressible (1) Height of A – Line = 0.73 ( = 14.6% Since of soil is more than 14.6%, it falls above A – line, Hence it is clay (C) Soil is C I

5.

[Ans. D]

Dry density, w 2.

[Ans. A] We know that

And

* *

e e e +

w

+

th

th

th




Index Properties and Soil Classification CE – 2005 1. A soil mass contains 40% gravel, 50% sand 10% silt. This soil can be classified as (A) Silty sandy gravel having coefficient of uniformity less than 60. (B) Silty gravelly sand having coefficient of uniformity equal to 10. (C) Gravelly silty sand having coefficient of uniformity greater than 60. (D) Gravelly silty sand and its coefficient of uniformity cannot be determined. CE – 2006 Statement for Linked Answer question 2&3 Laboratory sieve analysis was carried out on a soil sample using a complete set of standard IS sieves. Out of 500g of soil used in the test, 200g was retained on IS 600 sieve, 250g was retained on IS 500 sieve and the remaining 50g was retained on IS 425 sieve. 2. The coefficient of uniformity of the soil is (A) 0.9 (C) 1.1 (B) 1.0 (D) 1.2 3.

The classification of the soil is (A) SP (C) GP (B) SW (D) GW

CE – 2007 4. Sieve analysis on a dry soil sample of mass 1000 g showed that 980 g and 270 g of soil pass through 4.75 mm and 0.075 mm sieve, respectively. The liquid limit and plastic limits of the soil fraction passing through 425 sieves are 40% and 18%, respectively. The soil may be classified as (A) SC (C) CI (B) MI (D) SM

CE – 2008 5. Group symbols assigned to silty sand and clayey sand are respectively (A) SS and CS (C) SM and SC (B) SM and CS (D) MS and CS 6.

The liquid limit (LL), plastic limit (PL) and shrinkage limit (SL) of a cohesive soil satisfy the relation (A) LL > PL < SL (C) LL < PL < SL (B) LL > PL > SL (D) LL < PL > SL

CE – 2009 7. The laboratory test results of a soil sample are given below: Percentage finer than 4.75 mm = 60 Percentage finer than 0.075 mm = 30 Liquid Limit = 35% Plastic Limit = 27% The soil classification is (A) GM (C) GC (B) SM (D) ML-MI CE – 2010 8. A fine grained soil has liquid limit of 60 and plastic limit of 20. As per the plasticity chart, according to IS classification, the soil is represented by the letter symbols (A) CL (C) CH (B) CI (D) CL-ML CE – 2011 9. The results for sieve analysis carried out for three types of sand, P, Q and R, are given in the adjoining figure. If the fineness modulus values of the three sands are given as FMP, FMQ and FMR it can be stated that

th

th

th


.

Percentage passing

GATE QUESTION BANK Q

Geotechnical Engineering Torque

P Top view (plan)

R

Vane rod

Vane blade

Torque Vane rod

Sieve size

(A) (B) (C) (D) 10.

11.

FMQ = √ FMQ = 0.5 (FMP + FMR) FMP> FMQ> FMR FMP< FMQ< FMR

A soil is composed of solid spherical grains of identical specific gravity and diameter between 0.075mm and 0.0075mm. If the terminal velocity of the largest particle falling through water without flocculation is 0.5 mm/s, that for the smallest particle would be (A) 0.005 mm/s (C) 5 mm/s (B) 0.05 mm/s (D) 50 mm/s A field vane testing instrument (shown alongside) was inserted completely into a deposit of soft, saturated silty clay with the vane rod vertical such that the top of the blades were 500mm below the ground surface. Upon application of a rapidly increasing torque about the vane rod, the soil was found to fail when the torque reached 4.6 Nm. Assuming mobilization of undrained shear strength on all failure surfaces to be uniform and the resistance mobilized on the surface of the vane rod to be negligible, what would be the peak undrained shear strength (rounded off to the nearest integer value of kPa) of the soil?

15 mm

Elevation

100 mm

50 mm

(A) 5 kPa (B) 10 kPa

(C) 15 kPa (D) 20 kPa

CE – 2013 12. In its natural condition, a soil sample has a mass of 1.980 kg and a volume of 0.001 . After being completely dried in an oven, the mass of the sample is 1.800 kg. Specific gravity G is 2.7. Unit weight of water is 10kN/ . The degree of saturation of the soil is : (A) 0.65 (C) 0.54 (B) 0.70 (D) 0.61 13.

Laplace equation for water flow in soil is given below. x y z Head H does not vary in y and z directions. Boundary conditions are: at x = 0, H = 5m, and What is the value of H at x = 1.2 ?_________

CE – 2014 14. A certain soil has the following properties: = 2.71, n = 40% and w = 20%. The degree of saturation of the soil (rounded off to the nearest percent) is __________

th

th

th



15.

As per Indian Standard Soil Classification System (IS: 1498 - 1970), an expression for A – line is (A) w (B) w (C) w (D) w

16.


A given cohesion less soil has e = 0.85 and e = 0.50. In the field, the soil is compacted to a mass density of 1800 g at a water content of 8%. Take the mass density of water as 1000 g and as 2.7. The relative density (in %) of the soil is (A) 56.43 (C) 62.87 (B) 60.25 (D) 65.71


2.

[Ans. D] As 50% of the soil is sand, the soil is sand with gravel and silt as ingredients. Also, there is no information regarding D60 and D10. Hence the soil is gravelly silty sand whose coefficient of uniformity cannot be determined. [Ans. D] S.No. Sieve Weight size Cum.weight retained % retained %N retained

1. 600 200 200

2. 500 250 450

3. 425 50 500

40 60

90 10

100 0

4.

[Ans. A]

5.

[Ans. C]

6.

[Ans. B] u e

w

[Ans. A] More than 50% of the soil pass through sieve which eans that even a greater percentage of the soil passes through 4.75 mm sieve. Hence, the soil is definitely sandy soil. Also, we know that for a sand to be considered well graded, should be greater than 6. Here, in the case Cu = 1.2. Thus, the soil is poorly graded sand.

w

Water content So, w w w 7.

[Ans. B] Plasticity index,

w

w

Coarse grained soils which contain more than 12% fines (<0.075 mm) are classified as GM or SM if the fines are silty in character meaning, the limits plot below the A-line on the plasticity chart. Coarse grained soils are those having 50% or more retained on the 0.075 mm sieve. They are designated as gravel (G) if 50% or more of the coarse fraction is retained on 4.75 mm sieve; otherwise they are designated as sand.

Now uniformity coefficient,

3.

w

8.

[Ans. C] Plasticity index,

w

w

Equation of A line is given by w th

th

th



i

Thus the given soil lies above the A-line. The liquid limit of the soil is more than 50, hence the soil will be CH. [Ans. A]

10.

[Ans. A] By t e’s aws

13.

( ) (



x Integrating again x At x = 0, H = 5

)

V2 = 0.005mm/sec

[Ans. B] When the top end of the vane completely shear the soil, then the torque at failure is give by T = cu x π *

At x = 0, From eq. (i) x At x = 1.2 m

+

4.6 = cu x π *

[Ans. 3.8] x Integrating both sides, we get

2



+

cu = 10040.40 N/m2 Undrained shear strength, R = cu = 10 kPa 12.

From (2)

s

Terminal velocity V =

11.

i s

9.

∝


14.

[Ans. *] Range 81.0 to 81.5 By definition WG = Se e

[Ans. C] Given m = 1.980 kg = 1.800 kg

S

g water c ntent

15.

[Ans. A]

16.

[Ans. D] RD (%) = Here, e =

r

r

g

w e

s

Now , r

e

r

and r

e

i

s r i

se e

e

From (1) th

th

th




Permeability and Seepage CE – 2005 1. In a constant head permeameter with cross section area of 10 cm2, when the flow was taking place under a hydraulic gradient of 0.5, the amount of water collected in 60 seconds is 600 cc. The permeability of the soil is (A) 0.002 cm/s (C) 0.2 cm/s (B) 0.02 cm/s (D) 2.0 cm/s 2.

Two observation wells penetrated into a confined aquifer and located 1.5 km apart in the direction of flow, indicate head of 45 m and 20 m. If the coefficient of permeability of the aquifer is 30 m/day and porosity is 0.25, the time of travel of an inert tracer from one well to another is (A) 416.7 days (C) 750 days (B) 500 days (D) 3000 days

5.

To provide safety against piping failure, with a factor of safety of 5, what should be the maximum permissible exit gradient for soil with specific gravity of 2.5 and porosity of 0.35? (A) 0.155 (C) 0.195 (B) 0.167 (D) 0.213

CE – 2007 Common Data for Questions 6 and 7 Water is flowing through the permeability apparatus as shown in the figure. The c efficient f per eabi ity f the s i is ‘ ’ m/s and the porosity of the soil sample is 0.50 0.4 m 0.8 m

SOIL

CE – 2006 3. Which of the following statement is NOT true in the context of capillary pressure in soils? (A) Water is under tension in capillary zone (B) Pore water pressure is negative in capillary zone (C) Effective stress increases due to capillary zone (D) Capillary pressure is more in coarse grained soils 4.

The range of void ratio between which quick sand condition occurs in cohesion less granular soil deposits is (A) 0.4-0.5 (C) 0.8-0.9 (B) 0.6-0.7 (D) 1.0-1.1

R 0.4 m

6.

The total head, elevation head and pressure head in metres of water at the point R shown in the figure are (A) 0.8,0.4,0.4 (C) 0.4,0,0.4 (B) 1.2,0.4,0.8 (D) 1.6,0.4, 1.2

7.

What are the discharge velocity and seepage velocity through the soil sample? (A) , (C) , (B)

,

(D)

,

CE – 2009 8. The relationship among specific yield ( ), specific retention and porosity of an aquifer is (A) (C) (B) (D) th

th

th



CE – 2010 9. Quick sand condition occurs when (A) The void ratio of the soil becomes 1.0 (B) The upward seepage pressure in soil becomes zero (C) The upward seepage pressure in soil becomes equal to the saturated unit weight of the soil (D) The upward seepage pressure in soil becomes equal to the submerged unit weight of the soil 10.

13.

An open ended steel barrel of 1 m height and 1 m diameter is filled with saturated fine sand having coefficient of ⁄s The barrel permeability of stands on a saturated bed of gravel. The time required for the water level in the barrel to drop by 0.75 m is (A) 58.9 s (C) 100 s (B) 75 s (D) 150 s

Steady state seepage is taking place through a soil element at Q, 2 m below the ground surface immediately downstream of the toe of an earthen dam as shown in the sketch. The water level in a piezometer installed at P, 500 mm above Q, is at the ground surface. The water level in a piezometer installed at R, 500 mm below Q, is 100 mm above the ground surface. The bulk saturated unit weight of the soil is 18 kN/ and the unit weight of water is 9.81 kN/ . The vertical effective stress (in kPa) at Q is

P Q R

(A) 14.42 (B) 15.89

(C) 16.38 (D) 18.34

Common Data Questions 14 and 15 The flow net around a sheet pile wall is shown in the sketch. The properties of the soil are: permeability coefficient = 0.09 m/day (isotropic), specific gravity = 2.70 and void ratio = 0.85. The sheet pile wall and the bottom of the soil are impermeable.

CE – 2011 11. For a saturated sand deposit, the void ratio and the specific gravity of solids are 0.70 and 2.67, respectively. The critical (upward) hydraulic gradient for the deposit would be (A) 0.54 (C) 1.02 (B) 0.98 (D) 1.87 CE – 2012 12. Two soil specimens with identical geometric dimensions were subjected to falling head permeability tests in the laboratory under identical conditions. The fall of water head was measured after an identical time interval. The ratio of initial to final water heads for the test involving the first specimen was 1.25. If the coefficient of permeability of the second specimen is 5 times that of the first, the ratio of initial to final water heads in the test involving the second specimen is (A) 3.05 (C) 4.00 (B) 3.80 (D) 6.25


10 m 1.5 m 3m

14.

th

The seepage loss (in per day per unit length of the wall) of water is (A) 0.33 (C) 0.43 (B) 0.38 (D) 0.54 th

th



The factor of safety against the occurrence of piping failure is (A) 3.55 (C) 2.60 (B) 2.93 (D) 0.39

CE – 2013 16. The ration Nf/Nd is known as shape factor, where Nf is the number of flow lines and Nd is the number of equipotential drops. Flow net is always drawn with a constant b/a ratio, where b and a are distance between two consecutive flow lines and equipotential lines, respectively. Assuming that b/a ratio remains the same, the shape factor of a flow net will change if the (A) Upstream and downstream heads are interchanged (B) Soil in the flow space is changed (C) Dimensions of the flow space are changed (D) Head difference causing the flow is changed 17.

18.

Following statements are made on compacted soils, where in DS stands for the soil compacted on dry side of optimum moisture content and WS stands for the soil compacted on wet side of optimum moisture content. Identify the incorrect statement. (A) Soil structure is flocculated on DS and dispersed on WS. (B) Construction of pore water pressure is low on DS and high on WS (C) On drying , shrinkage is high on DS and low on WS (D) On access to water, swelling is high on DS and low on WS

reading of + 10 m elevation. Assume that the piezometric head is uniform in the sand layer. The quantity of water (in /s) flowing into the lake from the sand layer through the silt layer per unit area of the lake bed is : + 10

Stand pipe

0 Lake Elevation (m)

15.


Silt(k =

Lake bottom

m/s)

Sand (under artesian pressure) Rock

(A) (B)

(C) (D)

CE – 2014 19. Water is flowing at a steady rate through a homogeneous and saturated horizontal soil strip of 10 m length. The strip is being subjected to a constant water head (H) of 5 m at the beginning and 1 m at the end. If the governing equation of flow in the soil strip is

(where x is the distance

along the soil strip), the value of H (in m) at the middle of the strip is ____________. 20.

The flow net constructed for the dam is shown in the figure below. Taking the coefficient of permeability as m/s, the quantity of flow (in c /s) under the dam per meter of dam is ______________ 50 m 1.6 m

6.3 m

The soil profile below a lake with water level at elevation = 0 m and lake bottom at elevation = 10 m is shown in the figure, where k is the permeability coefficient. A piezometer (stand pipe) installed in the sand layer shows a

9.4 m 17.2 m

Impermeable stratum

th

th

th





[Ans. D] i c

i 2.

7.

[Ans. A] Discharge velocity, v i head ss But i ength f s i ass v s v eepage ve city, v n

8.

[Ans. C]

9.

[Ans. D]

s

[Ans. C] The average flow velocity or discharge velocity, v i (

)

eepage ve city, v

day v n

day

rave ti e

h

days L

3.

4.

[Ans. D] Capillary rise is more in fine grained soils and as a result of this, the capillary pressure is also more than the coarse grained soils.

Soil

Quick sand occurs when net effective pressure is zero ie ̅ h But h

[Ans. B] The specific gravity of cohesion less granular soils (sands) does not vary much and for all practical purposes it is taken to be 2.65. Critical hydraulic gradient should be nearly 1 for quick sand condition

So quick sand condition occurs when upward seepage pressure in soil becomes equal to the submerged unit weight of the soil.

i e ,i

e From the above equation, the void ratio range is found be between 0.6 & 0.7 5.

[Ans. B] Time required s

[Ans. C] i

e axi u

6.

10.

[Ans. A]

n

11.

[Ans. B] Critical hydraulic gradient, i

per issib e exit gradient,

= 0.98 12.

th

[Ans. A] a h t g h h t∝ g h Since time interval is same th

th



g

h h

h h

g

18.

h , h ubstituting the ab ve va ues h h g g h h h g g h h h iven

13.


[Ans. D] Head causing flow = 10m This head will be lost in 20 m depth of silt i i ischarge per unit area, s

19.

[Ans. B] Average water pressure head at point Q is h u h

[Ans. 3] Given, equation of the flow of soil strip is d dx d dx x at x ,

a 14.

i

x at x

[Ans. B] Seepage head, H = 10 – 1.5 = 8.5m , , /d

,

/day/m x 15.

[Ans. C]

at x

Exit gradient, i

given in the fig Critical hydraulic gradient,

20.

i Factor

of

safety

against

piping,

[Ans. *] Range 7.10 to 7.85 Seepage loss, q per meter length of dam is f f w channe f e uip tentia dr ps q=K

16.

Here, H = 6.3m K=

[Ans. C] depends only on dimension of flow field

s

and is independent of soil & head parameters. Option C is the most appropriate answer. 17.

s c

[Ans. C] Shrinkage is low on DS and high on WS

th

th

s

th




Consolidation (iii) Unit weight of sand and clay below the water table (iv) Coefficient of volume compressibility

CE – 2005 1. Root time method is used to determine (A) , time factor (B) , coefficient of consolidation (C) a , coefficient of compressibility (D) , coefficient of volume compressibility

G.L. 5m Original water table

CE – 2006 Statement for Linked Answer Questions 2 and 3 The average effective overburden pressure on 10m thick homogeneous saturated clay layer is 150 kPa. Consolidation test on an undisturbed soil sample taken from the clay layer showed that the void ratio decreased from 0.6 to 0.5 by increasing the stress intensity from 100 kPa to 300 kPa (G = 2.65). 2. The initial void ratio of the clay layer is (A) 0.209 (C) 0.746 (B) 0.563 (D) 1.000 3.

The total consolidation settlement of the clay layer due to the construction of a structure imposing an additional stress intensity of 200 kPa is (A) 0.10 m (C) 0.35 m (B) 0.25 m (D) 0.50 m

CE – 2007 Statement for linked answer questions 4&5 The ground condition at a site are as shown in the figure. The water table at the site which was initially at a depth of 5 m below the ground level got permanently lowered to a depth of 15 m below the ground level due to pumping of water over a few years. Assume the following data: (i) Unit weight of water (ii) Unit weight of sand above water table

15 m

20 m

20 m

Lowered water table

Clay soil layer

5m

4.

What is the change in the effective stress in at mid-depth of the clay layer due to the lowering of the water table? (A) 0 (C) 80 (B) 20 (D) 100

5.

What is the compression of the clay layer in mm due to the lowering of the water table? (A) 125 (C) 25 (B) 100 (D) 0

CE – 2008 6. A saturated clay stratum draining both at the top and bottom undergoes 50 per cent consolidation in 16 years under an applied load. If an additional drainage layer were present at the middle of the clay stratum, 50 per cent consolidation would occur in (A) 2 years (C) 8 years (B) 4 years (D) 16 years

th

th

th



CE – 2009 Statement for Linked Answers Questions 7&8 7. A saturated undisturbed sample from a clay strata has moisture content of 22.22% and specific weight of 2.7. Assuming = 10 the void ratio and the saturated unit weight of the clay, respectively are (A) 0.6 and 16.875 kN/m3. (B) 0.3 and 20.625 kN/m3 (C) 0.6 and 20.625 kN/m3 (D) 0.3 and 16.975 kN/m3 8.


CE – 2011 10. Identical surcharges are placed at ground surface at sites X and Y, with soil conditions shown alongside and water table at ground surface. The silty clay layers at X and Y are identical. The thin sand layer at Y is continuous and freedraining with a very large discharge capacity. If primary consolidation at X is estimated to complete in 36 months, what would be the corresponding time of completion of primary consolidation at Y? Uniform surcharge

Site X

Using the properties of the clay layer derived from the above question, the consolidation settlement of the same clay layer under a square footing (neglecting its self weight) with additional data shown in the figure below (assume-the stress distribution as 1 H: 2V from the edge of the footing and = 10 ) is

Ground surface

Water Table

5m Isotropic saturated silty clay

aturated sand

1.0 m tiff

1.0 m

Bedrock

(A) 32.78 mm (B) 61.75 mm

(C) 79.5 mm (D) 131.13 mm

Void ratio. e

CE – 2010 9. The e-log p curve shown in the figure is representative of

log p

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

Normally consolidated clay Over consolidated clay Under consolidated clay Normally consolidated clayey sand.

Isotropic saturated silty clay Thin sand layer Isotropic saturated silty clay Bedrock

(A) 2.25 months (B) 4.5 months

pressi n ndex

ense sand

Water Table

10 m

5m 1.0 m

Site Y

(C) 9 months (D) 36 months

CE – 2012 11. A layer of normally consolidated, saturated silty clay of 1 m thickness is subjected to one dimensional consolidation under a pressure increment of 20 kPa. The properties of the soil are: specific gravity = 2.7, natural moisture content= 45%, compression index= 0.45, and recompression index =0.05. The initial average effective stress within the ayer is a ssu ing erzaghi’s theory to be applicable, the primary consolidation settlement (rounded off the nearest mm) is (A) 2mm (C) 14mm (B) 9mm (D) 16mm th

th

th




CE – 2014 12. The following data are given for the laboratory sample. a e a e If thickness of the clay specimen is 25 mm, the value of coefficient of volume compressibility is _____


[Ans. B]

2.

[Ans. B] efficient f c e e

5.

̅

(

)

g

[Ans. A] ̅

6.

e

[Ans. B] r ay r’s f r u a, we have t d Here and are constant t∝d t d ( ) t d

̅

(̅ ) e g

(

g

[

)

e 3.

[Ans. D] e (

4.

̅

pressi n,

(̅ )

e

̅

⁄

g g Also,

hange in effective stress

̅ ̅ ) g

[

(

t

]

t

]

[Ans. D] When the water table is 5 m below the ground level ̅ ⁄ When the water table is 15 m below the ground level, ̅ ( ) ⁄

)

t 7.

years

[Ans. C] e

w [

e

i is saturated]

e e (

e ) e

(

) ⁄

th

th

th



8.

[Ans. B] The consolidation settlement at the centre of clay layer is given by ̅ g [ ] e ̅

11.


[Ans. D] w e c

,

a, a

For NCC,

g g

12.

Load distribution dimensions at the centre of clay layer ncrease in stress due t

[

(

) ]

[Ans. *] Range 7.6 to 8.0 Co – efficient of volume compressibility is defined as e e

= 7.61

ad

⁄ Effective stress at the centre of clay layer ⁄ g

9.

10.

[Ans. B] This condition consolidated clay.

[

represents

]

over

[Ans. C] ay r’s f r u a t t ( ) ( )

 

t

nths

th

th

th




Compaction CE – 2012 1. Two series of compaction test were performed in the laboratory on an inorganic clayey soil employing two different levels of compaction energy per unit volume of soil. With regard to the above tests, the following two statements are made. I. The optimum moisture content is expected to be more for the tests with higher energy. II. The maximum dry density is expected to be more for the tests with higher energy. The correct option evaluating the above statement is. (A) Only I is TRUE (B) Only II is TRUE (C) Both I and II are TRUE (D) Neither I nor II is TRUE

Answer keys & Explanations 1. [Ans. B]

higher energy s ,

w

w isture c ntent

th

th

th




Stress Analysis CE – 2005 1. A clayey soil has maximum dry density of 16 and optimum moisture content of 12%. A contractor during the construction of core of an earth dam obtained the dry density 15.2 and water content 11%. This construction is acceptable because. (A) The density is less than the maximum dry density and water content is on dry side of optimum (B) The compaction density is very low and water content is less than 12% (C) The compaction is done on the dry side of the optimum (D) Both the dry density and water content of the compacted soil are within the desirable limits CE – 2006 2. In a standard proctor test, 1.8 kg of moist soil was filling the mould (volume = 944 cc) after compaction. A soil sample weighing 23 g was taken from the mould and over dried for 24 hours at a temperature of 110°C. Weight of the dry sample was found to be 20 g. Specific gravity of soil solids is G = 2.7. The theoretical maximum value of the dry unit weight of the soil at that water content is equal to (A) 4.67 kN/m3 (C) 16.26kN/m3 (B) 11.5kN/m3 (D) 18.85kN/m3 CE – 2007 3. The vertical stress at some depth below the corner of a 2 m × 3 m rectangular footing due to a certain load intensity is 100 ⁄ . What will be the vertical stress in ⁄ below the centre of a 4 m × 6 m rectangular footing at the same depth and same load intensity? (A) 25 (C) 200 (B) 100 (D) 400

CE – 2008 4. Compaction by vibratory roller is the best method of compaction in case of (A) Moist silty sand (B) Well graded dry sand (C) Clay of medium compressibility. (D) Silt of high compressibility 5.

A footing 2 m × 1 m exerts a uniform pressure of 150 kN/m2 on the soil. Assuming a load dispersion of 2 vertical to 1 horizontal, the average vertical stress (kN/m2) at 1.0 m below the footing is (A) 50 (C) 80 (B) 75 (D) 100 Statement for linked answer questions 6 & 7. The ground conditions at a site are shown in the figure below. GL

Sand 5m P

Water table is at ground level Water content = 20% Specific gravity of solids =2.7 Unit weight of water = 10

6.

The saturated unit weight of the sand is (A) 15 (C) 21 (B) 18 (D) 24

7.

The total stress, pore water pressure and effective stress (kN/m2) at the point P are, respectively (A) 75,50 and 25 (C) 105,50 and 55 (B) 90,50 and 40 (D) 120,50 and 70

CE – 2010 8. In a compaction test, G, w, S and e represent the specific gravity, water content, degree of saturation and void ratio of the soil sample, respectively. If represents the unit weight of water and th

th

th



represents the dry unit weight of the soil, the equation for zero air voids line is

9.

(A)

(C)

(B)

(D)

The vertical stress at point P1 due to the point load Q on the ground surface as shown in figure is . According to Boussinesq's equation, the vertical stress at point P2 shown in figure will be

10.

What would be the effective stress (rounded off to the nearest integer value of kPa) at 30 m depth into the sand layer'? (A) 77 kPa (C) 268 kPa (B) 273 kPa (D) 281 kPa

11.

What would be the change in the effective stress (rounded off to the nearest integer value of kPa) at 30 m depth into the sand layer if the sea water level permanently rises by 2 m? (A) 19 kPa (C) 21 kPa (B) 0 kPa (D) 22 kPa

z/2 z r/2

r

(A) (B)

(C) (D)

CE – 2011 Common Data for Questions 10 & 11 A sand layer found at sea floor under 20 m water depth is characterized with relative density = 40%. Maximum void ratio = 1.0, minimum void ratio = 0.5, and specific gravity of soil solids = 2.67. Assume the specific gravity of sea water to be 1.03 and the unit weight of fresh water to be 9.81


CE – 2014 12. For a saturated cohesive soil, a triaxial test yields the angle of internal friction φ as zer he c nducted test is (A) Consolidated Drained (CD) test (B) Consolidated Undrained (CU) test (C) Unconfined Compression (UC) test (D) Unconsolidated Undrained (UU) test

th

th

th





[Ans. A]

2.

[Ans. D] Bu

t eve , an area = 2 x 1 = 2m2 At level B – B, Plan area = {(2 + 2 x 0.5) (1 + 2 x 0.5)} = 6m2 So, pressure intensity at level B – B (1m. below the A – A)

density f the s i g ⁄cc

ater c ntent s i

(

)

= 150 x = 50 kN/m2 For, theoretical maximum dry unit weight degree of saturation should be 100% e w

6.

[Ans. C] e e

e

e

e The theoretical maximum dry unit weight may be given as ,

3.

e

and ( 7.

⁄

(

)

)

[Ans. C] ta stress ⁄

[Ans. D] For same load intensity and same depth,

re water pressure, u ⁄ ffective stress, ̅

u ⁄

⁄ 4.

[Ans. B]

5.

[Ans. A]

8.

[Ans. B] The bulk unit weight of soil is given by e e For dry unit weight, degree of saturation, S=0

1m

2m

e But we know that w

A 1 1m

1m

2

A

e

For zero air voids, degree of saturation

1 2

w B

B 0.5

1m

0.5

2.0

th

th

th



9.

[Ans. D] As per Boussinesq equation, the vertical stress at a point located at a depth z and a horizontal distance r from the point of application of point load Q is

Pore water pressure at the bottom of sand layer, u Effective stress at the bottom of sand layer, ̅ u

⁄

[

πz


] () Alternatively, ̅

For a horizontal distance and a depth . The stress will be ⁄

π( )

( )

[

]

11. ⁄

[

πz 10.

] ()

[Ans. D] Relative density (

[Ans. B] The effective stress will not alter due to change in sea water level. The total stress will increase due to increase in sea water level. The pore water pressure will also increase in the same proportion, thus nullifying the effect of each other. Sea water

)

(

22 m

) e

30 m

e Also,

u

Sea water 20 m

̅

30 m

(

u

Sand layer

12. (

Sand layer

)

[Ans. D] Mohr circle for UU test

)

ϕ=0

Total stress at the bottom of sand layer,

Which is only possible for UU test on saturated clays. th

th

th


GATE Supplement Book


Stress Analysis CE – 2005 1. Assuming that a river bed level does not change and the depth of water in river was 10m, 15m and 8m during the months of February, July and December respectively of a particular year. The average bulk density of the soil is 20 kN/m3. The density of water is 110 kN/m3. The effective stress at a depth of 10 m below the river bed during these months would be (A) 300 kN/m2 in February, 350 kN/m2 July and 320 kN/m2 in December (B) 100 kN/m2 in February, 100 kN/m2 July and 100 kN/m2 in December (C) 200 kN/m2 in February, 250 kN/m2 July and 180 kN/m2 in December (D) 300 kN/m2 in February, 350 kN/m2 July and 280 kN/m2 in December 2.

For a triaxial shear test conducted on a sand specimen at a confining pressure of 100 kN/m2 under drained conditions, resulted in a deviator stress at failure of 100 kN/m2. The angle of shearing resistance of the soil would be (A) (C) (B) (D)

CE – 2006 3. A sample of saturated cohesionless soil tested in a drained triaxial compression test showed an angle of internal friction of . The deviatoric stress at failure for the sample at a confining pressure of 200 kPa is equal to (A) 200 kPa (C) 600 kPa (B) 400 kPa (D) 800 kPa CE – 2007 4. A clay soil sample is tested in a triaxial apparatus in consolidated drained conditions at a cell pressure of 100kN/m2. What will be the pore water

pressure at a deviator stress of 40 kN/m2? (A) 0 (C) 40 kN/m2 2 (B) 20 kN/m (D) 60 kN/m2 5.

Match List – I with List – II and select the correct answer using the codes given below the lists: List – I List - II A. Constant head 1. Pile permeability test Foundation B. Consolidation test 2. Specific gravity C. Pycnometer test 3. Clay soil D. Negative skin friction Codes: A B (A) 4 3 (B) 4 2 (C) 3 4 (D) 4 1

4. Sand

C 2 3 2 2

D 1 1 1 3

CE – 2008 6. A direct shear test was conducted on a cohesion less soil (c=0) specimen under a normal stress of 200 kN/m2. The specimen failed at a shear stress of 100 . The angle of internal friction of the soil (degrees) is (A) 26.6 (C) 30.0 (B) 29.5 (D) 32.6 CE – 2010 Statement for linked Answer Question 7and 8 The unconfined compressive strength of a saturated clay sample is 54 kPa. 7.

The value of cohesion for the clay is (A) Zero (C) 27 kPa (B) 13.5 kPa (D) 54 kPa

th

th

th



8.

If a square footing of size 4 m × 4 m is resting on the surface of a deposit of the above clay, the ultimate bearing capacity f the f ting as per erzaghi’s e uati n is (A) 1600 kPa (C) 200 kPa (B) 316 kPa (D) 100 kPa


CE – 2012 9. The effective stress friction angle of a saturated, cohesionless soil is . The ratio of shear stress to normal effective stress on the failure plane is (A) 0.781 (C) 0.488 (B) 0.616 (D) 0.438


[Ans. B] Since the river bed level does not change and the depth of water is changing only above the river bed, the effective stress below the river bed will not change. Total stress at a depth of 10 m below river bed, Pore water Pressure, u Effective stress, ̅ u Thus effective stress will be 100 kN/m2 each in the month of February, July and December.

2.

4.

[Ans. A] In the consolidated drained test, the soil sample is first consolidated under an appropriate cell pressure. With the cell pressure kept at the same value, the soil sample is then sheared by applying the deviator stress so slowly that excess pore water pressure does not develop during the test. Thus, at any stage of the test, the total stresses are the effective stresses.

5.

[Ans. A]

6.

[Ans. A] ] c ̅ tan ϕ [ c tan ϕ ϕ tan ϕ

7.

[Ans. C]

[Ans. B] We know that, sin ϕ sin ϕ

hensi n f c ay, c

a

sin ϕ 8.

sin ϕ ϕ

sin

[Ans. C] The ultimate bearing capacity of a square footing in clay as per erzaghi’s bearing capacity equation is given by ] c [ f r c ay

( )

ϕ 3.

[Ans. B] tan ( tan ( tan (

ϕ

)

ϕ

c tan ( )

)

a 9.

[Ans. A] tan ϕ

)

tan ϕ Deviatoric stress at failure

tan a th

th

th




Surface Investigations CE – 2005 1. During the subsurface investigations for design of foundations, a standard penetration test was conducted at 4.5m below the ground surface. The record of number of blows is given below. Penetration Number of depth (m) blows 0 - 7.5 3 7.5 - 15 3 15 - 22.5 6 22.5 - 30 6 30 - 37.5 8 37.5 - 45 7 Assuming the water table at ground level, soil as fine sand and correction factor for verburden as , the c rrected ‘ ’ va ue for the soil would be (A) 18 (C) 21 (B) 19 (D) 33

(C) Saturated silt/fine sand and N value of SPT > 15 after the overburden correction (D) Coarse sand under dry condition and N value of SPT < 10 after the overburden correction CE – 2014 4. The degree of disturbance of the sample collected by the sampler is expressed by a term called the "area ratio". If the outer diameter and inner diameter of the sampler are and respectively, the area ratio is given by

5.

(A)

(C)

(B)

(D)

Group I enlists in-situ field tests carried out for soil exploration, while Group II provides a list of parameters for sub-soil strength characterization. Match the type of tests with the characterization parameters. Group I Group II P. Pressuremeter Test 1. enard’s du us (PMT) (Em) Q. Static Cone 2. Number of blows Penetration Test (N) (SCPT) R. Standard 3. Skin resistance (fc) Penetration Test (SPT) S. Vane Shear Test 4. Undrained (VST) cohesion (cu) (A) (B) (C) (D)

CE – 2007 2. The no. of blows observed in a standard penetration test (SPT) for different penetration depths are given as follows: Penetration of Number of sampler blows 0 - 150 mm 6 150 - 300 mm 8 300 - 450 mm 10 The observed N value is (A) 8 (C) 18 (B) 14 (D) 24 CE – 2009 3. Dilatancy correction is required when a strata is (A) Cohesive and saturated and also has N value of SPT > 15 (B) Saturated silt/fine sand and N value of SPT < 10 after overburden correction

th

th

th





[Ans. C] The number of blows required for the first 150 mm of penetration is disregarded, and only the number of blows required for the last 300 mm of penetration is added together number of blows,

3.

[Ans. C] Saturated Dilatancy correction is applied for fine sand/silt if N > 15

4.

[Ans. A]

(i) Correction for overburden pressure Corrected value f bserved ’ (ii) Correction for dilatancy It is to be applied when number of blows obtained after overburden correction exceeds 15 in saturated fine sands and silts.

2.

[Ans. C] The number of blows for the first 150 mm penetration of the sampler is disregarded. The number of blows for the next 300 mm penetration is recorded as the observed N value Observed va ue

Area ratio = Inner clearance = Outer – clearance = 5.

[Ans. A]

th

th

th




Earth Pressure CE – 2005 1. A 3m high retaining wall is supporting a saturated sand (saturated due to capillary action) of bulk density 18 kN/m3 and angle of shearing resistance . The change in magnitude of active earth pressure at the base due to rise in ground water table from the base of the footing to 2 the gr und surface sha w = 10 kN/m ) (A) Increase by 20 kN/m2 (B) Decrease by 20 kN/m2 (C) Increase by 30 kN/m2 (D) Decrease by 30 kN/m2 CE – 2006 2. Figure given below shows a smooth vertical gravity retaining wall with cohesionless soil backfill having an angle of internal friction ϕ. β

CE – 2010 4. If , , and represent the total horizontal stress, total vertical stress, effective horizontal stress and effective vertical stress on a soil elements, respectively, the co-efficient of earth pressure at rest is given by (A) (C) (D)

(B)

CE – 2012 5. A smooth rigid retaining wall moves as shown in the sketch causing the backfill material to fail. The backfill material is homogeneous and isotropic, and obeys the Mohr – Coulomb failure criterion. The major principal stress is Initial wall position Final wall position

hr’s enve pe Sand H

Dry, granular, cohesionless backfill with horizontal top surface

P Ground line O

In the graphical representation of an ine’s active earth pressure f r the retaining wall shown in figure, length OP represents (A) Vertical stress at the base (B) Vertical stress at a height H/3 from the base (C) Lateral earth pressure at the base (D) Lateral earth pressure at a height H/3 from the base CE – 2008 3. When a retaining wall moves away from the backfill, the pressure exerted on the wall is termed as (A) passive earth pressure (B) swelling pressure (C) pore pressure (D) active earth pressure

(A) Parallel to the wall face and acting downwards (B) Normal to the wall face (C) Oblique to the wall face acting downwards (D) Oblique to the wall face acting upwards CE – 2013 6. Two different soil types (soil 1 and soil 2) are used as backfill behind a retaining wall as shown in the figure, where is total unit weight, and c and ϕ are effective cohesion and effective angle of shearing resistance. The resultant active earth force per unit length ( in kN/m) acting on the wall is : th

th

th




Retaining wall

2m

Soil 1: =15kN/

c =0;ϕ

Soil 2: 2m

(A) 31.7 (B) 35.2

=20kN/

c =0;ϕ

(C) 51.8 (D) 57.0


[Ans. A] When water table is at the base of footing

to the vertical stress at a specified depth. Ko = But for a saturated soil, 5.

[Ans. B] In passive earth pressure, major principle stress is horizontal.

6.

[Ans. A] sin sin

When water table rises upto the surface thrust due to water will be added

So change in pressure =

(increase)

= 10kN/m 2.

[Ans. A]

3.

[Ans. D]

4.

[Ans. B] Coefficient of earth pressure at rest is the ratio of intensity of earth pressure at rest

=13.02kN/m

th

th

th




Stability of Slopes CE – 2005 1. For two infinite slopes (one in dry condition and other in submerged condition) in a sand deposit having the angle of shearing resistance , factor of safety was determined as 1.5 (for both slopes). The slope angles would have been. (A) for dry slope and for submerged slope (B) for dry slope and for submerged slope (C) for dry slope and for submerged slope (D) for dry slope and for submerged slope CE – 2006 2. List-I below gives the possible types of failure for a finite soil slope and List-II gives the reasons for these different types of failure. Match the items in List-I with the items in List-II and select the correct answer from the codes given below the lists: List-I A. Base failure B. Face failure C. Toe failure List-II 1. Soils above and below the toe have same strength 2. Soil above the toe is comparatively weaker 3. Soil above the toe is comparatively stronger Codes: A B C (A) 1 2 3 (B) 2 3 1 (C) 2 1 3 (D) 3 2 1

CE – 2007 3. The factor safety of an infinite soil slope shown in the figure having the properties c = 0, ϕ = , = 16 kN/m3, and 3 = 20 kN/m is approximately equal to

(A) 0.7 (B) 0.8

(C) 1.0 (D) 1.2

CE – 2013 4. The soil profile above the rock surface for a infinite slope is shown in the figure, where s is the undrained shear strength and is total unit weight. The slip will occur at a depth of

5m 5m

(A) 8.83 m (B) 9.79 m

(C) 7.83 m (D) 6.53 m

CE – 2014 5. A long slope is formed in a soil with shear strength parameters: c' = 0 and ϕ ' = 34°. A firm stratum lies below the slope and it is assumed that the water table may occasionally rise to the surface, with seepage taking place parallel to the slope. Use = 18 kN/ and = 10 kN/ . th

th

th



The maximum slope angle (in degrees) to ensure a factor of safety of 1.5, assuming a potential failure surface parallel to the slope, would be (A) 45.3 (C) 12.3 (B) 44.7 (D) 11.3

6.


An infinitely long slope is made up of a c-φ s i having the pr perties: c hesi n (c) = 20 kPa and dry unit weight ( ) = 16 kN/ . The angle of inclination and critical height of the slope are 40° and 5 m, respectively. To maintain the limiting equilibrium, the angle of internal friction of the soil (in degrees) is _______________


[Ans. A] For dry or submerged slopes,

The slip will occur when shear stress is greater than or equal to shear strength.

Factor of safety =

z

=) tan β =

z

sin β c s β z sin β

=) β = 21.05 z 2.

3.

[Ans. D] Face failure or slope failure can occur when the s pe ang e β is very high and the soil close to the toe is quite strong or the soil in the upper part of slope is relatively weak. Base failure can occur when the soil below the toe is relatively weak and soft and the slope is flat. Toe failure occurs in steep slopes when the soil mass above the base and below the base is homogeneous.

z epth f s ip 5.

[Ans. D] For submerged case F.O.S = For seepage parallel to slope F.O.S = (2) is worst case since O f case < FOS of case (1) as

[Ans. A] Factor of safety,

*

Considering (2)

+

Assuming, * 4.

sin

tan tan i

O +

i 6.

[Ans. A]

[Ans. *] Range 21.0 to 23.0

Slip surface

tan ϕ

z

tanϕ

z β

β th

th

th




Bearing Capacity CE – 2005 1. The strip footing (8, wide) is designed for a total settlement of 40 mm. The safe bearing capacity (shear) was 150 kN/ and safe allowable soil pressure was 100 kN/ . Due to importance of the structure, now the footing is to be redesigned for total settlement of 25 mm. The new width of the footing will be (A) 5 m (C) 12 m (B) 8 m (D) 12.8m CE – 2007 2. The bearing capacity of a rectangular footing of plan dimensions 1.5m × 3m resting on the surface of a sand deposit was estimated as 600 kN/m2 when the water table is far below the base of the footing. The bearing capacities in kN/m2 when the water level rises to depth of 3m, 1.5m and 0.5m below the base of the footing are (A) 600, 600, 400 (C) 600, 500, 250 (B) 600, 450, 350 (D) 600, 400, 250 CE – 2008 Statement for Linked Answer Questions 3 and 4 A column is supported on footing as shown in the figure below. The water table is at a depth of 10m below the base of the footing Column

GL 1.0 m

Footing 1.5×3m

Sand = 18 kN/ = 24 = 20

3.

The net ultimate bearing capacity (kN/ ) of the footing based on erzaghi’s bearing capacity e uati n is (A) 216 (C) 630 (B) 432 (D) 846

4.

The safe load (kN) that the footing can be with a factor of safety 3 is (A) 282 (C) 945 (B) 648 (D) 1269

5.

A test plate 30 cm × 30 cm resting on a sand deposit settles by 10 mm under a certain loading intensity. A footing 150 cm × 200 cm resting on the same sand deposit and loaded to the same load intensity settles by (A) 2.0 mm (C) 30.2 mm (B) 27.8mm (D) 50.0mm

CE – 2009 6. A plate load test is carried out on a 300 mm × 300 mm plate placed at 2 m below the ground level to determine the bearing capacity of a 2 m × 2 m footing placed at same depth of 2 m on a homogeneous sand deposit extending 10 m below ground. The ground water table is 3 m below the ground level. Which of the following factors does not require a correction to the bearing capacity determined based on the load test? (A) Absence of the overburden pressure during the test (B) Size of the plate is much smaller than the footing size (C) Influence of the ground water table (D) Settlement is recorded only over a limited period of one or two days

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Common Data for Questions 7 and 8: Examine the test arrangement and the soil properties given below; 5m

11.

5m


Two geometrically identical isolated footings, X (linear elastic) and Y (rigid), are loaded identically (shown below). The soil reaction will Uniform pressure Footing X: Linear elastic

Rigid Steel Beam Concrete block 1.5×1.0×0.6m high

Saturated dense sand 5m ⁄ ϕ c a ⁄

7.

8.

G.W.T 500 mm diameter bored pile Angle of friction Earth pressure coefficient (K)= 1.5

Uniform pressure

The maximum pressure that can be applied with a factor of safety of 3 through the concrete block, ensuring no bearing capacity failure in soil using erzaghi’s bearing capacity e uati n without considering the shape factor, depth factor and inclination factor is (A) 26.67 kPa (C) 90 kPa (B) 60 kPa (D) 120 kPa The maximum resistance offered by the soil through skin friction while pulling out the pile from the ground is (A) 104.9 kN (C) 236 kN (B) 209.8 kN (D) 472 kN

CE – 2011 9. Likelihood of general shear failure for an isolated footing in sand decreases with (A) Decreasing footing depth (B) Decreasing inter-granular packing of the sand (C) Increasing footing width (D) Decreasing soil grain compressibility 10.

Isotropic linear elastic soil

For a sample of dry, cohesionless soil with friction angle, ϕ, the failure plane will be inclined to the major principal plane by an angle equal to (A) ϕ (C) ϕ⁄ (B) 45° (D) ϕ⁄

Footing Y: Rigid Isotropic linear elastic soil

(A) be uniformly distributed for Y but not for X (B) be uniformly distributed for X but not for Y (C) be uniformly distributed for both X and Y (D) not be uniformly distributed for both X and Y CE – 2012 12. An embankment is to be constructed with a granular soil (bulk unit weight = 20kN/ ) on a saturated clayey silt deposit (undrained shear strength = 25 kPa). Assuming undrained general shear failure and bearing capacity factor of 5.7, the maximum height (in m) of the embankment at the point of failure is (A) 7.1 (C) 4.5 (B) 5.0 (D) 2.5

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CE – 2013 Statement for Linked Answer Questions 13 & 14 A multistory building with a basement is to be constructed. The top 4m consists of loose silt, below which dense sand layer is present up to a great depth. Ground water table is at the surface. The foundation consists of the basement slab of 6 m width which will rest on the top of dense sand as shown in the figure. For dense sand, saturated unit weight = / , and bearing capacity factors and . For loose silt, saturated unit weight = 18kN/ , and . ffective c hesi n c’ is zer f r both soils. Unit weight of water is 10kN/ . Neglect shape factor and depth factor. verage e astic du us and iss n’s ratio of dense and is / and 0.3 respectively. Ground surface

Loose silt 4 m

Basement Foundation slab

CE – 2014 15. Group I contains representative loadsettlement curves for different modes of bearing capacity failures of sandy soil. Group II enlists the various failure characteristics. Match the load-settlement curves with the corresponding failure characteristics. Load J

Group I P. Curve J Q. Curve K R. Curve L (A) (B) (C) (D)

Loose silt

14.

Dense sand Using factor of safety =3, the net safe bearing capacity (in kN/ ) of the foundation is : (A) 610 (C) 983 (B) 320 (D) 693

L

K

Settlement

Group II 1. No apparent heaving of soil around the footing an ine’s passive z ne develops imperfectly 3. Well defined slip surface extends to ground surface , , , , , , , ,

16.

The contact pressure for a rigid footing resting on clay at the centre and the edges are respectively (A) maximum and zero (B) maximum and minimum (C) zero and maximum (D) minimum and maximum

17.

A circular raft foundation of 20 m diameter and 1.6 m thick is provided for a tank that applies a bearing pressure of 110 kPa on sandy soil with Young's modulus, ' = 30 MPa and Poisson's ratio, = 0.3. The raft is made of concrete ( = 30 GPa and = 0.15). Considering the raft as rigid, the elastic settlement (in mm) is (A) 50.96 (C) 63.72 (B) 53.36 (D) 66.71

6m

13.


The foundation slab is subjected to vertical downward stresses equal to net safe bearing capacity derived in the above question. Using influence factor =2.0, and neglecting embedment depth and rigidity corrections, the immediate settlement of the dense sand layer will be: (A) 58 mm (C) 126 mm (B) 111 mm (D) 179 mm

th

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[Ans. A] The allowable soil pressure for 25 mm settlement =

3.

[Ans. B] [

B(

+

[Ans. C]

7.

[Ans. A] The bearing capacity equation for rectangular footing given by Terzaghi is given by B c [ ]

where *

)

c

8.

[Ans. C] Using net Ultimate nearing capacity as 630 kN/ ad

act r f safety here are tw c ncrete b c s] a

[Ans. A] Ultimate (maximum) skin friction resistance, is given by ̅ tan Where, K=lateral earth pressure coefficient =1.5 ̅ Average effective overburden Pressure over the embedded length of pile,

= 432 + 243 = 675 KN/ The net ultimate bearing capacity, = = 675 18 1 = 657 kN/

afe

B

⁄

1.5

20

+ and *

Thus, neglecting shape factors in the above equation we get,

For sand, c= 0 (

]

+ areShape factors.

[ 18

B

[

+

+

]

6.

) =

] for sandy soils

=5m

[Ans. C] Ultimate bearing capacity of the footing as per erzaghi’s e uati n f r rectangu ar footing is given as )c

)

[

[Ans. A] The bearing capacity of the sand deposit will remain same until the water table rises to a depth less than the width of the footing. Thus, when water table is 3.0 below footing, the bearing capacity is 600 , similarly when water table is 1.5 m below footing, the bearing capacity remaining same i.e. 600 . But when the water table rises to a depth of 0.5 m below the footing, the sand deposit will be saturated for a depth of 1.5 – 0.5 = 1 m. thus, the bearing capacity below the base of footing will decrease due to submergence of soil (1 m).

=(

4.

(

100 = 62.5 kN/

Thus new width of footing = 2.

5.

= 945 kN

th

th

th



14.


[Ans. B]

⁄ = Surface area of pile in contact with soil π tan

[

]

Nearest option (B) 9.

[Ans. B]

10.

[Ans. D]

2a

ϕ

We know,

11.

[Ans. A]

12.

[Ans. A]

15.

[Ans. A] L: General shear failure J: punching shear failure K: Local shear failure

16.

[Ans. D]

17.

[Ans. B] Elastic settlement of rigid footing, B s Shape factor for circular raft footing

ϕ

[

]

⁄ Ultimate bearing capacity, Maximum vertical stress due embankment = Where = unit wt of embankment and H = Height Equating,

to

Height of embankment, H = 13.

[Ans. A] Safe bearing capacity ( (y

(

) )

B By

)

Nearest option (A)

th

th

th




Pile Foundation CE – 2005 1. Negative skin friction in a soil is considered when the pile is constructed through a (A) fill material (B) dense coarse sand (C) over consolidated stiff clay (D) dense fine sand CE – 2006 2. For the soil profile shown in figure below, the minimum number of precast concrete piles of 300 mm diameter required to safely carry the load for given factor of safety of 2.5 (assuming 100% efficiency for the pile group ) is equal to 5000 kN

Medium stiff clay

10 m

= 100k Pa = 0.57 Stiff clay a

(A) 10 (B) 15

(C) 20 (D) 25

CE – 2007 3. What is the ultimate capacity in kN of the pile group shown in the figure assuming the group to fail as a single block? 0.4 m diameter piles 10 m

Clay soil = 40kN/

ϕ = 1.2 m c/c 1.2 m c/c

(A) 921.6 (B) 1177.6

(C) 2438.6 (D) 3481.6

CE – 2008 4. A pile of 0.5m diameter and of length 10m is embedded in a deposit of clay. The undrained strength parameters of the clay are cohesion = 60 kN/ and the angle of internal friction = 0. The skin friction capacity (kN) of the pile for an adhesion factor 0.6 is (A) 671 (C) 283 (B) 565 (D) 106 CE – 2009 5. A precast concrete pile is driven with a 50 kN hammer falling through a height of 1.0 with an efficiency of 0.6. The set value observed is 4 mm per blow and the combined temporary compression of the pile, cushion and the ground is 6 mm. As per modified Hiley formula, the ultimate resistance of the pile is (A) 3000 kN (C) 8333 kN (B) 4285.7 kN (D) 11905 kN CE – 2010 6. The ultimate load capacity of a 10 m long concrete pile of square cross section 500 mm × 500 mm driven into a homogeneous clay layer having undrained cohesion value of 40 kPa is 700 kN. If the cross section of the pile is reduced to 250 mm × 250 mm and the length of the pile is increased to 20 m, the ultimate load capacity will be (A) 350 kN (C) 722.5 kN (B) 632.5 kN (D) 1400 kN CE – 2011 7. A singly under-reamed, 8-m long, RCC pile (shown in the adjoining figure) weighing 20 kN with 350 mm shaft diameter and 750 mm under-ream diameter is installed within stiff, saturated silty clay (undrained shear strength is 50 kPa, adhesion factor is 0.3, and the applicable bearing capacity th

th

th



factor is 9) to counteract the impact of soil swelling on a structure constructed above. Neglecting suction and the contribution of the under-ream to the adhesive shaft capacity, what would be the estimated ultimate tensile capacity (rounded off to the nearest integer value of kN) of the pile?

10.

8000 mm


A single vertical friction pile of diameter 500 mm and length 20 m is subjected to a vertical compressive load. The pile is embedded in a homogeneous sandy stratum where: ng e f interna fricti n φ , Dry unit weight ( ) = 20 kN/ and ang e f wa fricti n φ Considering the coefficient of lateral earth pressure (K) = 2.7 and the bearing capacity factor ( ) = 25, the ultimate bearing capacity of the pile (in kN) is _______________

400 mm

350 mm 750 mm

(A) 132 kN (B) 156 kN

(C) 287 kN (D) 301 kN

CE – 2013 8. Four columns of a building are to be located within a plot size of . The expected load on each columns is 4000 kN.allowable bearing capacity of the soil deposit is 100 kN/ . The type of foundation best suited is (A) Isolated foundation (B) Raft foundation (C) Pile foundation (D) Combined foundation CE – 2014 9. The action of negative skin friction on the pile is to (A) increase the ultimate load on the pile (B) reduce the allowable load on the pile (C) maintain the working load on the pile (D) reduce the settlement of the pile

th

th

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2.

[Ans. A] Negative skin friction is experienced when the soil around the pile settles at a faster rate than pile. Thus, piles installed in freshly prepared fills of soft compressive deposits are subject to a downward drag. This downward drag on the pile surface, where the soil moves down relative to the pile, adds to the structural loads and is called negative skin friction.

Again,

7.

[Ans. B] The tensile resistance of a pile is given as Where shaft resistance weight f pi e c π

[Ans. C] The ultimate load capacity of piles,

c

The load capacity of single pile, c c̅ π 8.

[Ans. C] Area of footing of each column required, ad ⁄ Beading capacity Which is very large to be provided in field size. r Shallow foundation will not be feasille Pile foundation is best choice

9.

[Ans. B] Negative skin function is downward acting on pile due to downward movement of surrounding compressible soil relative to pile. It reduces allowable load on pile.

10.

[Ans. *] Range 6150 to 6190 For friction pile f

π number of piles 3.

[Ans. D] The ultimate load capacity of pile group by block failure = + L = 40 × 9 10 × 40 = 3481.6 kN

4.

5.

+4

×

[Ans. B] The skin friction capacity of a pile is given by, = π = 565.48 kN = 565 kN

here, f

̅

tan

[Ans. B] R=

=

tan

= 4285.7 kN

tan ( ϕ)

tan (

)

,f 6.

[Ans. B] The ultimate load of a pile in clay is given by c c

,

π π

th

th

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GATE QUESTION BANK

Transportation & Surveying

Introduction to Transportation CE - 2005 1. PradhanMantri Gram SadakYojna (PMGSY), launched in the year 2000, aims to provide rural connectivity with allweather roads. It is proposed to connect the habitations in plain areas of population more than 500 persons by the year (A) 2005 (C) 2010 (B) 2007 (D) 2012

CE - 2014 2. On a section of a highway the speeddensity relationship is linear and is given by v * k+; where v is in km/h and k is in veh/km. The capacity (in veh/h) of this section of the highway would be (A) 1200 (C) 4800 (B) 2400 (D) 9600


[Ans. B]

2.

[Ans. B] v = 80 Capacity, q = v k = 80 k k For q to be maximum dq dk dq k dk k Max. capacity, q

(

)

th

th

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GATE QUESTION BANK


Geometric design of highway CE – 2005 1. The length of summit curve a two lane two way highway depends upon (A) Allowable rate of change of centrifugal acceleration (B) Coefficient of lateral friction (C) Required stopping sight distance (D) Required overtaking sight distance 2.

On an urban road, the free mean speed was measured as 70 kmph and the average spacing between the vehicles under jam condition as 7.0 m. The speedflow- density equation is given by [

At a horizontal curve portion of a 4 lane undivided carriageway, a transition curve is to be introduced to attain required super elevation. The design speed is 60 kmph and radius of the curve is 245m. Assume length of wheel base of a longest vehicle as 6m, super elevation rate as 5% and rate of introduction of this super elevation as 1 in 150. The length of the transition curve (m) required, if the pavement is rotated about inner edge is (A) 81.4 (C) 91.5 (B) 85.0 (D) 110.2

] nd q

k

Where, U = space-mean speed (kmph); Usf =free mean speed (kmph); k = density (veh/km); kj= jam density (veh/km); q = flow (veh/hr). The maximum flow (veh/hr) per lane for this condition is equal to (A) 2000 (C) 3000 (B) 2500 (D) None of these

A road is having a horizontal curve of 400 m radius on which a super-elevation of 0.07 is provided. The coefficient of lateral friction mobilized the curve when a vehicle is travelling at 100 kmph is (A) 0.07 (C) 0.15 (B) 0.13 (D) 0.4

CE - 2006 3. A vehicle moving at 60 kmph on an ascending gradient of a highway has to come to stop position to avoid collision with a stationary object. The ratio of lag to brake distance is 6:5. Considering total reaction time of the driver as 2.5 seconds and the coefficient of longitudinal frication as 0.36, the value of as ascending gradient (%) is. (A) 3.3 (C) 5.3 (B) 4.8 (D) 6.8 4.

5.

CE - 2007 6. The extra widening required for a twolane national highway at a horizontal curve of 300 m radius, considering a wheel base of 8m and a design speed of 100kmph is (A) 0.42m (C) 0.82m (B) 0.62m (D) 0.92m 7.

While designing a hill road with a ruling gradient of 6%, if a sharp horizontal curve of 50 m radius is encountered, the compensated gradient at the curve as per the Indian roads congress specifications should be. (A) 4.4% (C) 5.0% (B) 4.75% (D) 5.25%

8.

The design speed on a road is 60kmph. Assuming the driver reaction time of 2.5 seconds and coefficient of friction of pavement surface as 0.35, the required stopped distance for two-way traffic on a single lane road is. (A) 82.1m (C) 164.2m (B) 102.4m (D) 186.4m

th

th

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GATE QUESTION BANK

CE - 2008 9. A road is provided with a horizontal circular curve having deflection angle 55 and centre line radius of 250m. a transition curve is to be provided at each end of the circular curve of such a length that the rate of gain of radial acceleration is 0.3m/s3 at a speed of 50 km per hour . Length of the transition curve required at each of the ends is (A) 2.57m (C) 35.73 m (B) 33.33 m (D) 1666.67 m

10.

11.

Common data for Q no 10 and Q 11 A horizontal circular curve with a centre line radius of 200m is provided on a 2lane, 2-way SH section. The width of the 2-lane road is 7.0m. Design speed for this section is 80 km per hour. The brake reaction time is 2.4s, and the coefficients of friction in longitudinal and lateral directions are 0.3555 and 0.15, respectively. The safe stopping sight distance on the section is (A) 221 m (C) 125 m (B) 195 m (D) 65 m

13.

respectively. the curve length (which is less than stopping sight distance) to be provided is (A) 120m (C) 163m (B) 152m (D) 240m CE - 2010 14. Consider the following statements in the context of geometric design of roads. I. A simple parabolic curve is an acceptable shape for summit curves. II. Comfort to passengers is an important consideration in the design of summit curves The correct option evaluating the above statements and their relationship is (A) I is true , II is false (B) I is true, II is true, and II is the correct reason for I (C) I is true, II is true and II is Not the correct reason for I (D) I is false, II is true 15.

The set-back distance from the centre line of the inner lane is (A) 7.93m (C) 9.60m (B) 8.10m (D) 9.77m

CE - 2009 12. The value of lateral friction or side friction used in the design of horizontal curve as per Indian roads congress guidelines is (A) 0.40 (C) 0.24 (B) 0.35 (D) 0.15 A crest vertical curve joins two gradients of +3% and 2% design speed of 80 km/h and the corresponding stopping sight distance of 120m.The height of driver’s eye nd the object bove the ro d surface are 1.20m and 0.15m


The design speed for a two-lane road is 80kmph. When a design vehicle with a wheelbase of 6.6m is negotiating a horizontal curve on that road, the offtracking is measured as 0.096m,. the required widening of carriage way of the two-lane road on the curve is approximately (A) 0.55 m (C) 0.75 m (B) 0.65 m (D) 0.85 m

CE - 2011 16. A vehicle negotiates a transition curve with uniform speed v. If the radius of the horizontal curve and the allowable jerk are R and J, respectively, the minimum length of the transition curve is ⁄(v ) (A) (C) v ⁄ (B) ⁄( v) (D) v ⁄( ) 17.

th

If v is the initial speed of a vehicle, g is the gravitational acceleration, G is the upward longitudinal slope of the road and f is the coefficient of rolling friction th

th


GATE QUESTION BANK

during braking, the braking distance (measured horizontally) for the vehicle to stop is (C) ( (A) ) (

(B)

Assume total reaction time = 2.5 seconds; coefficient of longitudinal friction of the pavement = 0.35; height of head light of the vehicle = 0.75 m; and beam angle =

)

(

)

(D)

(


)

CE - 2012 18. The following data are related to a horizontal curved portion of a two – lane highway: length of curve = 200 m, radius of curve = 300 m and width of pavement = 7.5 m. In order to provide a stopping sight distance (SSD) of 80 m, the set back distance (in m) required from the centre line of the inner lane of the pavement is (A) 2.54 m (C) 7.10 m (B) 4.55 m (D) 7.96 m CE - 2013 19. The percent voids in mineral aggregate (VMA) and percent air voids ( ) in a compacted cylindrical bituminous mix specimen are 15 and 4.5 respectively. The percent voids filled with bitumen (VFB) for this specimen is : (A) 24 (C) 54 (B) 30 (D) 70

20.

What is the length of valley curve (in m) based on the head light sight distance conditions? _______________________

21.

What is the length of valley curve (in m) based on the comfort condition?_________

CE - 2014 22. The perception-reaction time for a vehicle travelling at 90 km/h, given the coefficient of longitudinal friction of 0.35 and the stopping sight distance of 170 m (assume g = 9.81 m s ), is _____________ seconds. 23.

A road is being designed for a speed of 110 km/hr on a horizontal curve with a super elevation of 8%. If the coefficient of side friction is 0.10, the minimum radius of the curve (in m) required for safe vehicular movement is (A) 115.0 (C) 264.3 (B) 152.3 (D) 528.5

Common Data for Questions 20 and 21 For a portion of national highway where a descending gradient of 1 in 25 meets an ascending gradient of 1 in 20, a valley curve needs to be designed for a vehicle travelling at 90 kmph based on the following conditions. (1) Headlight sight distance equal to the stopping sight distance (SSD) of a level terrain considering length of valley curve > SSD. (2) Comfort condition with allowable rate of change of centrifugal acceleration is 0.5 m/sec

th

th

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GATE QUESTION BANK


Answer Keys and Explanations 1.

2.

[Ans. D] Length of summit curve depends upon: a. SSD for single lane two way highway b. OSD for two lane two way highway

from

formul

[Ans. B]

( (e

e

f 5. (

)

m

The length of transition curve will be 110.22m

f)

f

3.

Now length of transition curve as per rate of introduction of super elevation x m

[Ans. B] Traffic volume = density × speed q= k k q ( )k k k q (k ) k m ensity

)

[Ans. B] g dist nce

t m

r ke dist nce l g dist nce br ke dist nce

m[

r king dist nce

(f

for m ximum tr ffic volume

)

( ( )

k ) k

(

n)

k

[Ans. D] When the pavement is rotated about the inner edge

ximum tr ffic volume q

(

k

k (

(

) ) k ( ) veh⁄hr

W+EW

)e ise x ( king width of e ch l ne *

6.

[Ans. C] The extra widening w given by n

m] √

[ x

k

)

e

[

)

[ u

k

x

K

dq dk

k k

n 4.

km⁄hr

nd

n) (

n

ver ge sp cing between vehicles

]

+

( )

] m

√ m

√ th

th

th


GATE QUESTION BANK

7.

[Ans. A]

11.


[Ans. C]

r de compens tion 1m

ulling gr deient ompens ted gr dient

8.

7.0m R

[Ans. C] topping ist nce

vt

cos

et b ck dist nce m

v f

cos (

m)

cos ( m But the traffic is two way therefore the stopping distance = 2 × 82.1= 164.2m 9.

[Ans. C] If is the rate of change of radial acceleration, the radial acceleration (a) attained during the time the vehicle passes over the transition curve is given by

)

cos (

)

m 12.

[Ans. D]

13.

[Ans. B] When the curve length is less than stopping sight distance then it is given by,

t ( m

di l cceler tion

(

14.

[Ans. A]

15.

[Ans. C]

)

ff tr cking

l

m

m 10.

m

[Ans. C] fe stopping sight dist nce, v vt f (

nl

xtr widening

√ √

) m

)

m

16.

[Ans. D]

17.

[Ans. B]

th

th

th

m


GATE QUESTION BANK

18.

[Ans. B] Length of curve, Radius of curve, R = 300 m Width of pavement, w = 7.5 m SSD, S = 80 m

22.

[Ans. *] (Range 3.1 to 3.2) t

)

(

(

19.

)

) cos(

)

) g

f (e

[Ans. D]

( 528.5 m

20.

)

s

[Ans. D] ( e f e

(

)t

t 23.

(

gf

(

distance, m = R – (R-d) cos ⁄


f) )

[Ans. *] t

f

m s (

21.

)

m

[Ans. *] (

c

)

(

)

Where V is in m s ( [

m⁄hr

(

)

) ms

]

m

th

th

th


GATE QUESTION BANK


Traffic Characteristics CE - 2005 1. A transport company operates a scheduled daily truck service between city P and city Q. One-way journey time between these two cities is 85 hours. A minimum layover time of 5 hours is to be provided at each city. How many trucks are required to provide this service? (A) 4 (C) 7 (B) 6 (D) 8 2.

A single lane unidirectional highway has a design speed of 65 kmph. The perceptionbrake-reaction time of drivers is 2.5 seconds and the average length of vehicles is 5 m. The coefficient of longitudinal friction of the pavement is 0.4. The capacity of road in terms of vehicles per hour per lane is (A) 1440 (C) 710 (B) 750 (D) 680

CE - 2006 3. Name the traffic survey data which is plotted by me ns of “ esire lines” (A) Accident (B) Classified volume (C) Origin and Destination (D) Speed and Delay CE - 2007 4. If a two-lane national highway and a two–lane state highway intersect at right angles, the number of potential conflict points at the intersection, assuming that both the roads are two-way is (A) 11 (C) 24 (B) 17 (D) 32 CE - 2008 5. The capacities of sidewalk (persons way 2-lane urban with no frontage

“One-way 1.5m wide per hour)" and "One road (PCU per hour, access, no standing

vehicles nd very little cross tr ffic)” re respectively (A) 1200 and 2400 (C) 1200 and 1500 (B) 1800 and 2000 (D) 2000 and 1200 6. A roundabout is provided with an average entry width of 8.4m, width of weaving section as 14m, and length of the weaving section between channelizing islands as 35m. The crossing traffic and total traffic on the weaving section are 1000 and 2000 PCU per hour respectively. The nearest rounded capacity of the roundabout (in PCU per hour) is (A) 3300 (C) 4500 (B) 3700 (D) 5200 7. A linear relationship is observed between speed and density on a certain section of a highway. The free flow speed is observed to be 80 km per hour and the jam density is estimated as 100 vehicles per km length. Based on the above relationship, the maximum flow expected on this section and the speed at the maximum flow will respectively be (A) 8000 vehicles per hour and 80km per hour (B) 8000 vehicles per hour and 25km per hour (C) 2000 vehicles per hour 80km per hour. (D) 2000 vehicles per hour and 40km per hour CE - 2009 8. On a specific highway, the speed-density relationship follows the Greenberg's model [v=vf log (kj/k)], where vf and kj are the free flow speed and jam density respectively. When the highway is operating at capacity, the density obtained as per this model is (A) e.kj (C) kj/2 (B) kj (D) kj/e th

th

th


GATE QUESTION BANK

CE - 2011 9. If the jam density given as k and the free flow speed is given as u the maximum flow for a linear traffic speed density model is given by which if the following options?

10.

(A)

k

u

(C)

k

u

(B)

k

u

(D)

k

u

The probability that k number of vehicles arrive (i. e. cross a predefined line) in time t is given as ( t) e k where is the average vehicle arrival rate. What is the probability that the time headway is greater than or equal to time t (A) e (C) e (B) e (D) e

CE - 2013 11. It was observed that 150 vehicles crossed a particular location of a highway in a duration of 30 minutes. Assuming that vehicle arrival follows a negative exponential distribution, find out of number of time headways greater than 5 seconds in the above observation?____ 12.


CE - 2014 13. The minimum value of 15 minute peak hour factor on a section of a road is (A) 0.10 (C) 0.25 (B) 0.20 (D) 0.33 14.

The speed-density (u k) rel tionship on a single lane road with unidirectional flow is u = 70 0.7k, where u is in km/hr and k is in veh/km. The capacity of the road (in veh/hr) is ___________

15.

A student riding a bicycle on a 5 km oneway street takes 40 minutes to reach home. The student stopped for 15 minutes during this ride. 60 vehicles overtook the student (assume the number of vehicles overtaken by the student is zero) during the ride and 45 vehicles while the student stopped. The speed of vehicle stream on that road (in km/hr) is (A) 7.5 (C) 40 (B) 12 (D) 60

16.

An observer counts 240 vehicle/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-seconds time interval is

For two major roads with divided carriageway crossing at right angle, a full clover leaf interchange with four indirect ramps in provided. Following statements are made on turning movements of vehiclesto all directions from both roads. Identify the correct statements: (A) Merging from left is possible, but diverging to left is not possible (B) Both merging from left and diverging to left are possible. (C) Merging from left is not possible, but diverging to left is possible (D) Neither merging left nor diverging to left is possible.

th

th

th





[Ans. D]

6.

2.

[Ans. C] Space headway, t

f

(

[Ans. B] Width of the weaving section, W =14m Average entry width, e = 8.4 m Length of weaving section between channelizing islands L = 35 m Proportion of weaving traffic is given by

)

m Capacity

The capacity the round about may be given as: w(

3.

[Ans. C]

4.

[Ans. C] On a right angled road intersection with two way traffic the total number of conflict points merging and diverging conflicts are considered as minor conflicts, numbering 4 in each case.

)(

)

(

)(

(

)

)

per hour

7.

[Ans. D] The maximum flow may be calculated as. free flow speed j m density q

vehicles⁄hr Maximum flow occurs when the speed becomes half of the free flow speed i.e., free flow speed peed t q

5.

[Ans. A] Tentative capacity values of urban roads suggested by IRC are: a. One-way two lane road with no frontage access, no standing vehicles and very little cross traffic = 2400 PCU per hour b. One way traffic two lane road with frontage access, but no standing vehicle and high capacity intersections – 1500 PCU/hr. c. One-way two lane road with free from frontage access, parked vehicles and heavy cross traffic -1200 PCU/hr

km⁄hr

8.

[Ans. D] Given k log ( ) k But capacity C = traffic density × speed =k×V k k log ( ) k d for m ximum c p city dk k d * log ( )+ dk k 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.comPage 335

GATE QUESTION BANK

k k d d log ( ) log ( ) k dk k k dk k k k k ( ) log k k k k log k k e k k k e

15.

9.

[Ans. A]

16.

10.

[Ans. D]

11.

[Ans. *] Insufficient data (Declared by IIT)

k


[Ans. D] Velocity =

ehicle⁄min ) el tive speed of vehicle

(

(

ehicle min ) el tive speed of vehicle

x x x = 60 km/hr [Ans. *]Range (0.25 to 0.28) ( t) e (n t) n ere no of vehicles vehicle km (

12.

[Ans. B] On cloverleaf intersection, merging & diverging from both direction is possible.

13.

[Ans. C] 15 min peak factor is used for traffic intersection division

V = peak hourly volume (

km hr

)

e

(

)

e

)

m x min volume within the pe k hr Max value = 1, value = 0.25 Normal range = 0.7 – 0.98

14.

[Ans. 1750] u = 70 – 0.7 k u p city

k

u free velocity k At k u ok tk

j m density

eh km u

o c p city

u

km hr veh hr

th

th

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GATE QUESTION BANK


Traffic Signs and Signal Design CE - 2006 1. For designing a 2-phase fixed type signal at an intersection having North – South and East – West road where only straight ahead traffic is permitted, the following data is available. Parameter North South East West Design Hour 1000 700 900 550 Flow (PCU/hr) Saturation 2500 2500 3000 3000 Flow (PCU/hr) Total time lost per cycle is 12 seconds. The cycle length (seconds) as per ebster’s ppro ch is (A) 67 (C) 87 (B) 77 (D) 91 CE - 2007 2. In signal design as per Indian Roads Congress specification, if the sum of the ratios of normal flows to saturation flow of two directional traffic flow is 0.50 and the total lost time per cycle is 10 seconds, the optimum cycle length in seconds is (A) 100 (C) 60 (B) 80 (D) 40 CE - 2008 3. The shape of the STOP sign according to IRC:67-2001 is (A) Circular (C) Octagonal (B) Triangular (D) Rectangular CE - 2009 4. A three-phase traffic signal at an intersection is designed for flows shown in the figure below. There are six groups of flows identified by the numbers 1 through 6. Among these 1, 3, 4 and 6 are through flows and 2 and 5 are right turning. Which phasing scheme is not feasible?

Combination choice P Q R S (A) P (B) Q

Phase Phase I II 1, 4 2, 5 1, 2 4, 5 2, 5 1, 3 1, 4 2, 6 (C) R (D) S

Phase III 3, 6 3, 6 4, 6 3, 5

CE - 2010 5. As per IRC: 67-2001, a traffic sign indicating the speed limit on a road should be of (A) Circular Shape with White Background and Red Border (B) Triangular Shape with White Background and Red Border (C) Triangular Shape with Red Background and White Border (D) Circular Shape with Red Background and White Border CE – 2012 6. A two – lane urban road with one – way traffic has a maximum capacity of 1800 vehicle/ hour. Under the jam condition, the average length occupied by the vehicles is 5.0 m. The speed versus density relationship is linear. For a traffic volume of 1000 vehicles/hour, the dencity (in vehicles / km) is (A) 70 (C) 71.11 (B) 69.10 (D) 75 th

th

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GATE QUESTION BANK

CE – 2014 7. An isolated three-phase traffic signal is designed by Webster's method. The critical flow ratios for three phases are 0.20, 0.30, and 0.25 respectively, and lost time per phase is 4 seconds. The optimum cycle length (in seconds) is ___________ 8.

9.


A pre-timed four phase signal has critical lane flow rate for the first three phases as 200, 187 and 210 veh/hr with saturation flow rate of 1800 veh/hr/lane for all phases. The lost time is given as 4 seconds for each phase. If the cycle length is 60 seconds, the effective green time (in seconds) of the fourth phase is ______________

The average spacing between vehicles in a traffic stream is 50 m, then the density (in veh/km) of the stream is _____________


[Ans. B] For N – S road and E – W road the higher traffic volume will be taken i.e., q nd q s nd s q y s q y s y y y

6.

[Ans. C] And (

m s Traffic capacity = Speed × density 1000 = 14.06 × density density

optimum cycle time (

2.

seconds

)

)

7.

[Ans. *] Range 90 to 95 Total time lost in a cycle, sec

[Ans. D] iven y

sec

y

optimum cycle length

( s

)

sec 8. 3.

[Ans. C] The stop sign is intended to stop the vehicle on a road way. It is octagonal in shape and red in colour with a white border. Stop sign is a regulatory or mandatory sign.

4.

[Ans. C] Under phase II, 1 and 3 cannot move simultaneously. It is also the case under phase III for 4 and 6.

5.

[Ans. A]

[Ans. 20] Density = = = 20 veh/km

9.

[Ans. *] Range 14.0 to 18.0 Given flow rates q eh hr q eh hr q eh hr Saturation flow rate = 1800 Veh/hr/lane

th

th

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GATE QUESTION BANK

Lost time length of cycle q y s q y s q y s


sec sec

C= 60 = 60 = y = 0.517 And y = y

y

y

y

y y

y ( y

) (

) sec

th

th

th


GATE QUESTION BANK


Intersection Design CE - 2011 4. The cumulative arrival departure curve of one cycle of an approach lane of a signalized intersection is shown in the adjoining figure. The cycle time is 50s and the effective red time is 30s and the effective green time is 20s. What is the average delay? Cumulative arrival or departure (No. of vehicles)

CE – 2007 1. Two straight line intersect at an angle of . The radius of a curve joining the two straight lines is 600 m. The length of long chord and mid-ordinates in meters of the curve are (A) 80.4, 600.0 (C) 600.0, 39.89 (B) 600.0, 80.4 (D) 49.89, 300.0 CE - 2008 2. Design parameter for a signalized intersection are shown in the figure below. The green time calculated for major and minor roads are 34 and 18s, respectively. The critical lane volume on the major road changes to 440 vehicles per hour per lane and the critical lane volume on the minor road remains unchanged. The green time will. inor o d l ne jor Road 4-lane divided

m

Vehicles Per hour

(A) Increase for the major road and remain same for the minor road. (B) Increase for the major road and decrease for the minor road (C) Decrease for both the roads. (D) Remain unchanged for both the roads. CE - 2010 3. Aggregate impact value indicates the following property of aggregates (A) Durability (C) Hardness (B) Toughness (D) Strength

Cumulative Arrival

30

20

10

0

urns prohibited

m

40

(A) 15 s

0

10

Cumulative departure 20 30 50 40 Time (s)

(B) 25 s

(C) 35 s

(D) 45 s

CE - 2012 5. Two major roads with two lanes each are crossing in an urban area to from an un – controlled intersection. The number of conflict points when both roads are one – w y is “X” nd when both ro ds re two – w y is “Y” the r tio of X to Y is (A) 0.25 (B) 0.33 (C) 0.50 (D) 0.75 CE - 2014 6. The chainage of the intersection point of two straight is 1585.60 m and the angle of intersection is If the radius of a circular curve is 600.00 m, the tangent distance (in m) and length of the curve (in m), respectively (A) 418.88 and 1466.08 (B) 218.38 and 1648.49 (C) 218.38 and 418.88 (D) 418.88 and 218.38

th

th

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GATE QUESTION BANK



[Ans. B]

6.

[Ans. C] (

(

)

)

(

)

R

600 m O

Length of long chord, sin ( ) sin( ) ( ) m Length of mid-ordinate [ cos( )] [ cos( )] m 2.

[Ans. A]

3.

[Ans. B]

4.

[Ans. A]

5.

[Ans. A] No. of points of contra flexure when both roads one way x=6 No of points of conflicts when both roads two way y=24

ength of the curve m Tangent distance (T) is the distance between P-C to P.I (also the distance from P.I to P.T) t n

t n

t n m

th

th

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GATE QUESTION BANK


Testing and Specifications of Paving Materials CE - 2005 1. Group I contains some properties of bitumen. Group II gives a list of Laboratory Test conducted on bitumen to determine the properties. Match the property with the corresponding test Group I P. Resistance to flow Q. Ability to deform under load R. safety Group II 1. Ductility test 2. Penetration test 3. Flash and fire point test (A) (B) (C) (D) 2.

Bituminous concrete is a mix comprising of (A) Fine aggregate, filler and bitumen (B) Fine aggregate and bitumen (C) Coarse aggregate, fine aggregate, filter and bitumen (D) Coarse aggregate, filter and bitumen

CE - 2006 3. If aggregate size of 50-40 mm is to be tested for finding out the portion of elongated aggregates using length gauge, the slot length of the gauge (A) 81 mm (C) 53mm (B) 45mm (D) 90mm 4.

CE - 2007 5. The consistency and flow resistance of bitumen can be determined from the following. (A) Ductility test (B) Penetration test (C) Softening point Test (D) Viscosity Test 6.

Match the following test on aggregate and its properties Test Property P. Crushing test 1. Hardness Q. Los Angles abrasion test 2. Weathering R. Soundness test 3. Shape S. Angularity test 4. Strength (A) (B) (C) (D)

CE - 2008 7. The specific gravity of paving bitumen as per IS:73-1992 lies between (A) 1.10 and 1.06 (C) 1.02 and 0.97 (B) 1.06 and 1.02 (D) 0.97 and 0.92 8.

A combined value of flakiness and elongation index is to be determined for a sample of aggregates. The sequence in which the two tests are conducted is (A) Elongation index test followed by flakiness index test on the whole sample (B) Flakiness index test followed by elongation index test on the whole sample. (C) Flakiness index test followed by elongation index test on the nonflaky aggregates. (D) Elongation index test followed by flakiness index test on non-elongated aggregates.

A subgrade soil sample was tested using standard CBR Apparatus and the observation are given Load ,Kg Penetration, mm 60.5 2.5 80.5 5.0 Assuming that the load-penetration curve is convex throughout, the CBR value (%) of the sample is (A) 6.5 (C) 4.4 (B) 5.5 (D) 3.9 th

th

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GATE QUESTION BANK

CE - 2009 9. During a CBR test, the load sustained by a remolded soil specimen at 5.0 mm penetration is 50 kg. the CBR value of the soil will be. (A) 10.0% (C) 3.6% (B) 5.0% (D) 2.4% CE - 2011 10. In marshal testing of bituminous mixes, as the bitumen content increases the flow value (A) Remains constant (B) Decreases first and then increases (C) Increase monotonically (D) Increases first and then decreases CE - 2012 11. Road roughness is measured using (A) Benkelman beam (B) Bump integrator (C) Dynamic cone penetrometer (D) Falling weight deflectomerter

12.


Two bitumen samples X and Y have softening points and , respectively. Consider the following I. Viscosity of X will be greater than that of Y at the same temperature II. Penetration value of X is lesser than that of Y under standard conditions The correct option evaluating the above statement is (A) Both I and II are true (B) I is false and II is true (C) Both are false (D) I true and II false

CE - 2014 13. In a Marshall sample, the bulk specific gravity of mix and aggregates are 2.324 and 2.546 respectively. The sample includes 5% of bitumen (by total weight of mix) of specific gravity 1.10. The theoretical maximum specific gravity of mix is 2.441. The void filled with bitumen (VFB) in the Marshall sample (in %) is ___________

th

th

th


GATE QUESTION BANK



[Ans. A]

8.

[Ans. B]

2.

[Ans. C]

9.

[Ans. D]

3.

[Ans. A] Slot length for elongated aggregate, me n dimension (

)

mm or fl kiness index the slot size (

)

mm

Thus, the higher value of CBR which is obtained at 2.5 mm penetration is adopted i.e. 4.4%. 4.

( )

10.

[Ans. C ]

11.

[Ans. B ] Road roughness is indicated roughometer or bump integrator,

12.

[Ans. C] Lower the softening, lower the viscosity, and higher the penetration value. So both statements wrong

13.

[Ans. *] Range 62 to 66

[Ans. C] lo d sust ined by specimen t mm penetr tion o d sust ined by st nd rd ggreg tes t mm penetr tion

by

Bitumen % by total wt of min = 5

Also, et tot l volume of min tot l weight

m kg

weight of bitumen Thus the higher value of CBR which is obtained as 2.5 mm penetration is adopted i.e. 4.4% 5.

[Ans. D]

6.

[Ans. D]

7.

[Ans. C]

= 116.2 kg Value of bitumen

= 68%

th

th

th


GATE QUESTION BANK


Design of Rigid Flexible Pavements CE - 2005 1. For a 25 cm thick cement concrete pavement, analysis of stresses gives the following values: Wheel load stress due to corner loading: 30 kg/cm2. Wheel load stress due to edge loading: 32 kg/cm2. Warping load stress at corner region during summer: 9 kg/cm2. Warping load stress at corner region during winter: 7 kg/cm2. Warping load stress at edge region during summer: 8 kg/cm2. Warping load stress at edge region during winter: 6 kg/cm2. Frictional stress during summer: 5 kg/cm2. Frictional stress during winter: 4 kg/cm2. The most critical stress value for this pavement is (A) 40 kg/cm2. (C) 44 kg/cm2. 2 (B) 42 kg/cm . (D) 45 kg/cm2. 2.

The following observation were made of an axle load survey on a road Axle Load (kN) Repetitions per day 35-45 800 75-85 400 The standard axle-load is 80 kN. Equivalent daily number of repetitions for the standard axle-load are (A) 450 (C) 800 (B) 480 (D) 1200

CE - 2006 3. In case of governing equations for calculating wheel load stress using esterg rd’s ppro ch the following statements are made. i. Load stress are inversely proportional to wheel load ii. Modules of subgrade reaction is useful for load stress calculation.

(A) (B) (C) (D) 4.

Both statement are True I is True and II is False Both statements are False I is False and II is True

Using IRC: 37“ uidelines for the design of flexible p vements” nd the following data chose the total thickness of the pavement. No of commercial vehicles when construction is completed = 2723veh/day Annual growth rate of the traffic = 5.0% design life of the pavement = 10 years Vehicle damage factor = 2.4 CBR value of the subgrade soil = 5% Data for 5% CBR value No. of standard Total thickness, Axles, msa mm 20 620 25 640 30 670 40 700 (A) 620 mm (C) 670 mm (B) 640 mm (D) 700 mm

CE - 2007 5. The width of the expansion joint is 20 mm in a cement concrete pavement. The laying temperature is and the maximum slab temperature in summer is . The coefficient of thermal expansion of concrete is mm mm and the joint filter compresses upto 50% of the thickness. The spacing between expansion joints should be (A) 20m (C) 30m (B) 25m (D) 40m 6.

The following data pertains to the number of commercial, vehicles per day for the design of a flexible pavement for a national highway as per IRC: 37-1984 th

th

th


GATE QUESTION BANK

Type of commercial vehicle

Number of vehicle per day considering the number of lanes 2000

Vehicle Damage Factor

Two axle 5 trucks Tandem axle 200 0 trucks Assuming a traffic growth factor of 7.5% per annum for both the types of vehicles. The cumulative number of standard axle standard axle load repetitions (in million) for a design life ten years is (A) 44.6 (C) 62.4 (B) 57.8 (D) 78.7 CE - 2008 7. It is proposed widen and strength an existing 2-lane NH section as a divided highway. The existing traffic in one direction is 2500 commercial vehicles (CV) per day. The construction will take 1 year. The design CBR of soil subgrade is found to be 5 percent. Given: traffic growth rate for CV = 8 percent, vehicle damage factor = 3.5 (standard axles per CV). Design life = 10 years and traffic distribution factor = 0.75. The cumulative standard axles (msa) computed are (A) 35 (B)37 (C) 65 (D) 70 CE - 2009 8. Which if the following stress combinations are appropriate in identifying the critical condition for the design of concrete pavements? Type of stress Location P. Load 1. Corner Q. Temperature 2. edge 3. interior (A) (C) (B) (D)


CE - 2010 9. Consider the following statements in the context of cement concrete pavements. 1. Warping stresses in cement concrete pavements are caused by the seasonal variation in temperature, 2. Tie bars are generally provided across transverse joints of cement concrete pavements The correct option evaluating the above statements is (A) 1: True and 2: False (B) 1: False and 2:True (C) 1: True and 2: True (D) 1: False and 2: False CE - 2012 10. A pavement designer has arrived at a design traffic of 100 million standard axles for a newly developing national highway as per IRC: 37 guidelines using the following data: life = 15 years, commercial vehicle count before pavement construction = 4500 vehicles/day, annual traffic growth rate = 8%. The vehicle damage factor used in the calculation was (A) 1.53 (C) 3.66 (B) 2.24 (D) 4.14 CE - 2013 11. select the strength parameter of concrete used in design of plain jointed cement concrete pavements from the following choices: (A) tensile strength (B) compressive strength (C) flexural strength (D) shear strength CE - 2014 12. The following statements are related to temperature stresses developed in concrete pavement slabs with free edges (without any restraint):

th

th

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GATE QUESTION BANK

P.

The temperature stresses will be zero during both day and night times if the pavement slab is considered weightless Q. The temperature stresses will be compressive at the bottom of the slab during night time if the selfweight of the pavement slab is considered R. The temperature stresses will be compressive at the bottom of the slab during day if the self - weight of the pavement slab is considered The TRUE statement(s) is (are) (A) P only (C) P and Q only (B) Q only (D) P and R only

13.


A traffic survey conducted on a road yields an average daily traffic count of 5000 vehicles. The axle load distribution on the same road is given in the following table: Axle load (tonnes) Frequency of traffic (%) 18 10 14 20 10 35 8 15 6 20 The design period of the road is 15 years, the yearly traffic growth rate is 7.5% and the load safety factor (LSF) is 1.3. If the vehicle damage factor (VDF) is calculated from the above data, the design traffic (in million standard axle load, MSA is) _______________


[Ans. B] There are four cases which should be considered i. During summer, critical condition combinations of stress at edges = wheel load stress due to edge loading + warping stress at edge region during winter frictional stress during summer = 32+ = 35 kg/cm2 ii.

During winter, critical combination of stresses at edges = Wheel load stress due to edge loading + warping stress at edge region during winter + frictional stress during winter. = 32 + 6 + 4 = 42 kg/cm2

iii. During summer, critical combination of stresses at corner regions = wheel load stress at corner loading + warping stress at corner region during summer = 30+ 9 = 39 kg/cm2

iv.

During winter, critical combination of stress at corner region = Wheel load stress at corner loading + warping stress at winter region during winter = 30 +7 = 37kg/cm2 Hence, to critical combination (II) i.e., 42 kg/cm2 2.

[Ans. A]

3.

[Ans. D] Load stresses are proportional to wheel load. Modules of subgrade reaction (K) is used to calculate the radius of relative stiffness (L) which eventually is used in load stress calculations.

4.

[Ans. C] Number of standard axles in msa is given by [(

)

]

[(

th

th

)

th

]


GATE QUESTION BANK

ms Now for 5% CBR value, the total thickness of the pavement for 30 msa is given as 670 mm. 5.

8.

[Ans. A]

9.

[Ans. D] Warping stress in cement concrete pavements are caused by the daily variation in temperature Tie bars are generally provided across longitudinal joints of cement concrete pavements

10.

[Ans. B]

[Ans. B] The spacing of expansion joint given by )

c( Given,

idth of exp nsion joint

mm

( 6.

)


[(

m

]

r) r

[(

[Ans. B] [(

)

]

]

r) r

[( (

)

]

]

r) r

[(

7.

[Ans. C] While designing rigid concrete pavement flexural st of concrete is considered

12.

[Ans. C] Since slab has free edges, no warping stresses shall be introduced if slab is considered weightless. During night time, temperature of lower surface is higher t tries to exp nd resulting in development of compressive stresses t bottom Hence correct option is C

13.

[Ans. *] Range 307 to 310 Vehicle damage factor

) [(

(

11.

)

]

)

[Ans. D] Number of commercial vehicles per day A = Number lane × existing traffic × traffic distribution factor = 2 × 2500 × 0.75 = 3750 Annual growth rate of commercial vehicles = 8% Vehicle damage factor, F = 3.5 Design life, n = 10 years Thus, the cumulative standards axles (msa) may be calculated as [( ] r) r [(

)

(

)

( )

( ( )

[(

o

th

(

)

( )

( )

( )

]

r) r

]

)

)

[(

]

esign tr ffic

th

th


GATE QUESTION BANK


Introduction CE - 2007 1. The plan of map was photo-copied to a reduced size such that a line originally 100mm was measured 90mm. The original scale of the plan was 1:1000. The revised scale is (A) 1:900 (C) 1:1121 (B) 1:1111 (D) 1:1221 CE - 2008 2. The plan of survey plotted to scale of 10m to 1cm is reduced in such a way that line originally 10cm long now measures 9cm. The area of reduced plan is measured as 81cm2. The actual area (m2) of survey is (A) 10000 (C) 1000 (B) 6561 (D) 656 CE - 2010 3. The local mean time at a place located in Longitude when the standard time is 6 hours and 30 minutes and the standard meridian is is (A) 5 hours, 2 minutes and 40 seconds (B) 5 hours, 47 minutes and 20 seconds (C) 6 hours and 30 minutes (D) 7 hours, 02 minutes and 40 seconds

5.

Following bearings are observed while traversing with a compass. Line Fore Bearing Back Bearing AB BC CD DE EA After applying the correction due to local attraction, the corrected fore bearing of line BC will be : (A) (C) (B) (D)

CE - 2014 6. The survey carried out to delineate natural features, such as hills, rivers, forests and man-made features, such as towns, villages, buildings, roads, transmission lines and canals is classified as (A) Engineering survey (B) Geological survey (C) Land survey (D) Topographic survey

CE - 2013 4. The latitude and departure of a line AB are + 78 m and 45.1 m, respectively. The whole circle bearing of the line AB is : (A) (C) (B) (D)

th

th

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GATE QUESTION BANK



4.

[Ans. B]

[Ans. D] N

Reduction factor = Revided scale = original scale

B

R.F

=

2.

3.

W

[Ans. A] Shrinkage factor = 9/10 = 0.9 =S.F Reduced plan are = (S.F)2 actual plan area = 81 = ( ) Actual plan area = Actual plan area = 100cm Actual area of survey in ( ) m [Ans. D] Difference between standard meridian and the longitude of the given pl ce The longitude of the given place is more than the standard meridian. Hence the local mean time of the given place will be ahead of the standard time.

S

+ve latitude means either I quadrant or 4th quadrant (casein is +ve in 1st or 4th ) -ve departure means either 3rd of 4th quadrant [ dep rture l sin ] should lie in 4th quadrant.

5.

[Ans. D] Angle difference between back bearing and fore bearing as line DE station D and E are free from local attraction of (correct) of But observed BB of

hours hours nd

minutes nd seconds

correct

minutes hours nd minutes nd seconds Time for a difference of minutes nd seconds Local mean time = standard time + 32 minutes and 40 seconds = 6 hours and 30 minutes + 32 minutes and 40 seconds = 7 hours, 02 minutes and 40 seconds

E

A

of b

correct of So correction applied correct

6.

of

[Ans. D]

th

th

th




Measurement of Distance & Direction CE - 2006 1. Consider the following figure, which is an extract from a contour map (scale = 1:20, 000) of an area. An alignment of a road at a ruling gradient of 4% is to be fixed from the point O and beyond. What should be the radius of the arc with O as the center to get the point of alignment of the next contour on the map?

CE - 2008 2. A light house of 120m height is just visible above the horizon from a ship. The correct distance (m) between the ship and the light house considering combined correction for curvature and refraction, is (A) 39.098 (C) 39098 (B) 42.226 (D) 42226 CE - 2014 3. The Reduced Levels (RLs) of the points P and Q are +49.600 m and +51.870 m respectively. Distance PQ is 20 m. The distance (in m from P) at which the +51.000 m contour cuts the line PQ is (A) 15.00 (C) 3.52 (B) 12.33 (D) 2.27

O

10 m 30 m

(A) 0025 cm (B) 0.25 cm

50 m 70 m

(C) 2.5 cm (D) 5.0 cm


m dius of rc

2.

= 0.0673

[Ans. C] The contour interval = 20 m For 4% gradient, the length needed from one contour to another

d d d

3.

d2

√ 1000 m

[Ans. B] For change in elevation of 2.27m, Distance = 20m For (+51.000 – 49.600) = 1.4m

cm

[Ans. D] Correction due to curvature, Cc = 0.0785d2 Correction due to refraction, Cr = +0.0112d2 omposite correction C = +0.0112d2 +0.0112d2 Where C is in metres and d is in kilometers Here, C = 120m; d =?

ist nce

th

th

m

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GATE QUESTION BANK


Theodolite & Traversing CE - 2006 1. In the figure given below, the lengths PQ (WCB: ) and OR (WCB: ) respectively up to three places of decimal are

4.

The magnetic bearing of a line AB is S E and the declination is west. The true bearing of the line AB is (A) S E (C) S E (D) S W (B) S E

N

R(1000N, 1000E)

0

P(100N, 200E)

(A) (B) (C) (D) 2.

E

273.205, 938.186 273.205, 551.815 551.815, 551.815 551.815, 938.186

CE - 2008 5. The lengths and bearings of a closed traverse PQRSP are given below. Line Length(m) Bearing(WCB) PQ 200 QR 1000 RS 907 SP ? ? The missing length and bearing, respectively of the line SP are (A) 207m and (C) 707m and (B) 707m and (D) 907m and 6.

The focal length of the object glass of a tachometer is 200 mm, the distance between the vertical axis of the tachometer and the optical centre of the object glass is 100 mm and the spacing between the upper and lower line of the diaphragm axis is 4 mm. with the line of collimation perfectly horizontal, the staff intercepts are 1 m (top), 2 m (middle), and 3 m (bottom). The horizontal distance (m) between the staff and the instrument station is (A) 100. 3 (C) 150.0 (B) 103.0 (D) 153.0

The observed magnetic bearing of a line OE was found to be . It was later discovered that station O had a local attraction of + . The true bearing of the line OE, considering a magnetic declination of E shall be (A) (C) (B) (D)

CE - 2007 3. The following table gives data of consecutive coordinates in respect of a closed theodolite traverse PQRSP. P Q R S Station Northing, m 400.75 100.25 300.0 Southing, m 199.0 199.25 399.75 Easting, m 200.5 Westing, m 300.5 The magnitude and direction of error of closure in whole circle bearing are. (A) 2.0m and (B) 2.0m and (C) 2.82m and (D) 3.42m and

CE - 2009 7. In quadrantal bearing system, bearing of a line varies from (A) to (C) to (B) to (D) to 8.

The magnetic bearing of a line AB was N in the year 1967, when the declination was . If the present declination is the whole circle bearing of the line is (A) (C) (B) (D) th

th

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GATE QUESTION BANK

CE - 2011 9. The observations from a closed loop traverse around an obstacle are Segment Observation Length Azimuth from (m) (clockwise from station magnetic north) PQ P Missing QR Q 300.00 RS R 354.524 ST S 450.000 TP T 268.000 What is the value of the missing measurement (rounded off to the nearest 10 mm)? (A) 396.86 m (C) 396.05 m (B) 396.79 m (D) 396.94 m CE - 2013 10. A theodolite is set up at station A and a 3 m long staff is held vertically at station B. The depression angle reading at 2.5 m marking on the staff is . The horizontal distance between A and B is 2200 m. Height of instrument at station A is 1.1 m and R.L. of A is 880.88 m. Apply the curvature and refraction correction, and determine the R.L, of B ( in m)._______


CE - 2014 11. Group I lists tool/instrument while Group II lists the method of surveying. Match the tool/instrument with the corresponding method of surveying. Group I Group II P. Alidade 1. Chain surveying Q. Arrow 2. Levelling R. Bubble tude 3. Plain table surveying S. Stadia hair 4. Theodolite surveying (A) P – 3; Q – 2; R – 1; S – 4 (B) P – 2; Q – 4; R – 3; S – 1 (C) P – 1; Q – 2; R – 4; S – 3 (D) P – 3; Q – 1; R – 2; S – 4 12.

A tacheometer was placed at point P to estimate the horizontal distances PQ and PR. The corresponding stadia intercepts with the telescope kept horizontal, are 0.320 m and 0.210 m, respectively. The ∠ is me sured to be 30' 30". If the stadia multiplication constant = 100 and stadia addition constant = 0.10 m, the horizontal distance (in m) between the points Q and R is _________________ Q P R


[Ans. A] Let, length of OQ and QR are L1 and L2 respectively. From P1 coordinates of Q = (100+ cos sin ) From R, Coordinates of Q = (1000 L2 cos , 1000 L2 sin .) But, 100+ L1 cos = 1000 L2 cos . √

L1 +

√

√

2.

[Ans. B] Observed magnetic bearing of OE Local attraction at O = Correction magnetic bearing of

L2 = 900 ----------- ------- (i)

And 200+L1 sin L1 +

Solving (i) and (ii) we get L1 = 273.205m and L2 = 938.186m

= 1000 L2 sin

.

= 800 --------------------- (ii) th

th

th


GATE QUESTION BANK TM

True bearing of AB, ( ) of gnetic declin tion

MM

5.

E

[Ans. B] For a closed traverse sum of latitudes and departures should be zero respectively i.e. ∑ cos + 1000 cos + 907 cos 0 180 cos cos 0.10678 --- (i) ∑ =0 sin + 1000 sin + 907 sin sin sin 707.10678 ---(ii) Dividing (ii) by (i), we get t n

MB

True bearing of OE= Corrected magnetic bearing of OE+ Magnetic declination

3.

[Ans. C] ∆L = 400.75 + 100.25 199.0 300.0 =2 ∆D = 300.5 + 199.25 + 299.75* 200.5 = 2 [* This value is given as 399.75 in the original question paper which is incorrect]. rror of closure e

√(∆ )

(∆ )

= √( ) m t n

(

m 6.

[Ans. A]

)

Vertical axis

b

∆

Or

Diaphrag m

( )

focus

a

d

[Ans. C] N

(

)

(

B

) m

E

T.B

S

u

f

The horizontal distance (D) between the vertical axis and staff may be given as D=u+d f ut u ( ) s f i f ( )s f d i

45 =

W

D objective A s

∆

t n

4.


7.

[Ans. C]

MB

T.B = M.B ± Declination ( e st west)

th

th

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GATE QUESTION BANK

8.


[Ans. B] m

A

m

RL = 808.88 m

ye r

2200 m

True staff reading at station, m (t n ) now =237.701 m R.L. of station B = R.L. of Plane of Collimation – true staff reading –

True Bearing of AB, = Magnetic beaing of AB magnetic declination

Present magnetic bearing of AB = True bearing AB Magnetic declination

m 11.

[Ans. D]

12.

[Ans. *]Range 28.0 to 29.0

Q P

Whole circle bearing of

R 9.

10.

[Ans. B] In a closed loop travers, the algebraic sum of all the latitudes should be equal to zero i.e. ∑ cos cos cos cos cos cos m

s ( ) m ks = 100 (0.21) + 0.1 = 21.1 m Applying the cosine rule (

√

(

√

)( )(

) cos ) cos

m

[Ans. *] Range 641.9 to 642.3 R.L. of A = 880.88 m R.L. of Plane of Collimation = 880.88 + 1.2 = 882.08 m

th

th

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GATE QUESTION BANK


Leveling CE - 2006 1. A bench marck (BM) with Reduced Level (RL) = 155.305m has been established at the floor of a room. It is required to find out the RL of the underside of the roof (R) of the room using Spirit Leveling. The Back Sight (BS) to the BM has been observed as 1.500m wheareas the fore sight (FS) to R has been observed as 0.575m (Staff held inverted). The RL (m) of R will be (A) 155.880 (C) 157.380 (B) 156.230 (D) 157.860

CE - 2009 5. Consider the following statements: Assertion (A): Curvature correction must be applied when the sights are long. Reason (R): Line of collimation is not a level line but is tangential to the level line. Of these statements (A) both A and R are true and R is the correct explanation of A (B) both A and R are true but R is not a correct explanation of A (C) A is true but R is false (D) A is false but R is true

2.

CE - 2010 6. A bench mark has been established at the soffit of an ornamental arch at the known elevation of 100.0 m above mean sea level. The back sight used to establish height of instrument is an inverted staff reading of 2.105 m. A forward sight reading with normally held staff of 1.105 m is taken on a recently constructed plinth. The elevation of the plinth is (A) 103.210 m (C) 99.000 m (B) 101.000 m (D) 96.790 m

During a leveling work along a falling gradient using a Dumpy level and a staff of 3m length, following successive readings were taken. 1.785, 2.935, 0.360, 1.320. What will be the correct order of booking these four readings, in a level book?(BS: Beck sight, IS: Intermediate Sight, FS: Fore sight) (A) BS, FS, BS, FS (C) BS, IS, IS, FS (B) BS, IS, FS, FS (D) BS, IS, BS, FS

CE - 2007 3. The following measurements were made during testing a leveling instrument. Instrument at Staff reading At P1 Q1 P 2.800m 1.700m Q 2.700m 1.800m P1 is close to P and Q1 is close to Q. if the reduced level of station P is 100.000m, the reduced level of station Q is (A) 99.000m (C) 101.000m (B) 100.000m (D) 102.000m. CE - 2008 4. The type of surveying in which the curvature of the earth is taken into account is called (A) Geodetic surveying (B) Plane surveying (C) Preliminary surveying (D) Topographical surveying

CE - 2011 7. Curvature correction to a staff reading in a differential leveling survey is (A) Always subtractive (B) Always zero (C) Always additive (D) Dependent on latitude CE – 2012 8. The horizontal distance between two stations P and Q is 100 m. The vertical angles from P and Q to the top of a vertical tower at T are and above horizontal, respectively. The vertical angles from P and Q to the base of the tower are and below horizontal, respectively. Stations P, Q and the tower are in the same th

th

th


GATE QUESTION BANK


vertical plane with P and Q being on the s me side of neglecting e rth’s curvature and atmospheric refraction, the height (in m) of the tower is (A) 6.972 (C) 12.540 (B) 12.387 (D) 128.745 9.

Which of the following errors can be eliminated by reciprocal measurements in differential leveling? I. rror due to e rth’s curv ture II. Error due to atmospheric refraction (A) Both I and II (B) I only (C) II only (D) Neither I nor II

CE - 2014 10. A levelling is carried out to establish the Reduced Levels (RL) of point R with respect to the Bench Mark (BM) at P. The staff readings taken are given below. Staff P Q R Station BS 1.655 m m IS FS m 0.750 m RL 100.00m ? If RL of P is + 100.00 m, then RL (in m) of R is (A) 103.355 (C) 101.455 (B) 103.155 (D) 100.355

th

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GATE QUESTION BANK



2.

3.

[Ans. C] Height of instrument = RL of bench mark + BS = 155.305 + 1.500 = 156.805m Now, the staff is held inverted and the foresight (FS) is 0.575m of eight of instrument = 156.805+ 0.575 = 157.380m [Ans. A] The first reading is always BS. The second reading is foresight because the instrument has been shifted after this reading which is evident from the low value of third reading. Since the instrument has been shifted the third reading is backlight again. Successively the fourth reading is foresight.

Height of instrument, ( ) m lev tion of plinth eight of instrument m 7.

[Ans. A]

8.

[Ans. B]

X

(X )t n Xt n (X )t n Xt n (i) (ii) X(t n t n ) (X [t n t n ] X m eight of tower [t n t n ] = 12.387 m

[Ans. C] h h

(

)

(

)

h educed level of st tion educed level of h m 4.

9.

[Ans. A]

10.

[Ans. C]

[Ans. A] Geodetic surveying is that type of surveying in which the shape of the earth is taken into account. Geodetic surveying involves spherical trigonometry

5.

[Ans. A]

6.

[Ans. D] Height of instrument = BM + Back sight Since the staff is inverted, the back sight will be negative.

ore sight

(i) (ii) )

nd Staff P Q station BS 1.655 0.95 IS FS 1.5 RL HI 101.655 102.205 RL 100 103.155 RL of R = 101.455 m

th

th

th

R 0.75 101.455


GQB_CE.pdf

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