Physics and Mathematics

Chapter # 2

Physics & Mathematics

SOLVED EXAMPLES 1. Sol.

Two vectors having equal magnitudes A make an angle  with each other. Find the magnitude and direction of the resultant. The magnitude of the resultant will be B= =

A 2  A 2  2AA cos 

2A 2 (1  cos ) =

4 A 2 cos 2

 2

 2 The resultant will make an angle  with the first vector where = 2A cos

A sin  tan  = = A  A cos 

  2 A sin cos  2 2 = tan 2  2 2 A cos 2

 . 2 Thus, the resultant of two equal vectors bisects the angle between them. or,

2.

=

Two vectors of equal magnitude 5 unit have an angle 60º between them. Find the magnitude of (a) the sum of the vectors and (b) the difference of the vectors.

B 60º B

B A+

B

A

120º B A+

A

Sol.

    Figure shows the construction of the sum A  B and the difference A  B .     (a) A  B is the sum of A and B . Both have a magnitude of 5 unit and the angle between them is 60º. Thus the magnitude of the sum is   | A  B | 5 2  5 2  2  5  5 cos 60º = 2 × 5 cos 30º = 5 3 unit.       (b) A  B is the sum of A and (– B ). As shown in the figure, the anlge between A and (– B ) is 120º.   The magnitudes of both A and (– B ) is 5 unit. So,   | A  B | 5 2  5 2  2  5  5 cos120º = 2 × 5 cos 60º = 5 unit.

3. Sol.

A force of 10.5 acts on a particle along a direction making an angle of 37º with the vertical. Find the component of the force in the vertical direction. The component of the force in the vertical direction will be F1 = F cos  = (10.5 N) cos 37º) = (10.5 N)

4.

4 = 8.40 N 5

   The work done by a force F during a displacement r is given by F  r . Suppose a force of 12 N acts on a particle in vertically upward direction and the particle is displaced through 2.0 m in vertically downward direction. Find the work done by the force during this displacement. manishkumarphysics.in

Page # 1

Chapter # 2 Sol.

5.

Sol.


  The angle between the force F and the displacement r is 180º. Thus, the work done is  W = F  r = Fr cos  = (12 N) (2.0 m) (cos 180º) = –24 N – m = – 24 J.    The vector A has a magnitude of 5 unit, B has a magnitude of 6 unit and the cross product of A and    B has a magnitude of 15 unit. Find the angle between A and B .   If the angle between A and B is , the cross product will have a magnitude   | A × B | = AB sin  or, 15 = 5 × 6 sin  1 2  = 30º or, 150º,

sin  =

or, Thus,

6.

From the curve given in figure find

dy at x = 2, 6 and 10. dx

P

6-

D

42C –2

Sol.

2

1 0

2

4

6

8

10 12

14

The tangent to the curve at x = 2 is AC. It slope is tan1 =

AB 5 = BC 4

dy 5 = at x = 2 dx 4 The tangent to the curve at x = 6 is parallel to the X-axis.

Thus,

dy = tan  = 0 at x = 6. dx

Thus,

The tangent to the curve at x = 10 is DF. Its slope is tan 2 = dy 5 = – at x = 10. dx 4

Thus,

7. Sol.

dy if y = ex sin x dx y = ex sin x

Find

So

8.

DE 5 =– . EF 4

dy d d d = (ex sin x) = ex (sin x) + sin x (ex) dx dx dx dx ex cos x + ex sin x = ex (cos x + sin x).

The height reached in time t by a particle thrown upward with a speed u is given by

1 2 gt 2 where g = 9.8 m/s 2 is a constant. Find the time taken in reaching the maximum height. h = ut –

manishkumarphysics.in

Page # 2

Chapter # 2 Sol.


The height h is a function of time. Thus, h will be maximum when h = ut =

dh = 0. We have, dt

1 2 gt 2

dh d d 1 2  gt   (ut) – dt dt dt  2 

or,

=u

dt d 2 1 – g (t ) dt dt 2

1 g (2t) = u – gt. 2

=u– For maximum h,

dh =0 dt

or,

u – gt = 0

or,

t=

u . g

6

9.

Evaluate

 (2x

2

 3 x  5) dx .

3

 ( 2x

Sol.

=

2

 3 x  5) dx

 2x dx   3x dx   5 dx 2







= 2 x 2 dx  3 x dx  5 x 0 dx = 2

= 6

Thus,

 3

x3 x2 x 3 5 3 2 1

2 3 3 2 x  x  5x 3 2 6

2 3 3 2  (2x 2  3 x  5) dx =  x  x  5 x  2 3 3

2 2 (216 – 27) + (36 – 9)+ 5(6 – 3) 3 3 = 126 + 40.5 +15 = 181.5.

=

10. Sol.

11.

Round off the following numbers to three significant digits (a) 15462, (b) 14.745, (c) 14.750 and (d) 14.650 × 1012. (a) The third significant digit is 4. This digit is to be rounded. The digit next to it is 6 which is greater than 5. The third digit should, therefore, be increasesd by 1. The digits to be dropped should be replaced by zeroes because they appear to the left of the decimal. Thus, 15462 becomes 15500 on rounding to three significant digits. (b) The third significant digit in 14.745 is 7. The number next to it is less than 5. So 14.745 becomes 14.7 on rounding to three significant digits. (c) 14.750 will become 14.8 because the digit to be rounded is odd and the digit next to it is 5. (d) 14.650 × 1012 will become 14.6 × 1012 becausethe digit to be rounded is even and the digit next to it is 5.

Evaluate

25.2  1374 . All the digits in this expression are significant. 33.3


Page # 3

Chapter # 2 Sol.


25.2  1374 = 1039.7838... 33.3 Out of the three numbers given in the expression 25.2 and 33.3 have 3 significant digits and 1374 has four. The answer should have three significant digits, it becomes 1040. Thus, we write

We have

25.2  1374 = 1040. 33.3

Ex. Sol.

Evaluate 24.36 + 0.0623 + 256.2 24.36 0. 0623 256.2 –––––––––– 280.7 The sum is 280.7.

13.

The focal length of a concave mirror obtained by a student in repeated experiments are given below. Find the average focal lengthwith uncertainty in ±  limit. No. of observation 1 2 3 4 5 6 7 8 9 10

Sol.

 1 The average focal length f  10

focal length in cm 25.4 25.2 26.6 25.1 25.2 25.5 25.4 25.4 25.3 25.7 10

f

i

= 25.37 = 25.4

i 1

The calculation of  is shown in the table below :



i

fi cm

fi – cm

(f1 – )2 cm2

 ( f1  f ) 2 cm2

1 2 3 4 5 6 7 8 9 10

25.4 25.2 25.6 25.1 25.3 25.2 25.5 25.4 25.3 25.7

0.0 -0.2 0.2 0.3 -0.1 -0.2 0.1 0.0 -0.1 0.3

0.00 0.04 0.04 0.09 0.01 0.04 0.01 0.00 0.01 0.09

0.33

 1 ( f1  f ) 2  0.033 cm 2 = 0.18 cm. 10

= 0.2 cm. Thus, the focal length is likely to be within (25.4 ± 0.2 cm) and we write f = (25.4 ± 0.2) cm.


Page # 4

Chapter # 2


QUESTIONS

FOR

SHORT

ANSWER

1.

Is a vector necessarily changed if it is rotated through an angle?

2.

Is it possible to add two vectors of unequal magnitudes and get zero? It is possible to add three vectors of equal magnitudes and get zero?

3.

Does the phrase "direction of zero vector" have physical significance? Discuss in terms of velocity, force etc.

4.

Can you add three unit vectors to get a unit vector? Does your answer change if two unit vectors are along the co-ordinate axes?

5.

Can we have physical quantities having magnitude and direction which are not vectors?

6.

Which of the following two statements is more appropriate? (a) Two forces are added using triangle rule because force is a vector quantity. (b) Force is a vector quantity because two forces are added using triangle rule.

7.

Can you add two vectors representing physical quantities having different dimensions? Can you multiply two vectors representing physical quantities having different dimensions?

8.

Can a vector have zero component along a line and still have nonzero magnitude ?

9.

  Let 1 and 2 be the angles made by A and  A with the positive X-axis. Show that tan1 = tan2. Thus, giving  tan does not uniquely determine the direction of A .

10.

  Is the vector sum of the unit vectors i and j a unit vector? If no, can you multiply this sum by a scalar number to get a unit vector?

11.

    Let A  3î  4ˆj . Write four vectors B such that A  B but A = B.

12.

    Can you have A  B  A  B with A  0 and B  0? What if one of the two vectors is zero?

13.

      If A  B  0 , can you say that (a) A  B , (b) A  B ?

14.

         B Let A  5 i  4 j and B  7.5 i  6 j . Do we have B  k A ? Can we say   k ? A

Objective - I 1.

A vector is not changed if (A) it is rotated through an arbitary angle (C) it is cross multiplied by a unit vector

(B) it is multiplied by an arbitary scalar (D*) it is slid parallel to itself

dksbZz lfn'k ifjofrZr ugha gksrk gS] ;fn & (A) bldks fdlh LoSfPNd dks.k ls ?kqek;k tk;sA (B) bldks fdlh LoSfPNd vfn'k jkf'k ls xq.kk fd;k tk;sA (C) bldk fdlh ,dkad lfn'k ls otz&xq.ku fd;k tk;sA (D) bldks Lo;a ds lekukUrj foLFkkfir fd;k tk;sA 2.

Which of the sets given below may represent the magnitudes of three vectors adding to zero ? (A) 2, 4, 8 (B) 4, 8, 16 (C*) 1, 2, 1 (D) 0.5, 1, 2

fuEu esa ls dkSulk leqP; O;Dr djrk gS fd blds rhu lfn'kksa ds ;ksx dk ifj.kkeh 'kwU; izkIr gksxk & (A) 2, 4, 8

(B) 4, 8, 16

(C) 1, 2, 1


(D) 0.5, 1, 2 Page # 5

Chapter # 2 3.

(A)  <    A rFkk B (A)  <  4.

5.


    The result of A and B makes an angle  with A and  with B , dk ifj.kkeh]

(C*)  < if A > B

(D)  < if A = B

ls  dks.k cukrk gS & (C)  <  ;fn A > B

The component of a vector is (A) always less than its magnitude (C) always equal to its magnitude

(D)  >  ;fn A = B

(B) always greater than its magnitude (D*) none of these

fdlh lfn'k dk ?kVd & (A) lnSo blds ifjek.k ls de gksrk gS (B) lnSo blds ifjek.k ls vf/kd gksrk gS (C) lnSo blds ifjek.k ds cjkcj gksrk gS (D) buesa ls dksbZ ugha     A vector A points vertically upward and B points towards north. The vector product A x B is (A*) along west  lfn'k A m/okZ/kj Åij (A) if'pe dh vksj

6.

(B)  < if A < B   A ls  dks.k rFkk B (B)  <  ;fn A < B

(B) along east

dh vksj bafxr gS rFkk (B) iwoZ dh vksj

(C) zero

 B

mÙkj dh vksjA lfn'k xq.ku (C) 'kwU;

(D) vertically downward   A x B gS & (D) m/okz/Z kj

The radius of a circle is stated as 2.12 cm. Its area should be written as (A) 14 cm2 (B*) 14.1 cm2 (C) 14.11 cm2 fdlh o`r dh f=kT;k 2.12 lseh crkbZ tkrh gSA bldk {ks=kQy n'kkZ;k tk;sxk & (A) 14 lseh2 (B) 14.1 lseh2 (C) 14.11 lseh2

Åij dh vksj

(D) 14.1124 cm2 (D) 14.1124 lseh2

Objective - II 1.

2.

A situation may be described by using different sets of co-ordinate axes having different orientations. Which of the following do not depend on the orientation of the axes ? (A*) the value of a scalar (B) component of a vector (C*) a vector (D*) the magnitude of a vector

ifjfLFkfr funs'Z kkad v{kksa ds fofHkUu >qdko okys dbZ funs'Z k v{kksa ds fofHkUu leqP;ksa }kjk O;Dr dh tkrh gSA fuEu esa ls dkSulh v{kksa ds >qdko ij fuHkZj ugha djrh gS & (A) fdlh vfn'k dk eku (B) fdlh lfn'k dk ?kVd (C) dksbZ lfn'k (D) fdlh lfn'k dk ifjek.k    Let C = A + B       (A) | C | is always greater than | A | (B*) It is possible to have | C | < | A | and | C |< | B | (C) C is always equal to A + B    ekukfd C  A  B   (A) | C | lnSo | A | ls cM+k gksxk

3.

(D) C is never equal to A + B (B) ;g









lEHko gS fd | C |  | A | rFkk | C |  | B | gksA (C) C lnSo A + B ds rqY; gksxk (D) C dHkh Hkh A + B ds cjkcj ugha gksxk    Let the angle between two nonzero vectors A and B be 120o and its resultant be C . (A) C must be equal to | A - B| (B) C must be less than |A - B| (C*) C must be greater than | A - B | (D) C may be equal to | A - B |    ekukfd nks v'kwU; lfn'kksa A rFkk B ds e/; dks.k 120° gS rFkk bldk ifj.kkeh C gS & (A) C fuf'pr :i ls |A – B| ds cjkcj gksxk (B) C fuf'pr :i ls |A – B| ls de (C) C fuf'pr

4.

:i ls |A – B| ls vf/kd gksxk

gksxk (D) C, |A – B| ds cjkcj gks ldrk gSA

The x-component of the resultant of several vectors (A*) is equal to the sum of the x-component of the vectors (B*) may be smaller than the sum of the magnitudes of the vectors (C) may be greater than the sum of the magnitudes of the vectors (D*) may be equal to the sum of the magnitudes of the vectors. dqN lfn'kksa ds ifj.kkeh dk x-?kVd & (A) lfn'kksa ds x-funsZ'kkadksa ds ;ksx ds rqY; gksxk (B) lfn'kksa ds ;ksx (C) lfn'kksa

ds ;ksx ds ifjek.k ds cjkcj gks ldrk gS

ds ifjek.k ds ;ksx ls de gks ldrk gS (D) lfn'kksa ds ;ksx ds ifjek.k ds cjkcj gks ldrk gSA


Page # 6

Chapter # 2




5.



The magnitudes of the vectors product of two vectors A and B may be (A) greater than AB (B*) equal to AB (C*) less than AB

(D*) equal to zero.

(A) AB ls

(D) 'kwU;

  nks lfn'kksa A rFkk B ds lfn'k xq.ku dk ifjek.k gks ldrk gS &

vf/kd

(B) AB ds

cjkcj

(C) AB ls

de

ds cjkcj

WORKED OUT EXAMPLES 1. Sol.

A vector has component along the X-axis equal to 25unit and along the Y-axis equal to 60 unit. Find the magnitude and direction of the vector. The given vector is the resultant of two perpendicular vectors, one along the X-axis of magnitude 25 unit and the other along the Y-axis of magnitude 60 unit. The resultant has a magnitude A given by

A  (25 )2  (60 )2  2  25  60 cos 90 º =

(25 )2  (60 )2 = 65.

The angle  between this vector and the X-axis is given by tan  = 2.

60 . 25

Find the resulatant of the three vectors shown in figure Y

2.0m m 5.0

3.0m X

Sol.

Take the axes as shown in the figure, The x-component of the 5.0 vector = 5.0 m cos 37º = 4.0 m, the x-component of the 3.0 m vector = 3.0 m and the x-component of the 2.0 m vector = 2.0 m cos 90º = 0. Hence, the x-component of the resultant = 4.0 m + 3.0 m + 0 = 7.0 m. The y-component of the 5.0 m vector = 5.0 m sin 37º = 3.0 m, the y-component of the 3.0 m + 0 = 7.0 m. The y-component of the 3.0 vector = 0 and the y-component of the resultant = 3.0 m + 0 + 2.0 m = 5.0 m. The magnitude of the resultant vector =

(7.0 m)  (5.0 m)2

= 8.6 m. If the angle made by the resultant with the X-axis is , then tan  =

y - component 5 .0 = or, q = 35.5º, x  component 7 .0


Page # 7

Chapter # 2 3. Sol.


The sum of the three vectors shown in figure is zero. Find the magnitudes of the vectors OB and OC . Take the axes as shown in the figure. Y

C

45º X

0

B 5m

A

The x-component of OA = (OA) cos 90º = 0. The x-component of OB = (OB) cos 0º = 0.

The y-component of OC = (OC) cos 135º = –

1

OC.

2

Hence, the x-component of the resultant = OB –

1

OC. ....(i) 2 It is given that the resultant is zero and hence is x-component is also zero. From (i), OB =

1

OC.

....(ii)

2 The y-component of OA = OA cos 180º = – OA. The y-component of OB = OB cos 90º = 0.

1

The y-component of OC = OC cos 45º =

2

OC.

Hence, the y-component of the resultant

1

=

OC – OA .....(iii) 2 As the resultant is zero, so is its y-component. From (iiii),

1 2 From (ii), OB =

4.

1 2

OC = OA, or OC =

2 OA = 5 2 m.

OC = 5 m.

The magnitude of vectors OA , OB and OC in figure are equal. Find the direction of OA + OB – OC , Y

C

A 45º

30º 0

X

60º 5m B

Sol.

Let OA = OB = OC = F. manishkumarphysics.in

Page # 8

Chapter # 2


3 . 2

x-component of OA = F cos 30º = F

F 2

x-component of OB = F cos 60º =

F x-component of OC = F cos 135º = –

2

x-component of OA + OB – OC

F 3  F  F         =    2  2  2  =

F ( 3 +1+ 2

2 ).

y-component of OA = F cos 30º =

F . 2

y-component of OB = F cos 150º =

F 3 2 F

y-component of OC = F cos 45º = –

2

y-component of OA + OB – OC

 F   F 3   F   =  2     2         2 =

F (1 – 2

3 –

2 ).

Angle of OA + OB – OC with the X-axis

= tan

5.

1

F (1  3  2 ) (1  3  2 ) 2 = tan–1 (1  3  2 ) F (1  3  2 ) 2

Find the resultant of the free vectors OA , OB and OC shown in figure. Radius of the circle is R. C B 45º 45º 0

Sol.

A

OA = OC.

OA + OC is along OB (bisector) and its magnitude is 2R cos 45º = R/2. ( OA + OC ) + OB is along OB and its magnitude is R 2 + R = R(1 +

2 ).


Page # 9

Chapter # 2 6.


The resultant of vectors OA and OB is perpendicular to OA (figure). Find the angle AOB. Y B 6m

 0

Sol.

4m

X A

Take the dotted lines as X, Y axes, x-component of OA = 4 m, x-component of

OB = 6m cos  x-component of the resultant = (4 + 6 cos ) m But it is given that the resultant is along Y-axis. Thus, the x-component of the resultant = 0 4 + 6 cos  = 0

7. Sol.

 + 30 i

Sol.

9. Sol.

cos = –

2 3

    Write the unit vector in the direction of A = 5 i + j – 2 k .  | A | = 5 2  12  ( 2)2 = 30 .  A The required unit vector is  |A| 5

8.

or ,

1 30

 j –

2

 . 30 k

      If | a + b | = | a – b | show that a  b .       We have | a + b |2 = ( a + b ).( a – b )         = a .a + a .b +b .a + b .b   = a2 + b2 = 2 a . b . Similarly,       | a – b |2 = ( a – b ).( a – b )   = a2 + b2 – 2 a . b     If |a + b |= |a – b | ,     a2 + b2 + 2 a . b = a2 + b2 – 2 a . b   or , a .b = 0   or , a  b.

          If a = 2 i + 3 j + 4 k and b = 4 i + 3 j + 2 k , find the angle beween a and b .   We have a . b = ab cos  a.b or , cos = ab   where  is the angle between a and b .   Now a . b = axbx + ayby + azbz = 2 × 4 + 3 × 3 + 4 × 2 = 25. Also

a= =

2

2

ax  ay  az

4  9  16 =

2

29


Page # 10

Chapter # 2


and

b=

Thus ,

cos =

Sol.

2

2

=

16  9  4 =

29 .

25 29

 25  .  = cos–1   29 

or ,

10.

2

bx  by  bz

             If A = 2 i – j + 7 k , B = i + 2 k and C = j – k , find A .( B × C ).       B × C = ( i + 2k ) × ( j – k )       = i × ( j – k ) + 2k × ( j – k )         = i × j – i × k + 2k × j – 2k × k       = k + j –2i –0=–2i + j + k          A .( B × C ) = (2 i – 3 j + 7 k ) . (– 2 i + j + k ) = (2) (–2) + (–3)(1) + (7)(1) = 0.

11.

The volume of a sphere is given by 4 R3 3 where R is radius of the sphere. (a) Find the rate of change of volume with respect to R. (b) Find the change in volume of the sphere as the radius is increased from 20.0 cm to 20.1 cm . Assume that the rate does not appreciably change between R = 20.0 cm to R = 20.1 cm.

V=

Sol.

(a)

V=

4 R3 3

dV 4 d 4 =  (R)3 = . 3R2 = 4R2. dR 3 dr 3 (b) At R = 20 cm , the rate of change of volume with the radius is

or ,

dV = 4R2 = 4 (400 cm2) dR = 1600  cm2. The change in volume as the radius changes from 20.0 cm to 20.1 cm is dV R dR = (1600  cm2) (0.1 cm) = 160  cm3 .

V =

12.

Find the derivative of the following functions with respect to x (a) y = x2 sinx , (b) y =

Sol.

(a) y = x2 sinx

sin x and (c) y = sin (x)2. x

dy d d 2 = x2 (sinx) + (sinx) (x ) ] dx dx dx = x2 cosx + (sinx) (2x) = x (2 sinx + x cosx).

(b) y =

sin x x

d  dx  x (sin x )  sin x  dy dx  dx  = dx 2 x manishkumarphysics.in

Page # 11

Chapter # 2

Physics & Mathematics =

x2

.

d dy d( x 2 ) 2 = (sinx ). dx dx 2 dx 2 = cosx (2x) = 2x cosx2.

(b)

13.

x cos x  sin x

Find the maximum or minimum values of the function y = x +

Sol.

y=x+

1 for x > 0. x

1 x

dy d d –1 = (x) + (x ) dx dx dx = 1 + (– x–2) 1 . x2 For y to be maximum or minnimum ,

=1–

dy =0 dx

or , Thus

1–

1

=0 x2 x = 1 or – 1.

For x > 0 the only possible maximum or minimum is at x = 1 , At x = 1 , y = x +

1 = 2. x

1 is very large because of the term x. Thus , x = 1 must correspond to a minimum. Thus, x y has only a minimum for x > 0. This minimum ocurs at x = 1 and the minimum value of y is y = 2. Near x = 0 , y = x +

14.

Figure shows the curve y = x2. Find the area of the shaded part between x = 0 and x = 6.

x

x Sol.

The area can be divided into strips by drawing ordinates between x = 0 and x = 6 at a regular interval of dx. Consider the strip between the ordinates at x and x + dx. The height of this strip is y = x2. The area of this strip is dA = ydx = x2 dx. The total area of the shaded part is obtained by summing up these strips -areas x varying from 0 to 6. Thus, 6

A=

 x dx 2

0

6

 x3  216  0 = 3  = = 72 . 3   0 t

15.

Evaluate

 A sin t dt where A and  are constants. 0

t

Sol.

 A sin t dt 0


Page # 12

Chapter # 2

Physics & Mathematics t

  cos t  A =A  = (1 – cost).   0  

16.

The velocity v and displacement x of a particle executing simple harmonic motion are related as

Sol.

dv = – 2 x. dx At x = 0 , v = v0 . Find the velocity v when the displacemet becomes x. We have

v

dv = – 2 x dx v dv = – 2 x dx

v

or ,

v

or ,



v0

x

v dv

=

 

2

x dx

.....(i)

0

When summation is made on – w2 x dx the quantity to be varied is x. When summation is made on vdv the quantity to be varied is v. As x varies from 0 to x the velocity varies from v0 to v. Therefore , on the left the limits of integration are from v0 to v and on the right they are from 0 to x. Simplifying (i) , v

v

1 2 1 2  2 v  = – 2  2 v    v0   v0 or , or ,

1 2 2 x2 (v –v0 ) = – 2 2 2 v2 = v02 – 2 x2

or ,

v 0 2  2 x 2 .

v=

17.

The charge flown through a circuit in the time interval between t and t + dt is given by dq = e– t /  dt where  is a constant. Find the total charge flown through the circuit between t = 0 to t = .

Sol.

The total charge flown is the sum of all the dq's for t varying from t = 0 to t = t0 . Thus , the total charge flown is 

Q=

e

t / 

dt

0



 et /   1  =   1/   =   1   .  e   0 18. Sol.

Evaluate (21.6002 + 234 + 2732.10) × 13.

21.6002 234 2732.10

22 234 2732 2988

The three numbers are arranged with their decimal points aligned (shown on the left part above). The column just left to the decimal has 4 as the doubtful digit. Thus , all the numbers are rounded to this column. The rounded numbers are shown on the rigid part above. The required expression is 2988 × 13 = 38844. As 13 has only two significant digits the product should be rounded off after two signigicant digits. Thus the result is 39000.


Page # 13

Chapter # 2


EXERCISE

1.

  A vector A makes an angle of 20º and B makes an angle of 110º with the X-axis. The magnitudes of these vectors are 3 m and 4m respectively. Find the resultant.



x-v{k



ls ,d lfn'k A , 20° dks.k cukrk gS rFkk nwljk lfn'k B , 110° dks.k cukrk gSA bu lfn'kksa ds ifjek.k Øe'k% 3 o 4 gSA ifj.kkeh Kkr dhft;sA 2.

  Let A and B be the two vectors of magnitude of 100 unit and inclined to the X-axis at angles 30º and 60º respectively, find the resultant.   ekukfd nks lfn'kksa A o B ds izR;sd ifjek.k 10 bdkbZ gSA ;fn budk x-v{k ls >qdko Øe'k% 30° rFkk 60° gS] ifj.kkeh

Kkr dhft;sA 3.

4.

5.

   Add vectors A , B and C each, having magnitude of 100 unit and inclined to the X-aixs at angles 45°, 135° and 315° respectively.    A , B rFkk C izR;sd dk ifjek.k 100 bdkbZ rFkk buds x-v{k ls >qdko Øe'k% 45°, 135° rFkk 315° gS] budk ;ksx dhft;sA

  Let a  4 î  3ˆj and b  3 î  4ˆj (a) Find the magnitudes of   ekukfd a  4î  3ˆj rFkk b  3î  4ˆj gSA Kkr dhft;s &     (A) a (B) b (C) a  b

  (D) a  b 

Refer to figure. Find (a) the magnitude, be (b) x and y components and (c) the angle with the resultant of OA , 



BC anda DE .

fp=k ds vk/kkj ij Kkr dhft;s & (a) ifjek.k (b) x rFkk y ?kVd 





(c) OA , BC ,oa DE 6.

ds ifj.kkeh dk x-v{k ls dks.k

Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle between them if the magnitude of the resultant is (a)1 unit, (b) 5 unit and (c) 7 unit

nks lfn'kksa ds ifjek.k Øe'k% 3 bdkbZ o 4 bdkbZ gSA buds e/; dks.k fdruk gksxk ;fn budk ifj.kkeh gS & (a) 1 bdkbZ (b) 5 bdkbZ (c) 7 bdkbZ 7.

A spy report about a suspected car reads as follows. "The car moved 2.00 km towards east, made a perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped". Find the displacement of the car.

,d tklwl lwpuk nsrk gS fd langs okyh dkj dh xfrfof/k bl izdkj gS] dkj iwoZ dh vksj 2.00 fdeh- pyh] ck;ha vksj ,d yEcor~ ?kqeko fy;k] 500 eh- nkSM+h] nk;ha vksj ledks.k ij ?kqeh] 4.00 fdeh pyh rFkk :d x;hA dkj dk foLFkkiu Kkr dhft;sA 8.

A carrom board (4 ft × 4ft square) has the queen at the centre. The queen, hit by the striker moves to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen (a) from the centre to the front edge, (b) from the front edge to the hole and (c) from the centre to the hole, ,d dsjecksMZ (4 QqV x 4 QqV oxkZdkj) esa dsUnz ij Dohu gSA LVªkbdj ls pksV yxus ij] Dohu lkeus okys fdukjs dh vksj

xfr'khy gksrh gS] iyVrh gS rFkk LVªkbd js[kk ds ihNs fLFkr Nsn esa pyh tkrh gSA Dohu ds foLFkkiu dk ifjek.k Kkr dhft;sA (a) dsUnz fcUnq ls lkeus okys fdukjs rd (b) lkeus okys fdukjs ls Nsn rd rFkk manishkumarphysics.in

Page # 14

Chapter # 2 (c) dsUnz

ls Nsn rd

2 4  x 2x 2 – x = 2x 3x = 2

x 2-x S1

2 x= 3

 2 ˆ + 2 ˆj S1 = 3 i |S1| =

>

S3

S2

>

>

Sol.


4

4

4 2 4 = 10 Ans. 9 3

 2 | S 2 | = (2– ) î – 4 ˆj 3 =

4 ˆ – 4 ˆj 3 i



4 10 3

   2 4 | S 3 | = S1  S 2 = î  2ˆj  î – 4 ˆj 3 3  | S 3 | = 2 î – 2 ˆj  | S3 | = 2 2 9.

A mosquito net over a 7 ft × 4 ft bed is 3 ft high. The net has a hole at one corner of the bed through which a mosquito enters the net. It flies and sits at the diagonally opposite upper corner of the net.(a) Find the magnitude of the displacement of the mosquito. (b) Taking the hole as the origin, the length of the bed as the X-axis, its width as the Y-axis, and vertically up as the Z-axis, write the components of the displacement vector. 7 QqV x 4 QqV vkdkj okys iyax ij 3 QqV Å¡ph ePNjnkuh yxh gqbZ gSA iyax ds ,d dksus ij ePNj nkuh esa ,d Nsn

gS] ftlls ,d ePNj] tkyh esa izfo"V gks tkrk gSa ;g mM+rk gS rFkk tkyh ds fod.kZr% foijhr dksus ij cSB tkrk gSA (a) ePNj ds foLFkkiu dk ifjek.k Kkr dhft;sA (b) Nsn dks ewy fcUnq] iyax dh yEckbZ dks x-v{k rFkk pkSMk+ bZ dks y-v{k ,oa m/oZ/kj Åij dh vksj z-v{k ekurs gq,] foLFkkiu lfn'k ds ?kVd fyf[k;sA 10.

11.

   Suppose a is a vector of magnitude 4.5 unit due north. What is the vector (a) 3 a , (b) –4 a ? ekukfd lfn'k a dk ifjek.k 4.5 bdkbZ] mÙkj dh vksj gSA lfn'k D;k gksxsa \   (A) 3a (B) 4a Two vectors have magnitudes 2m and 3m. The angle between them is 60º. Find (a) the scalar product of the two vectors, (b) the magnitude of their vector product.

nks lfn'kksa ds ifjek.k 2 eh- o 3 eh- gSA buds e/; dks.k 60° gSA Kkr dhft;sA (a) nksuksa lfn'kksa dk vfn'k xq.ku (B) buds lfn'k xq.ku dk ifjek.k 12.

Let A1, A2, A3, A4, A5, A6, A1 be a regular hexagon. Write the x-components of the vectors represented by the six sides taken in order. Use the fact that the resultant of these six vectors is zero, to prove that cos 0 + cos  2 3 4 5 + cos + cos + cos + cos = 0 . Use the known cosine values to verify the result. 3 3 3 3 3

ekukfd A1, A2, A3, A4, A5, A6, A1 ,d fu;fer "kV~Hkqt gSA bldh N% Hkqtkvksa dks Øe esa j[krs gq,] Hkqtkvksa ls O;Dr lfn'kksa ds x-?kVd fyf[k;sA ;g rF; iz;D q r djrs gq, bu N% lfn'kksa dk ifj.kkeh 'kwU; gS] fl) dhft;s fd & cos 0 + cos

 2 3 4 5 + cos + cos + cos + cos =0 3 3 3 3 3


Page # 15

Chapter # 2


ifj.kke dks dksT;kvksa ds eku dk mi;ksx djds lqfuf'pr dhft;sA 13.

14.

15.

16.

17.

  Let a  2î  3ˆj  4kˆ and b  3î  4ˆj  5kˆ . Find the angle between them.   ekukfd a  2î  3ˆj  4kˆ rFkk b  3î  4ˆj  5kˆ buds e/; dks.k Kkr dhft;sA    Prove that A . ( A x B)  0    fl) dhft;s fd A . ( A x B)  0     If A  2 î  3ˆj  4kˆ and B  4 î  3ˆj  2kˆ , find A x B .     ;fn A  2 î  3ˆj  4kˆ rFkk B  4 î  3ˆj  2kˆ Kkr dhft;s A x B       If A, B, C are mutually perpendicular, show that C x ( A x B)  0 . Is the converse true?       ;fn A, B, C ijLij yEcor~ gS] O;Dr dhft;s fd C x ( A x B)  0 D;k bldk O;qRØe Hkh lR;

gS \

A particle moves on a given straight line with a constant speed u. At a certain time it is at a point P on its   straight line path. O is a fixed point. Show that OP x v is independent of the position?

,d d.k nh xbZ ljy js[kk ds vuqfn'k fu;r pky v ls xfr'khy gSA fdlh fuf'pr le; t ij ;g ljy js[kh; iFk ds fcUnq P ij 18.





gSA O ,d fLFkj fcUnq gSA O;Dr dhft;s fd OP x v , P dh fLFkfr ij fuHkZj ugha djrk gSA

     The force on a charged particle due to electric and magnetic fields is given by F  qE  qv x B . Suppose E  is along the X-axis and B along the Y-axis. In what direction and with what minimum speed u should a positively charged particle be sent so that the net force on it is zero?      fdlh vkosf'kr d.k j fo|qr ,oa pqEcdh; {ks=kksa ds dkj.k cy] F  qE  qv x B }kjk O;Dr fd;k tkrk gSA ekukfd E ] x v{k ds vuqfn'k gS rFkk B , y-v{k ds vuqfn'k gSA ,d /kukosf'kd d.k dks fdl fn'kk esa rFkk fdl U;wure pky v ls xfr djokbZ

tk;s fd bl j ifj.kkeh cy 'kwU; gks \ 19.

20.

      Give an example for which A . B  C . B but A  C       A . B  C . B fdUrq A  C dk ,d mnkgj.k nhft;sA Draw a graph from the following data. Draw tangents at x = 2, 4, 6 and 8. Find the slopes of these tangents. Verify that the curve drawn is y = 2x2 and the slope of tangent is tan =

dy = 4x. dx

fuEufyf[kr vkadM+ksa dk mi;ksx djds ys[kkfp=k cukb;sA x = 2, 4, 6 rFkk 8 ij Li'kZ js[kk,a [khaph;saA bu Li'kZ js[kkvksa ds
21.

1 2

2 8

3 18

4 32

5 50

6 72

A curve is represented by y = sin x. If x is changed from ,d

7 98

8 128

oØ y = sin x

9 162

dy = 4x gSA dx

10 200

   to + , find approximately 3 3 100

the change in y


Page # 16

Chapter # 2


}kjk O;Dr fd;k tkrk gSA ;fn x dk eku ifjofrZr djds

 3

ls

  3 100

dj fn;k tkrk gS] y ds eku esa vuqekfur ifjorZu

Kkr dhft;sA 22.

The electric current in a charging R-C circuit is given by i = io–t/RC where io, R and C are constant parameters of the circuit and t is time. Find the rate of change of current in at (a) t = 0, (b) t = RC, (c) t = 10 RC. R-C ifjiFk esa vkos'ku ds le; /kkjk i = i0e–t/RC }kjk O;Dr dh tkrh gS] tgk¡ i0, R rFkk C ifjiFk dh fu;r jkf'k;k¡ gS rFkk t le;

23.

gSA le; (a) t = 0, (b) t = RC, (c) t = 10 RC ij ifjiFk esa /kkjk ds eku esa ifjorZu dh nj Kkr dhft;sA

The electric current in a discharging R-C circuit is given by R-C i = i0 e–t/RC where io, R and C are constant parameters and t is time. Let io = 2.00 A, R = 6.00 × 105  and C = 0.500 F. (a) Find the current at t = 0.3s. (b) Find the rate of change of current at t = 0.3 s. (c) Find approximately the current at t = 0.31 s.

ifjiFk ds fy, foltZu dky esa fo|qr /kkjk i = i0 e–t/RC }kjk O;Dr dh tkrh gS] tgka i0, R rFkk C ifjiFk dh fu;r jkf'k;ka gS rFkk t le; gSA ekukfd i0 = 2.00 A, R = 6.00 x 105  rFkk C = 0.500 µF (a) t = 0.3 ls ij /kkjk Kkr dhft;sA (b) t = 0.3 ls ij /kkjk esa ifjorZu dh nj Kkr dhft;sA (c) t = 0.31 ls- ij /kkjk dk eku ¼yxHkx½ Kkr dhft;sA 24.

Find the area bounded under the curve y = 3x2 + 6x + 7 and the X-axis with the ordinates at x = 5 and x = 10. oØ y = 3x2 + 6x + 7 rFkk funsZ'kkadksa x = 5 rFkk x = 10 ds chp x-v{k }kjk ifjc) {ks=kQy Kkr dhft;sA

25.

Find the area enclosed by the curve y = sin x and the X-axis between x = 0 and x = .

oØ y = sin x rFkk x = 0 ,oa x =  ds chp x-v{k }kjk ifjc) {ks=kQy Kkr dhft;sA 26.

Find the area bounded by the curve y = e–x , the X-axis and the Y-axis oØ y = e–x, x-v{k rFkk y-v{k }kjk ifjc) {ks=kQy Kkr dhft;sA

27.

A rod of length L is placed along the X-axis between x = 0 and x = L. The linear density (mass/length) p of the rod varies with the distance x from the origin as p = a + bx. (a) Find the SI units of a and b. (b) Find the mass of the rod in terms of a,b,and L. L yEckbZ dh ,d NM+ x-v{k ds vuqfn'k x = 0 rFkk x = L ds e/; j[kh gqbZ gSA NM+ dk jSf[kd ?kuRo ¼nzO;eku@yEckbZ½ P

ewy fcUnq ls nwjh x ds lkFk P = a + bx ds vuqlkj ifjofrZr gksrk gSA (a) a rFkk b dh SI bdkbZ;ka Kkr dhft;sA (b) a, b rFkk L ds inksa esa NM+ dk nzO;eku Kkr dhft;sA 28.

The momentum p of a particle changes with time t according to the relation

dp = (10N) + (2N/s)t. If the dt

momentum is zero at t = 0, what will the momentum be at t = 10s ?

fdlh d.k dk laoxs ] le; ds lkis{k lw=k

dp = (10 U;wVu) + (2 U;wVu@ls-)t ds dt

vuqlkj ifjofrZr gksrk gSA ;fn t ds vuqlkj

ifjofrZr gksrk gSA ;fn t = 0 ij laosx 'kwU; gks] rks t = 10 ls- ij laosx fdruk gksxk \ 29.

The changes in a function y and the independent variable x are related as dy/dx = x2. Find y as a function of x. Qyu y esa ifjorZu rFkk Lora=k pj x esa lEcU/k dy/dx = x2 gSA x ds Qyu ds :i esa y Kkr dhft;sA

30.

Write the number of significant digits in

lkFkZd la[;k,a fyf[k;s & (a) 1001

(b) 100.1

(c) 100.10


(d) 0.001001

Page # 17

Chapter # 2 31.


A metre scale is graduated at every millimetre. How many significant digits will be there in a length measurement with this scale?

,d ehVj iSekuk izR;sd feyhehVj esa foHkkftr gSA bl iSekus }kjk ekih xbZ yEckbZ esa fdruh lkFkZd la[;k,a gS \ 32.

33.

Round the following numbers to 2 significant digits. fuEufyf[kr la[;kvksa dks 2 lkFkZd vadksa rd lhfer dhft;s & (a) 3472 (b) 84.16 (c) 2.55

(d) 28.5

The length and the radius of a cylinder measured with a slide callipers are found to be 4.54 cm and 1.75 cm respectively. Calculate the volume of the cylinder. LykbM dSfyijlZ dh lgk;rk ls ,d csyu dh yEckbZ rFkk f=kT;k ekih tkrh gS] buds eku Øe'k% 4.54 lseh ,oa 1.75 lseh

izkIr gksrs gSA csyu dh vk;ru dh x.kuk dhft;sA 34.

The thickness of a glass plate is measured to be 2.17 mm, 2.17 mm and 2.18 mm at three different places. Find the average thickness of the plate from this data. ,d dkap dh ifV~Vdk dh rhu fHkUu&fHkUu LFkkuksa ij ekih x;h eksVkbZ;ka 2.17 feeh- 2.17 feeh rFkk 2.18 feeh gSA bu vkadM+kas

dh lgk;rk ls ifV~Vdk dh vkSlr eksVkbZ Kkr dhft;sA 35.

The length of the string of a simple pendulum is measured with a metre scale to be 90.0 cm. The radius be 2.13 cm using measurements with a slide callipers. What is the effective length of the pendulum? (The effective length is length of the pendulum between the point of suspension and the centre of the bob.),d ehVj

iSekus dh lgk;rk ls ljy yksyd dh Mksjh dh yEckbZ 90.0 lseh ekih tkrh gSA LykbM dSfyilZ ls ekius ij ckWc ,oa gqd dh lfEefyr yEckbZ 2.13 lseh izkIr gksrh gSA yksyd dh izHkkoh yEckbZ fdruh gS \ ¼fuyEcu fcUnq rFkk ckWc ds dsUnz ds chp dh nwjh izHkkoh yEckbZ dgykrh gS½


Page # 18

Physics and Mathematics

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