Physics for You - June 2013 (Gnv64)

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PHYSICSVol. XXI No. 5 May 2013

Corporate Office:

Plot 99, Sector 44 Institutional area,

Gurgaon -122 003 (HR),Tel: 0124-4951200

Regd. Office

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Ring Road, New Delhi - 110029.

e-mai l : [email protected] website : www.mtg. in

Managing Editor : Mahabir Singh

Editor : Anil Ahlawat (BE, MBA)

Contents

JEE Advanced Practice Paper: 2013 6

Thought Provoking Problems 16

(Kinematics)

JEE Main Solved Paper: 2013 21

NCERT Xtract Questions for NEET 29

Target PMTs Practice Questions 37

Brain Map 48

Examiner's Mind 50

Challenging Problems 52

Exam prep : MCQs for Practise 58

AIIMS Practice Paper: 2013 69

Test Your Physics Apti tude 86

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New Delhi - 29 and printed by Personal Graphics and Advertisers (P) Ltd., Okhla

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enquiries before acting upon any advertisements published in this magazine. Focus/

Infocus feature s are marketing incen tives MTG does not vouch or subscri be to the claim s

and representations made by advertisers. All disputes are subject to Delhi jurisdiction

only.

Editor : Anil Ahlawat

Copyright© MTG Learning Media (P) Ltd.

All rights reserved. Reproduction in any form is prohibited.

edit Q rialExpanding Knowledge and In-depth

Research are not contradictory

Widening knowledge and intensive research are not contradictory but

complementary to each other. Taking examples from our "plus two"

education, one starts from Newton's Laws for our case study. One studies

the first law as if it is resting on a single pillar. But one does not understand

why something that is moving should continue to move in the same way.

But this is the first step. When teaching, we should also sow seeds of doubt

why it should be like that.

This prepares the ground for further expansion. The second law is a further

advance and final ly the third law. When one is prepared to attack problems,

supplementary concepts are thrown in with experimental studies. With

ideal strings and ideal pulleys one learns more.

Each part is a grid. First one masters the grid and then the partitionsdisappear to become a bigger grid. Still the concepts from electricity,

magnetism and other subjects look different. The method of the grid system

is continued. In the research method, a small grid is studied in depth. With

every available knowledge from every other field, one goes on attaching

the problem from various angles. This is also a grid problem but one tries to

deepen the grid or go deep into the problem with every skill.

Expanding knowledge is a different case. The pieces of puzzle are different.

Taking the various prices, for example, scattering, fundamental particles,

Einstein's theories and that of de-Broglie, Heisenberg's principle and

the advances made by scientists like Bohr are apparently different. To

study them in depth, and seek unity in diversity by removing apparent

contradictions from individual chapters and then putting them in shape is a

different research. This is on a higher plane.

For reaching the last stage, one has to prove one's credibility by first-solving

the grid problems in the great laboratories. Unless these grids of various

shapes are well polished, one cannot put them together.

The recently discovered Sun's magnetic "heartbeat" causing solar flares

can be solved by going back to the study of the connection of various

discoveries. Science has no barriers. This has to be fundamental concept.

There is also no alternative medicine for Hard Work!

Anil Ahlawat,

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14 PHYSICS FOR YOU I MAY '13

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PRACTICE PAPER 2 Q 1 3

AdvancedBy : Akhil Tewari, B-Tech, IT-BHU

SECTION

Straight Objective Type

This section contains 6 multiple choice questions numbered

1 to 6. Each questio n has 4 choices (a), (b), (c) and (d), ou t o fwhich ONLY ONE is correct.

1. Velocity of a particle moving along a straight linea

is related with position as v = ;— - . Here ,(V

a and b are positive constants. The appr oxima te

graph of acceleration versus position is

(a) > X

(b)

J k a^^

r(c)

2.

(d) non e of these

Assuming all the

surfaces to be

frictionless find

the magnitude of

net acceleration

of smaller block

m with respect to ground

M

S T

_ 0

7777777777777777777777777777777177

(a)

(c)

2V5 mg

(5 m + M)

7y[5mg(5m + M)

(b)2 mg

(5m + M)

(d) non e of these

3.

4.

5.

6.

Resultant of two vectors having same magnitude

forms an angle with any of the vectors. If the

magnitude of second vector is reduced to half

of initial magnitude without changing the angle

between the direction of new resultant vector and

first vector is also reduced to half, then the angle

between the two vectors is

(a) 120° (b) 60°

(c) 90° (d) 45°

A particle of mass m movi ng d ue east with a speed

v collides with another particle of same mass and

same speed moving due north. The two particles

coalesce on collision. Find the velocity of the new

particle ?

(a) v (b) 2v

v(c) —j= (d) Non e of these

V 2

To a man moving due north with a speed

5 m s_1 , the rain appears to fall vertically. When

the man doubles his speed, the rain appears to

fall at 60°. Find the actual speed of the rain andits direction.

(a) 10 m s"1,120° (b) 10 ms "1 ,180°

(c) 10 m s"1, 90° (d) lO ms "

1, 60°

A car is travelling at a velocity 10 km h _ 1 on a

straight road. The driver of the car throws a

10

parcel with a velocity —j= km h"1 when car is

passing by a man standing on the side of a road. If




parcel just reaches the man, find the angle which

the direction of throw makes an angle with the

direction of car?

(a) 135° (b) 45°

(a) The mi ni mu m time in whic h he can cross the

(c) tan _ 1 V2 (d) tan

SECTION - II

Multiple Correct Answer(s) Type


7 to 10. Each questi on has 4 choices (a), (b), (c) and (d), ou t o f

wh ich ONE or MORE THAN ONE is/are cor rect.

7.

8.

9.

In the figure, the pulley

P moves to the right

with a constant speed u.

The downward speed

of A is Vfy and the speed

of B to the right is Vg.

Then,(a) v A = v B

v B = u + v A

V B + U = V A

B

H k

(b)

(c)

(d)

10.

the two blocks have accelerations of the same

magnitude.

Two particles A and B start simultaneously from

the same point and move in a horizontal plane. A

has an initial velocity Wj due east and acceleration

a-) due north. B has an initial velocity u2 due north

and acceleration a2 due east.

(a) Their pat hs mu st intersect at som e point

(b) They mus t collide at some point

(c) They will collide only if a\U-[ = a2u2

(d) If Mi > u2 a nd a \ < a2 , the particles will have the

same speed at some point of time.

A large rectangular box

ABCD falls vertically with an

acceleration a. A toy gun fixed

at A an d aimed to war ds C fires

a particle P.

(a) P will hit C if a = g

(b) P will hit the roof BC if a > §

(c) P will hit the wall CD or the floor AD if a < g

(d) may be eithe r (a), (b) or (c), de pe nd in g on the

speed of projection of P

A man who can swim at a speed v relative to the

water wants to cross a river of width d, flowing

with a speed u. The point opposite him across the

river is P.

river is — .v

(b) He can reac h the poi nt P in time — .

(c) He can reac h the poi nt P in time

V^ 2 - U 2

(d) He cannot reach Pifu> v.

SECTION - III

Assertion Reason type


11 to 14. Each question contains Statement-1 (Assertion) and

Statement-2 (Reason). Each question has 4 choices (a), (b), (c)

and (d) out of wh ic h ONLY ONE is correct.

(a) State ment- 1 is True, Stat ement -2 is True; Statem ent-2 is a

correct explanation for Statement-1.

(b) State ment- 1 is True, Statem ent-2 is True; Statem ent-2 is

not a correct explanation for Statement-1.

(c) Sta tement -1 is True, Sta tement -2 is False.

(d) Sta tement -1 is False, Sta tement -2 is True.

11. Statement-1 : The ma xim um ra nge along the

inclined plane, when thrown downward is greater

than that when thrown upward along the same

inclined plane with constant velocity.

Statement-2 : The maximum range along inclined

plane is independent of angle of inclination.

12. Statement-1 : A parti cle is project ed at an angle

9 (< 90) to horizontal, with a velocity u. When

particle strikes the ground its speed is again u.

Statement-2 : Velocity along horizonta l d irec tion

remai ns same but velocity along vertical direc tion

is changed. When particle strikes the ground then

magnitude of final vertical velocity is equal to

magnitude of initial vertical velocity.

13. Statement-1 : A block of mas s m is placed on a

smooth inclined plane of inclination 0 with the

horizontal. The force exerted by the plane on the

block has a magnitude mgcosQ.Statement-2 : Nor mal rea ction always acts

perpendicular to the contact surface.

14. Statement-1: In high jump, it hurts less when an

athlete lands on a heap of sand.

Statement-2 : Because of grea ter dis tanc e a nd

hence greater time over which the motion of an

athlete is stopped, the athlete experience less

force when lands on heap of sand.




SECTION - IV

Linked Comprehension Type

This section contai ns 3 paragraphs. Based upon each parag raph,

3 multiple choice questions have to be answered. Each question

has 4 choices (a), (b), (c) and (d), ou t of whic h ONLY ONE is

correct.

P 1 5 _ 1 7 : Paragraph for Quest ion Nos. 15 to 17

When a particle is projected at some angle with thehorizontal, the path of the particle is parabo lic in nature.

In the process the horizontal velocity remain s constant

but the magnitude of vertical velocity changes. At any

instant during flight the acceleration of particle remains

g in vertically downward direction. During flight at any

point the path of particle can be considered as a part

of circle and radius of that circle is called the radius of

curvature of the path of particle.

Consider that a particle is projected with velocity

u = 10 m s_1 at an angle 0 = 60° wit h the horizontal the n

15. The radiu s of cur vature of pa th of parti cle at theinstant when the velocity vector of the particle

becomes perpendicular to initial velocity vector

(a)

(c)

20

3^3

40

3^3

10(b) WTm

(d)80

3 ^ 3

16. The magni tude of accele ration of parti cle at that

instant is

(a) 10 m s '2

(b) 5^ 3 m s~2

(c) 5 m s " 2 ( d ) l o V i m s - 2

17. Tangential acceleration of particle at that instant

will be

(a) 10 ms" 2 (b) 20 m s - 2

(c) 5 m s - 2 (d) 5V3 m s~2

P18-20 : Paragraph for Questi on Nos. 18 to 20

A particle is projec ted with

velocity u at an angle 0

with an inclined plane of

inclination 0 < 45° with

the horizontal.

18. The time taken w he n velocity of projectile

becomes parallel to the plane

(a)

(c)

wsinO

2wtan0

(b)

(d)

!(COt0

Mtan0

8

19. The net velocity at the time when velocity is

parallel to the plane is

, . ucos0 „ . t<sin20(a) (b)

(c)ucos20

COS0(d) u tan©

20. Radi us of curv atur e of the path of projectile whe n

velocity is parallel to the plane

(a)

(c)

w2cos 20

U COS0

(b)

(d)

M 2COS 220

gcos 30

..2

gcos3 0

P21-23 : Paragraph for Questi on Nos . 21 to 23

An object at rest remain s at rest and an object in mo tio n

will continue its motio n with a constant velocity unle ss

it experiences a net external force. But the magn itu de of

force given by New ton' s 2n d law and 3rd law represents

or gives the information about the nature of force.The second law gave a specific way of determining

how the velocity changes under different influences

called forces. There are so many forces calculated by

Newton's law such as normal force, tension, viscous

force, weight but Newton's laws are not applicable,

when velocity of an.object comparable to the velocity

of light and microscopic particle. If the system contains

large number of particles, then if we apply the Newt on's

laws, concept of centre of mass is included.

21. Pulley and strings are massl ess. The force acting

on the block of mass m(a) 2F

(b) F

(d) 4F

I

/ y ) / / ) / / / / ) / / / / / / / / / / / / /

22. A partic le of mass m moves along a circle of r adi us

R. The modulus of average value of force acting

on particle over the distance equal to a quarter

of circle, if the particle moves uniformly with

velocity v is

(a)

(c)

V2w

2\fln

k r

(b)

(d)

2V21mv

mv

23. If velocity of mova ble pull ey is v and velocities of

block of masses M] and M2 are vj and V2 then the

correct relation between them

10 PHYSICS FOR YOU | MAY'131



M

777777777777777777777£

(a) v = i>i +

(c) V = 2{V X + V 2 )

( n

M l U [ j M 2

a + l>i(b) l> = —

2

(d ) o =

SOLUTIONS

1. (a)

2. (a): Free body diagram for m

yr

N

r2a

... (i)

...(ii)

For m,

mg-T=m 2a

N=ma

Free body diagram for M

For M

2T-N = Ma ... (iii)

N'

j L

M

" T Mg

P 1

-N->r

On solving, a =2mg

(M + 5m)

Net acceleration of m,

am = = ^ a = jSmg_m (5m + M )

3. (a):

tana =

tana =

BsinO

A + Bcos6

sin0

1 + cosG...(i)

t a r i n 6

tan— = —2 , a a

a + - c o s G2

a sin9tan— =

2 (2 + cos0). . . ( i i )

2tana

t a n a =

1 - ta n2 a

sin0

sin0

2 + cosG1 + cos 0 _ si n2 9

(2 + cos0)

On solving, Cos 0 = —

/ . 0 = 120° 2

4. (c): Refer the given below.

H n

m # —

mvi+ mvj = 2 m(v' xi + v' y j)

2 A A A A

or — (vi + vj) = v' x i+ v' j

v , vor vY = - or v.. = -

2 y 2

2 "T2

5. (a): Let vr =vrxi + vry j and vm=5i (in 1 s tcase)

A A

Vrm = (V rx ~

vm)

i+vryi

VCase (i): ta n90 ° = — or vrx = 5 m s _1

vrx~

5

Case (ii): vrm = (5i -10 / ) + vry) (y vm = 10/)

tan 60° = or vrv = -5S5 - 1 0 r y

vr = 5i - 5>/3

5, = 10 m s"- 1

Z(j) = ta n'

6. (a)

- 5 ^ 3or d> = 120°

12 PHYSICS FOR YOU | MAY '13



7. (b, d): At any instant of time, let the length of the

string BP = Zj and the length PA = Z2. In a further

time f, let B move to the right by x and A move

down by y, while P moves to the right by ut. As

the length of the string must remain constant,

l1 + l2 = (l1-x + ut) + (l2 + y)

or x = ut + y or x=u + y

x = speed of B to the right =

y= downward speed of A = v A

••• v B = u + v A

Also z >b ~va

or flg = a a -

8. (a, c, d)

9. (a, b, c) : Superimpose an up war d acceleration

a on the system. The box becomes stationary.

The particle has an upward acceleration a and a

downward acceleration g. If a = g, the particle has

no acceleration and will hit C. If a > g, the particle

has a net upward acceleration, and if a < g, the

particle has a net downwa rd acceleration.

10. (a, c, d)

11. (c)

12. (a): u x = u cosO, a x = 0

v x = u x + a x t = ur = u cosO

u y = u sinO ; a y --

: / ( s in0-gf = ttsinO -

2usin0

v y = -u sinO.

13. (b): In the direction of normal reaction, net

acceleration is zero. Hence forces in this direction

will be balanced. Hence N = mgcosQ.

14. (a): As we know that, F = —At

If Af is more, then F will be less.

1 0 m s ~ i

1 5 - ( a ) : 5 ^ 3 m s"

5 m s

Time after which velocity vector becomes

perpendicular to initial velocity vector is

10 2t =

gsin0 10sin60° s

Let Vy be the vertical component of velocity at that

instant. Then,

= 5>/3-

5_

10x2

10v = —j= m s

V3

! =5ms

or gc os a =

R =

R

g cosa

B 2 0 R = — m

3v3

16. (a)

-217. (c): at = gsincc = 10 x ~ = 5 m s

18. (d) 19. (c) 20. (b)

21. (c): Equation of motion for pulley,

F - 2T = trip x a

Since pulley is massless i.e., mp = 0

F = 2T,

, T = —2

22. (c) : p = =dt At

For quarter of a circle,

Av = v\Fl and At = —2v

F =2n/2 mv2

nr

23. (b): Velocity of block of mass M\ is

V\ = V-V'

Velocity of block of mass M2 is

VI = V + v'

Adding equation (i) and (ii), we get

...(i)

...(ii)

- nV, + VR,


t



ThoughtProvokingKinematics

1. A train starts from a station with a constan t

acceleration of 0.40 m s'

2

. A passenger arrives ata station 6 s after the end of the train left the very

same point. What is the least constant speed at

which the passenger can run and catch the train?

2. A particl e is proje cted fro m a poin t at the foot

of a fixed plane, inclined at an angle 45° to the

horizontal, in the vertical plane containing

the line of greatest slope through the point. If

<)> (> 45°) is the inclination to the horizontal of the

initial direction of projection, for what value of

tan § will the partic le strike the plan e: (i) hor izon tal

(ii) at right angle?

3 A particle is proje cted fr om a point on the level

ground and its height is h when at horizontal

distances a and 2a from its point of projection.

Find the velocity of projection.

4. A bulle t is fired into a viscous liquid wit h a

velocity v0. The retarding force is proportional to

squa re of velocity, so tha t the acceleration be com es

a = -kv2.

(a) Derive an expr ess ion for the dist ance tr avell ed

in the liquid.

(b) What is the dist ance travelled in the

liquid wh en velocity is reduc ed to — and the

corresponding time ?

5. A small sphe re of mas s m is released from rest in

a large vessel filled with oil where it experiences

a resistive force proportional to its speed, i.e.,

F d = -kv.

By : Prof. Rajinder Singh Randhawa*

(a) Find the speed of the ball with which it

varies.(b) After a certain time the sphere reaches a

terminal speed, find it?

6. Water is ru nn in g out of a conical fu nn el at the

rate of a m m 3 s"1. If the radius of the base of the

funnel is R mm and the altitude is H mm, find,

the rate at which the water level is dropping

when it is h mm from the top?

7. A circular wire fra me is fixed in a vertical pl ane.

A smooth wire is slightly stretched between

points P x and P2. A bead slides from point Pu

the highest point of the circle. Determine (a) itsvelocity v when it arrives at P 2 (b) find the time

taken by it?

1.

SOLUTIONS

Assume the train is at x = 0 at t = 0, the equation

for train is

1 , 1 , xT =-aT t

2 = ±(0.40)f 2

The passenger reached x = 0 a t t = f 0 = 6 s, so his

coordinate at time tis xP = v,,(t -10).

For the passenger to catch the train, xT = xP.

~aT t 2=v p(t-t 0) or aT t

2-2v pt+2vpt 0=0

or t_v p± - 2aT v pt 0

The roots are real if Vp-2aT v p t Q > 0

v p >2aT t 0 = 2x 0.40 x 6 = 4.8 m s" 1.

Ra nd ha wa Ins tit ut e of Physic s, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Secon d FL, Sector-20C, Ch an di ga rh




2. (i) Along horizonta l direction,

h = (Mcos0)f

p / ,7MCOS(()

h

O Q

Along vertical direction,

0 = Msinij) - gt o r Msin<|> = gt

and h = Msin<j>f • V2 5

Using Eqs. (i) and (ii) in Eq. (iii), we get

1

(Mcos0)f = (Msin<j>)f - - (i/ sin t)>) f

tan 0 = 2

(ii) Along perpendicular to the plane,

0 = usin(<j)—45°)f - ~( ^c os 45°)f 2

or ti j l i

sin (0 - 45°).

45°- <»

ucos(<t>-45°) =V 2

^ ^ sin (<|) - 45°)

or MCOS(0 - 45°) = 2M sin(0 - 45°)

1 , , r o , tan 0 - 1or - =tan (0 - 4 5 ) =

1

2 1 + tan 0

Solving, we get

tan 0 = 3

™(i)

...(ii)

.(iii)

O Q

Along the plane, MCOS(0 - 45°) = (gsin45°)f

3. If u is the velocity of projection and 0 is the angle

of projection, the equation of trajectory isJ2.

•••(i)y = xtanO - - —2 m2

cos20

With origin at the point of projection,

gx2 - 2u2 x sin0 cos0 + 2 u2 cos20 h = 0 ...(ii)

Since the projectile passes through two points

(a, h) and (2a, h), then a and 2a must be roots of

Eq. (ii),

a + 2a =

a r i d ax2a =

2m sinO cos0

2M2 cos

2 0 xh

...(iii)

...(iv)

Divide (iii) by (iv), we get

3 a tan 0 „ 3 h—- = or ta n0 = —2a h 2 a

From Eq. (iv),

M 2 = ^ s e c 2 e = ^ [ 1 + t a n 2 e ] = S ^

h h 1 h1 +

9h~

4 a2

h•9/2

4. Acceleration of bullet, ~ = - k v 2

at

dv r jtor — --kdt

vOn integrating, we get,

" J F

• dv

...(i)

v 0 0

k dt or -kt

1 1

-ktV n V

dx 1 dtor v = — = — or dx = —

dt . 1 1kt + kt +

v,o v.

jdx=j

o o kt +

dt 1,—— or x = - In

f k

v n

o

kt +

...(ii)

...(iii)

. .( iv)

We calculate time t, when velocity is reduced to

From Eq. (ii)

vn v.0 kvn

Put t in Eq. (iv), we get

1 , „ 0.693 x = - n2 = .

k k




5. (a) Force acting on sphere = mg - kv

where k is constant.

Acceleration of ba ll , — = g- — vdt m

or-dv

mg-dt

m

...(i)

...( i i )

On integrating, we get In

If v = 0, f = 0, then

C = l n ^ £k

Put in Eq. (iii), we get

(—£ — p ) - In —£ = -— ;k m

mg= 1 + C ... (ii i)

m

InI k

or In

t r :

?• t d -- - -

s v 1

r 3 3 3

? f i fs v 1

r 3 3 3

? f i f 3 3 3

mg-v

mgk

,-TS. -t1)!l - e - ( i v )

o r

YHwhere , x = — is called time constant.

k

(b) When the sphere reaches terminal speed, the

acceleration of the sphere becomes zero. Then

mgmg = kvt or vt = — 1-.

V

V 1I - 7 H

V IAt any time water is at a height 'h' and radius r.

From similar A's, — = —

R H

hR

H

1 2

r = ~rr, volume of water, V = -j tr h.

T/ 1 (hRY, dV nR2h2 dhV = -71 — « or — =

3 {H J dt h2 dt

dh a3 H 2

(a) v2 = v2 + lax = 02 + 2(^cos9)(2Rcos6)

•Of = 2 Rg cos Q

t j ± j 4 R g c o s e = 2 I r

a gcos6 ]j g

20 PHYSICS FOR YOU | MA Y 137



«

SOLVED PAPER 2 Q 1 3

Main1. The anode voltage of a photoc ell is kep t fixed. The

wav ele ngth X of the light falling on the cath ode

is gradually changed. The plate current I of the

photocell varies as follows

(a)

3.

(c)

A circular loop of radius 0.3 cm lies parallel to a

much bigger circular loop of radius 20 cm. The

centre of the small loop is on the axis of the biggerloop. The distance between their centres is 15 cm.

If a current of 2.0 A flows through the smaller

loop, then the flux linked with bigger loop is

(a) 6.6 x 10~9 we be r (b) 9.1 x 10"11 weber

(c) 6 x 10"n we be r (d) 3.3 x 10"11 weber

The graph between angle of deviation (8) and

angle of incidence (i) for a triangular prism is

represented by

(a)

O O

(c) f (d) f

O O

4. In an ICR circuit as shown below both switches

are open initially. Now switch S. is closed, S2 kept

open, (q is charge on the ca paci tor a nd T = RC is

capacitive time constant). Which of the following

statement is correct?

V

R\

Si

c JS2

5.

L

(a) At f = 1 q = CV( 1 - e"1)

(b) Work don e by the batt ery is half of the energy

dissipated in the resistor

(c) At t = T, q = CV/2

(d) At t = 2x, q = CV( 1 - e2)

Two short bar ma gne ts of leng th 1 cm each

have magnetic moments 1.20 Am2 and 1.00 Am2

respectively. They are placed on a horizontaltable parallel to each other with their N poles

pointing towards the South. They have a common

magnetic equator and are separated by a distance

of 20.0 cm. The value of the resultant horizontal

magnetic induction at the mid-point O of the line

jo in ing thei r ce nt re s is close to

(Horiz ontal comp one nt of ear th ' s magnet ic

induction is 3.6 x 10"5 Wb/m2)

(a) 5.80 x 10"4 Wb /m (b)

(c) 2.56 x 10"4 Wb/m2 (d)

3.6 x 10~5 Wb/m2

3.50 x 10"4 Wb/m2

6. This question has Statement-I and Statement-II. Ofthe four choices given after the Statements, choose

the one that best describes the two Statements.

Statement-I : Hi ghe r the range , greate r is the

resistance of ammeter.

Statement-II: To increase the range of ammeter,

additional shunt needs to be used across it.

(a) State ment-I is false, Statement -II is true.

(b) S tate ment -I is true , Statemen t-II is true,

Statement-II is the correct explanation of

Statement-I.

PHYSICS FOR YOU I MAY 13 21



(c) Stat emen t-I is true, State ment- II is true ,

Statement-II is not the correct explanation

of Statement-I.

(d) Statemen t-I is true, Statemen t-II is false.

7. An ideal gas enclo sed in a vertica l cylindri cal

container supports a freely moving piston of

mass M. The piston and the cylinder have equal

cross sectional area A. When the piston is in

equilibrium, the volume of the gas is V 0 and itspressure is P0. The piston is slightly displaced

from the equilibrium position and released.

Assuming that the system is completely isolated

from its surrounding, the piston executes a simple

harmonic motion with frequency

1 AyP0(a)2K ]J AyPg

1 VqMPQ

2n A2 y

(b)2K V 0 M

(c) < d >2K V MV,o

8. Let [GQ] denot e the dimens ional form ula of the

permittivity of vacuum. If M = mass, L = length,

T = time and A = electric current, then

(a) [£0] = [M 1 L2 r 1 A]

(b) [e0] = [M"1 L"3 T 2 A]

(c) [ G 0 ] = [M"1 L"3 T4 A2]

(d) [GQ] = [M"1 L2 T"1 A'2]

2Pn

ii

i > 1 r

I o 2 v 0

The above p-v d i a g r a m r e p r e s e n t s t h e

ther modyna mic cycle of an engine, operat ing wit h

an ideal monoatomic gas. The amount of heat,

extracted from the source in a single cycle is

(a) 4p0»o

(b) p0v0

<o i f POV O

(d ) ( y J / W o

10. A projectile is given an initia l velocity of

(f+2/)m/s, where i is along the ground a nd j

is along the vertical. If g = 10 m/s2, the equation

of its trajectory is

(a) 4y = 2x - 25x2 (b) y = x- 5x

2

(c) y = 2x - 5x^ (d) 4y = 2x-5x z

11. A beam of unpolarised light of intensity I0 is

passed through a polaroid A and then through

another polaroid B which is oriented so that its

principal plane makes an angle of 45° relative to

that of A. The intensity of the emergent light is

(a) / 0 / 8 (b) I Q ( C ) I 0 /2 (d) /„/ 4

12. A diode detector is used to detect an amplitude

modulated wave of 60% modulation by using a

condenser of capacity 250 pico farad in parallelwith a load resistance 100 kilo ohm. Find the

maximum modulated frequency which could be

detected by it.

(a) 5.31 kH z (b) 10.62 MH z

(c) 10.62 kH z (d) 5.31 MH z

13. The supply voltage to a room is 120 V. The

resistance of the lead wires is 6 Q. A 60 W bulb

is already switched on. What is the decrease of

voltage across the bulb, when a 240 W heater is

switched on in parallel to the bulb?

(a) 10.04 Volt (b) ze ro Volt(c) 2.9 Volt (d) 13.3 Volt

14. A metallic rod of length '/' is tied to a string of

length 21 and made to rotate with angular speed

CO on a horizontal table with one end of the string

fixed. If there is a vertical magnetic field 'B' in the

region, the e.m.f. induced across the ends of the

rod is

(a)

(b)

(c)

5Bwl

2

2B(ol2

3 Bear (d)4Bco/

15. The magnetic field in a travelling electromagn etic

wave has a peak value of 20 nT. The peak value

of electric field strength is

(a) 12 V/m (b) 3 V/m

(c) 6 V/m (d) 9 V/m

16. A sonometer wire of length 1.5 m is made

of steel. The tension in it produces an elastic

strain of 1%. What is the fundamental frequency

of steel if density and elasticity of steel are

7.7 x 103 kg/m3 and 2.2 * 10" N/m 2 respectively?

(a) 770 Hz (b) 188.5 Hz

(c) 178.2 Hz (d) 200.5 Hz

17. This ques tion has Statement-I and Statement-II. Of

the four choices given after the Statements, choose

the one that best describes the two Statements.

Statement-I : A point particle of mass m moving

with speed v collides wit h stationary point particle




of mass M. If the maximum energy loss possible

is given as / [ - my * then / =m

M + m

Statement-II: Maximum energy loss occurs when

the particles get stuck together as a result of the

collision.

(a) Stateme nt-I is false, Statement -II is true.

(b) Statement-I is true, Statement-II is true,Statement-II is a correct explanation of

Statement-I.

(c) Statement-I is true, Statement-II is true,

Statement-II is not a correct explanation of

statement-I.

(d) Stateme nt-I is true, Statement -II is false.

18. A charge Q is uniformly distributed over a long

rod AB of length L as sho wn in the figure. The

electric potent ial at the point O lying at a dis tance

L from the end A is

AWfflBZMZSZmKo

(a)

(c)

Qln2

471 e 0L

3 Q

4K e0 L

(b)

(d)

Qln2

8n e0 L

Q

4jte 0L In 2

19. A uniform cylinder of leng th L and mas s M h avin g

cross-sectional area A is suspended, with its

length vertical, from a fixed point by a massless

spring, such that it is half submerged in a liquid

of density a at equilibrium position. The extension x0 of the spring when it is in equilibrium is

Mg(a) M

k I M

(c) LAc\

M J

(b)

(d) Mg(1 LAo)

k I 2 M j

20.

Si s?

D

points

(d) semi-circles

Screen

(Here k is spring constant)

Two coherent point

sources Sj and S2 are

separated by a small

d i s t ance ' d ' as shown.

The fringes obtained on

the screen will be

(a) concentr ic circles (b)

(c) straight lines

21. A hoop of rad ius r and mass m rotating with an

angular velocity co0 is placed on a rough horizontal

surface. The initial velocity of the centre of the

hoop is zero. What will be the velocity of the

centre of the hoop when it ceases to slip?

26.

(a) m 0 (b)r CON"0 ( c ) (d) r ( 0o

4 3 2

22. The amplitude of a damped oscillator decreases

to 0.9 times its original magnitude in 5 s. In

anot her 10 s it will decrea se to a time s its original

magnitude where a equals

(a) 0.6 (b) 0.7 (c) 0.81 (d) 0.729

23. Assume that a drop of liquid evaporates by decrease

in its surface energy, so that its temperature

remains unchanged. What should be the minimum

radius of the drop for this to be possible? The

surface tension is T, density of liquid is p and L

is its latent heat of vaporization.

^ (b) Bkp L T

£ (d) £VP L P

L

24. What is the minimum energy required to launcha satellite of mass m from the surface of a planet

of mass M and radius R in a circular orbit at an

altitude of 2R?

GmM .. , 5GmM

(a)

(c)

(a)

(c)

3 R

2 GmM

(b)

(d)

6 R

GmM

3 R 2R

25. In a hyd rog en like ato m electron mak es tr ansiti on

from an energy level with quantum number n to

another with quantum number (n - 1). If n > > 1,

the frequency of radiation emitted is proportionalto

1 1

,3/2(a) — (b) — (c) — (d)

n n n

Two charges, each equal to q, are kept at x = - a

and x = a on the x-axis. A particle of mass m and

charge q0 = ^ is placed at the origin. If ch arg e

q0 is given a small displacement (y < < a) along

the y-axis, the net force acting on the particle is

proportional to

(a) - I (b) y (c) - y (d) -y y

27. If a piece of metal is heated to temperature 0

and then allowed to cool in a room which is at

temperature 0O, the graph between the temperature

T of the metal and time t will be closed to

(a)

O

T

(b)

14 PHYSICS FOR YOU I MAY '1 3)



t

(C) 8t

O

(d) e0

o

28. Two capacitors C, and C2 are c harged to 120 V a nd

200 V respectively. It is fo un d that by conne cting

them together the potential on each one can bemade zero. Then

(a) 9C 1 = 4C2 (b) 5CI = 3C2

(c) 3Ci = 5C2 (d) 3Q + 5C2 = 0

29. The I-V characteristic of an LED is

(a)

(c) R

30. Dia met er of a pla no- conv ex lens is 6 cm a nd

thickness at the centre is 3 mm. If speed of light

in material of lens is 2 x 10s m/s, the focal length

of the lens is(a) 10 cm (b) 15 cm (c) 20 cm (d) 30 cm

SOLUTIONS

1. (a)

2. (b):

As field due to current loop 1 at an axial point

•>2

B 1 = -2 (d

2+R

2)

3' 2

Flux linked with smaller loop 2 due to B1 is

<t>2 = B 1A2 = " • nr

2 (d 2+R

2)

3' 2

The coefficient of mutual inductance between

the loops is

M _ $2 _ M 2 ™ " 2

= T ~ 2 (d 2 + R

2)

3' 2

Flux linked with bigger loop 1 is

= = TTK2(d + R ) '

Substituting the given values, we get

4TI x IP'7

x (20 x 10~z

)z

x 7t x (0.3 x 10"2

)2

x 2

2[(15 x 10~2)

2 + (20 x 10"

2)

2]

3/24>i =

((), = 9.1 x 10~n weber

3. (d): The graph between

angle of deviation(S) and

angle of incidence (i) for a

triangular prism is as shown

in the adjacent figure. O

4. (d):

VHi—

R

I KC

uooooff^-

L

As switch S, is closed and switch S 2 is kept open.

Now, capacitor is charging through a resistor R.

Charge on a capacitor at any time t is

q = q0(l - e~th

)

(j = CV(1

x

„-tlv [As q0 = CV]

At t =

q = CV(1 - e~ x/2x

) = CV(1 - e' V2

)

At t = x

q = CV(1 -

2x,

~ X,X

) = CV( 1-tT 1)

At t •

5.

q = CV(1 - e _2t /T

) = CV(1 - e~2)

(c) : The situation is as shown in the figure.

N

b h .

B,

•B 2

N

o

N

As the point O lies on broad-side position with

respect to both the magnets. Therefore,

The net magnetic field at point O is

Bnet = Bj + B 2 + B h

PHYSICS FOR YOU | MAY 13 25



H 0 n0 M 2

An r

= Ho

4jtr3(MJ + M 2 ) + B h


- - 4 7 1 X 1 0 ' [1.2 + 1] + 3.6 x 10~5

n e t 471 x (10 x 10~2)3

->-71 0 "

10,-3x 2.2+ 3.6x10-5

= 2.2 x 10"4 + 0.36 x 10"4 = 2.56 x 10^ Wb/m'

6. (a)

7. (d):M

FBD of piston at equilibrium

PgA

Mg

P3TM A + Mg = Pq A ...(i)

FBD of piston when piston is pushed down a

distance x

Mg (p0 + dP)A

(P0 + dP)A - (-Patm A + Mg) = M . . . ( i i )

As the system is completely isolated from its

surrounding therefore the change is adiabatic.

For an adiabatic process

PVT = constant

VdP + V^PdV = 0

yPdVor dP = —

dP =

V

yP0(Ax)

V n(v dV = Ax) ...(hi)

Using (i) and (iii) in (ii), we get

M ^ - ^ x ordt

2 v 0

d 2 x

dt 2

i M :

MV n

Comparing it with standard equation of SHM,

d 2 x

dt 2

We get

2= - (0 1

c o 2 = ^ I MV n

or co =Y ^

MVn

Frequency, x> = 1 yP0 A2co

271 2TT V MVro

8. (c) : According to Coulomb's law

r =_ 1

fill

4ne0 r 2

~~ F r 2

[AT] [AT]

1 E

0J _ 7 7

[MLT"2][L]2

9. (c): 4

2Po

Wt

= [M_1 L"3T4A2]

— ~ w —

D

t;2%

Heat is extracted from the source in path DA

and AB.

Along path DA, volume is constant. Hence,

AQ da = nC v AT = nC v(T A - T D)

According to ideal gas equation

pv

pv = nRT or T = nRFor a monoatomic gas, q = —R

v 2

2Po

vo Po

vo

nRAQ,•DA

nR ~ 2 Povo

Along the path AB, pressure is constant. Hence

AQab = nC p AT = nC p(T B - T A)5

For monoatomic gas, Cp = — R

2Po

2vo

2P0vo

nR nR

10= y W o

.•. The am ou nt of hea t extrac ted fr om the sou rce

in a single cycle is

A Q = A Q DA + A Q AB

3 10 13= 2 pov

o+ y ? W o = y W o

10. (c) : Given: u=i + 2j

A A

As u = u xi + u y j

:. ur = 1 and u„ = 2

2 6 PHYSICS FOR YOU | MAY '13 ?



i

Also x = u xt

1 2

an d y = Uyt -—gt

:. x = tan d 1/ = 2f - i x 10 x t

2 = 2f - 5f

2s 2

Equation of trajectory is

y = 2x - 5x2

.11. (d): Intensity of light after passing polaroid A

is

1 2

Now this light will pass through the second

pola roid B whose axis is inclined at an angle of

45° to the axis of polaroid A. So in accordance

with Malus law, the intensity of light emerging

from polaroid B is

\2

I2 = / j cos 45° =

-&)J _ - k

12. (c) : The max im um fre que ncy which can be

detected is

1"0 =

2nma x

where, x = CR

Here, C = 250 pico farad = 250 * 10~12

farad

R = 100 kilo ohm = 100 * 103 ohm

ma = 0.6

1U =

2;t x 0.6 x 250 x 10"12 x 100 x 103

= 10.61 x 103 Hz = 10.61 kHz.

13. (a): As P = Yl R

Heater

Here, the supply voltage is

taken as rated voltage.

.'. Resistance of bul b

_ 120 V x 120 VB ~ 60 W

Resistance of heater, RH

— W W6Q Bulb

H '

120 V

= 240 Q

120 V x 120 V = 60 Q240 W

Voltage across bulb before heater is switched

on,

120 V x 240 Q = 117.07 V240 Q + 6 Q

As bulb and heater are connected in parallel.

Their equivalent resistance is

(240Q)(60n ) = 4 8 nec

i 24 0n + 6 0 0

.•. Voltage across bulb after heater is switched

on

120 V .48112 48 Q + 6 Q

Decrease in the voltage across the bulb is

\V = V 1 - V 2 = 10.41 V = 10.04 VCO

21 ... IM-

14. (a):

Consider a element of length dx at a distance x

from the fixed end of the string,

e.m.f. induced in the element is

dz = B(u>x)dx

Hence, the e.m.f. induced across the ends of the

rod is

31 31

- ~~[(3')2 - (2/)2]21 2

e = J Bmxdx = But

21

_ 5B(ri2

2

15. (c) : In electromagnetic wave, the peak value of

electric field (E0) and peak value of magnetic field

(B0) are related by

E 0 = B0c

E 0 = (20 x 10""9 T) (3 x 10

8 m s"

1) = 6 V/m

16. (c) : Fundamental frequency of vibration of wire

is

JL ll2L\n

where L is the length of the wire, T is the tension

in the wire and ft is the mass per length of the

wire

As fi = pA

where p is the density of the material of the wire

and A is the area of cross-section of the wire.

•'• U - - ..2L^pA

Here tension is due to elasticity of wire

T = YA_ L _

. StressAs y = =

Strain

_ TL "

~ AAL_

1 lYAL

2L' pL

inHere, Y = 2.2 x 10" N/m", p = 7.7 x 10J kg/m

J

— = 0.01, L = 1.5 m L

Cont. on Page No. 82

PHYSICS FOR YO U | MAY 13 2 7



FOCUS

Engineering Aspirants 2013 -

Are you Prepared for JEE Advanced?(Here are the few Tips to get your optimum)

The JEE Advanced (formerly known as IIT-JEE) is an

annual entrance examination to get Admission in IITs. It

is also one of the toughest engineering entrance exams in the

world. Only 1.5 lac stude nts will be short listed fro m JEE

Main 2013 to appear for the JEE Advanced 2013 on June 2nd,

2013. A serious aspiran t ideally mus t have completed the

syllabus by now.

Schedu le of JEE (Ad van ced) , 20 13

The examination will be held on Sunday, June 02, 2013 as per

the schedule given below:

Paper 1 9:00 to 12:00 hrs. (1ST)Paper 2 14:00 to 17:00 hrs. (1ST)

EXAMINATION PATTERN:

There will be two question papers, each of three hours

duration. Both the question papers will consist of three

separate sections on Chemistry, Physics and Mathematics.

Questions will be of objective type, designed to test

comprehension, reasoning and analytical ability of Students.

All the questions will be Multiple Choice Type (MCQ)

Negative marking scheme will be followed in the checking

of examinations.

A Student can opt for question paper in any of the language

viz. English or Hindi.

SYLLABUS COVERAGE:

JEE Syllabus of Class XI & XII contri bute s a bou t 45% and 55%

of IIT-JEE question-papers respectively. While preparing all

the chapters of Physics, Chemistry and Mathematics, based

on our past experience stress may be given in particular on

the following topics:

Mathemati cs : Quadr atic Eq uations & Expressions, Comple x

Numbers, Probability, Vectors & 3D Geometry, Matrices

in Algebra; Circle, Parabola, Hyperbola in Coordinate

Geometry; Functions, Limits, Contin uity a nd Differentiability,

Application of Derivatives, Definite Integral in Ca lculus.

Physics: Mechanics, Fluids, Heat & Thermodynamics,

Waves and Sound, Capacitors & Electrostatics, Magnetics,Electromagnetic Induction, Optics and Modern Physics.

Chemistry: Qualitative Analysis, Coordination Chemistry &

Chemical Bonding in Inorganic Chemistry, Electrochemistry,

Thermodynamics, Chemical Equilibrium in Physical

Chemistry and Organic Chemistry Complete as a topic.

TIPS FOR JEE Adv anc ed, 2013 :

PHYSICS (please see Tips on Mathematics in 'Mathematics

Today' & on Chemistry in 'Chemistry Today')

1. Mechanics is one topic of Physics that is considered less

scoring by most experts. However to add to the dilemma

this is also the topic that forms the major portion of the

JEE (ADVANCE) in terms of marks. So this topic cannot be

neglected.

2. One must also try to concentrate on other scoring topics to

ensure a better performance, for example Optics, Electricity

and Magnetism, etc.

3. Kinematics and Particle dynamics are very important

topics of Mechanics that make regular appearance in the JEE

papers.

4. Accordi ng to the general tre nds, Mecha nics an d Electricity

and Magnetism are the most important topics in terms of the

number of questions asked in JEE of previous years.

5. In the decreasing order of the marks they carry are listed

different topics of Physics according to their appearance in

previous year's papers.

Mechanics and Electricity and Magnetism (Equal

importance)

Modern Physics

Optics

Heat and Thermodynamics and Waves and Sound

Measurement and errors

6. Thermodynamics is important from the terms of both

Physics and Chemistry so concentrate on that as well. It is

wise to cover Wave Optics first in 'Optics' topic. The reason

is that the portion is smaller compared to Ray Optics thus

quick to cover.

Crac king the JEE (ADVA NCED) 2013

"Stay focussed and maintain a positive attitude

"Develop speed. Refer to reputed mock-test series to build

a winning exam temperament. Solve the past year's IIT-JEE

papers. Focus on your weak areas and improve upon your

concepts.

"Practise of JEE level questions is.necessary as it improves

your reasoning and analytical ability.

"Remember it is quality of time spent and not the quantity

alone. Hence give short breaks of 5 to 10 minutes every 1-2

hours of serious study. Completely relax when you take abreak. Practice meditation to develop inner calm, poise,

confidence and power of concentration.

"Don't overstress yourself. Five to six hours of sleep every

night is a must, especially three-four days before IIT-JEE to

keep you physically and mentally fit. While short naps may

help to regain freshness, avoid over-sleeping during the day.

"Finally, don't be nervous if you find the paper tough since

it is the relative performance that counts. Put your best

analytical mind to work, and believe in your preparation.

Authored by Ramesh Batlish, FIITjEE Expert

mm




• • •

NCERTXtract1 • • / / • • •

uestions forNEETGRAVITATION

l . Which of the following statements is correct?

(a) Accelerat ion due to grav ity increases with

increasing altitude.

(b) Acceleration due to gravity increases with

increasing depth.

(c) Acceleration due to gravity increases wit h

increasing latitude.

(d) Acceleration due to gravity is independent

of the mass of the earth.

The potential energy of a system of four particles

each of mass m are placed at the vertices of a

square of side I is

sllGm2(a) -

(c) >/2G>»

•h

f ^ - i )

(b) - 2Gm 2 + J _

42

(d) I

2Gm A f ^ . 1

3.

4.

The earth is an approxima te sphere. If the interior

contained matter which is not of the same density

everywhere, then on the surface of the earth, the

acceleration due to gravity

(a) will be directed towards the centre but not

the same everywhere.

(b) will have the same value ever ywhere but no t

directed towards the centre.

(c) will be same everywhere in magnitude

directed towards the centre.

(d) cannot be zero at any point.

A planet orbits the sun in an elliptical path as

shown in the figure. Let vP an d v A be speed of

the planet when at perihelion and aphelion

respectively. Which of the following relations is

correct?

Planet

PPerihelion

>AAphelion

5.

P_ _

f A \ "A

The tim e pe ri od T of the mo on of pl ane t

mars (mass Mm) is related to its orbital radius

jR as (G = Gravitational constant)4TC

2R

3 nn2 4k 2GR3

(a)

(c)

T 2=:

T2=

GM m

In.R3G

M„,

(b)M„

(d) T2 = 4rcMmGR

3

6.

7.

Assuming that earth and mars move in circular

orbits around the sun, with the martian orbit

being 1.52 times the orbital radius of the earth.

The length of the martian year in days is

(a) (1.52)2/3 x 365 (b) (1.52)3'2 x 365

(c) (1.52)2 x 365 (d) (1.52)

3 x 365

A satellite of mass m is in a circular orbit of

radius 2RE about the earth. Energy required

to transfer it to a circular orbit of radius 4RE is

(where ME and R E is the mass and radius of the

earth respectively)

GMptn(b)

GM E ma) E

2 RE

(b)4 R E

GM Pm(d)

GM E mGM Pm(d)

16 R£

PHYSICS FOR YOU | MAY '13 2 9



8- Which of the foll owing stat emen ts is true?

(a) A geostatio nary satellite goes aroun d the

earth in east-west direction.

(b) A geostatio nary satellite goes aro und the

earth in west-east direction.

(c) The time-period of a geostationary satellite

is 48 hours.

(d) The angle bet ween the equat orial pla ne an d

the orbital plane of geostationary satellite is

90°.9. Two uni for m solid sphe res of equa l radii R, but

mass M and 4M have a centre to centre separation

6R, as shown in figure. A projectile of mass m is

projected fr om the surface of the sphe re of mass M

directly towards the centre of the second sphere.

The minimum speed of the projectile so that it

reaches the surface of the second sphere is

6 R

(a)4 GM

5 R

5 GM

V Vl lT

, x 3 GM( c )

I S I R(d)

5 GM

3 R

10. The time interval between two successive noo n

when sun passes through zenith point (meridian)

is known as

(a) sidereal day (b) me an solar da y(c) solar year (d) lunar mo nt h

11- An object of mass m is rais ed f ro m the surfac e of

the earth to a height equal to the radius of the

earth. The gain in its potential energy is

(where R E is the radius of the earth)

(a) m§R E

( c) ~ m g R E

(b) - m g R E

(d) | mgR E

12. A satellite is to be placed in equatorial geostationary

orbit around earth for communication. The height

of such a satellite is

[M£ = 6 x 1024 kg, R E = 6400 km, T = 24 h,

G = 6.67 x 10~n N m2 kg~2]

(a) 3.59 x 105 m (b) 3.59 x 10

6 m

(c) 3.59 x 107 m (d) 3.59 x 108 m

A system of fou r particles each of mass m is placed

at the vertices of a square of side I. The potential

at the centre of the square is

(a)

(c) - 2 ^

(b)

(d) -4V2Gm

~T14. The escape velocity from the surfac e of the earth

is

(where R E is the radius of the earth)

(a) (b) JgR(c) 2jgR E (d) fiK

15. A bod y weigh s 63 N on the surface of the earth.

What is the gravitational force on it due to the

earth at a height equal to half the radius of the

earth?

(a) 24 N (b) 28 N (c) 32 N (d) 36 N

16. In considering motion of an object under the

gravitational influence of another object. Which

of the following quantities is not conserved?

(a) Angular momentum(b) Mass of an object

(c) Total mech anical energ y

(d) Linear momentum

17. In our solar system, the inter-pl anetary region ha s

chunks of matter (much smaller in size compared

to planets) called asteroids. They

(a) will not move arou nd the sun since they ha ve

very small masses compared to sun.

(b) will move in an irregular way because of

their small masses and will drift away into

outer space.(c) will mov e arou nd the sun in closed orb its

but not obey Kepler's laws.

(d) will move in orbits like planets and obey

Kepler's laws.

18- A rocket is fired from the eart h towa rds the su n.

At what distance from the earth's centre is the

gravitational force on the rocket zero?

[Mass of the sun = 2 * 1030 kg, mass of the earth

= 6 x 1024

kg, orbital radius = 1.5* 10" m]

(a) 2.6 x 104 kg (b) 2.6 x 106 kg

(c) 2.6 x 108 kg (d) 2.6 x 1010 kg

19. Two sphe res each of mas s M and rad iu s R are

separated by a distance of r. The gravitational

potential at the midpoint of the line joining the

centres of the spheres is

, , GM(a)

r(b)

2GM

(c) -GM

2r(d) -

4GM




20.

21.

22.

24.

A satellite is in an elliptic orbit around the earth

with aphelion of 6RE and perihelion of 2RE, where

R E is the radiu s of the earth.

The eccentricity of the orbit is

1 1 1

3 (C )

J ( d )

6

In the question number 20, the ratio of the velocity

of the satellite at apogee and perigee is1

6

(a) ±v / 2 (b)

<b) 1v ' 3

(d)

The gravitational force between a hollow spherical

shell (of radius R and uniform density) and a

point mass is F.

Which of the following graphs represents the

variation of F with r where r is the distance of

the point from the centre of the hollow spherical

shell of uniform density.

F

t(a)

(c)

(b)

23. An astronaut experiences weightlessness in a

space satellite. It is because

(a) the gravita tion al force is small at that location

in space.

(b) the grav ita tional force is large at tha t location

in space.

(c) the ast ron aut experiences no gravity.

(d) the gravita tion al force is infini tely large at

that location in space.

Three masses each of

ma ss m are pl ace d

at the vertices of an

equi la tera l t r iangle

ABC of side I as sh own

in figure. The force

acting on a mass 2m

placed at the centroid

O of the triangle is

*-x

(a) zero

(c) -6Gm

2 r

(b)

(d)

6Gm2 o

Gm2 *

25. A satellite of mass m orbits the ear th at a height h

above the surfac e of the earth. Ho w muc h e nergy

must be expended to rocket the satellite out of

earth's gravitational influence?

(where M E an d R E be mass and radius of the earth

respectively)

(a)

(c)

GM E m

4 (R E +h)

GM E m

(b)

(d)

GM E m

2(R£+/i)

2 GM E m

26.

(. R E +h) ' ' (R

E+h

)

Two stars each of mass M are approaching each

other for a head-on collision. When they are at a

distance r, their spe eds are negligible. The ra dius

of each star is R(r >> R). The speed which they

collide is

(a)

(c)

(b)

(d)2 GM

R

27. Diff erent poin ts in earth are at slightly differe nt

distances from the sun and hence experience

different forces due to gravitation. For a rigid

body, we know that if various forces act at various

poi nts in it, the resu ltan t mot ion is as if a net force

acts on the centre of mass causing translation

and a net torque at the centre of mass causing

rotati on arou nd an axis thr oug h the centre of mass.

For the earth-sun system (approximating theearth as a uniform density sphere)

(a) the tor que is zero.

(b) the tor que causes the ear th to spin.

(c) the rigid bo dy resul t is not applicable since

the earth is not even approximately a rigid

body.

(d) the torque causes the earth to move aro und

the sun.

28. A comet orbits the sun in a highly elliptical orbit.

Which of the following quantities remains constantthroughout its orbit?

(i) Linea r spe ed

(ii) Angular speed

(iii) Angul ar mom ent um

(iv) Kinetic energy

(v) Potential energy

(vi) Total energy

(a) (i), (ii), (iii) (b)

(c) (iii) an d (vi) (d)

(iii), (iv), (v)

(ii), (iii) and (vi)

PHYSICS FOR YOU I MAY '13 3 1



29. Match the following.

For a satellite in circular orbit,

33.

Column I Column II

(A) Kinetic energy (P)GM E m

2 r

(B) Potential energy (q)GM E

V r

(C) Total energy (r)GM E M

r

(D) Orbital velocity (s)GM E m

2 r

(where M£ is the mass of the earth, m is mass of

the satellite and r is the radius of the orbit)

(a) A - r, B - s, C - q, D - p

(b) A - q, B - p, C - r, D - s

(c) A - p , B - q , C - s, D - r

(d) A - s, B - r, C - p, D - q

30. Which of the following statements is incorrect

regarding the gravitational force?

(a) The gravit ational forc e is de pe nd en t of the

intervening medium.

The gravitational force is a conservative

force.

The gravitational force is a central force,

(d) The gravitational force obeys the inverse

square law.

31. What is the angle betw een the equatoria l pl aneand the orbital plane of polar satellite?

(a) 0° (b) 45° (c) 90° (d) 180°

32. Particles of masses 2M, m and M are respectively

at points A, B and C with AB = 1/2(BC). M is

much-much smaller than M and at time t = 0,

they are all at rest. At subsequent times before

any collision takes place

A B C

(b)

(c)

2 M M

(a) m will remain at rest.(b) m will move towa rd s M.

(c) m will move towards 2M.

(d) m will have oscillatory motion.

The escape velocity of a body from the earth

depend on

(i) the mass of the body.

(ii) the location from wh er e it is pro ject ed.

(iii) the direction of projection.

(iv) the height of the location from where the

body is launched.

(a) (i) an d (ii) (b) (ii) and (iv)

(c) (i) an d (iii) (d) (iii) an d (iv)

34. The escape speed of a bo dy on the earth's sur face

is 11.2 km s"1. A body is projected with twice of

this speed. The speed of the body wh en it escapes

the gravitational pull of earth is

(a) 11.2\/3 km s_ 1 (b) 11.2 km s"1

11 2(c) II.2V 2 km s_ 1 (d) -4=- km s_1

V2

35. As observ ed fro m earth, the sun appe ars to mov e

in an approximate circular orbit. For the motion

of another planet like mercury as observed from

earth, this would

(a) be similarly true.

(b) not be true because the force bet ween earth

and mercury is not inverse square law.

(c) not be tru e becau se the maj or gravi tatio nal

force on mercury is due to sun.

(d) not be tru e becaus e mer cur y is infl uenced byforces other than gravitational forces.

SOLUTIONS

1. (c) : Acceleration due to gravity at a altitude h

above the earth's surface is

SH=I

2h

Rr...(i)

where g is the acceleration due to gravity on the

earth's surface and R E is the radius of the earth.

Eq. (i) shows that acceleration due to gravity

decreases with increasing altitude.Acceleration due to gravity at a depth d below

the earth's surface is

1 - i -R EJ

...(ii)

Eq. (ii) shows that acceleration due to gravity

decreases with increasing depth.

Acceleration due to grav ity at lat itude A.

gx = g - RE® 2 COS 2X ...(iii)

wh er e co is the an gu lar s peed of rotat ion of the

earth.

Eq. (iii) shows that acceleration due to gravity

increases with increasing latitude.

Acceleration due to gravity of body of mass m is

placed on the earth's surface is

...(iv)gm£

Rl

Eq. (iv) shows that acceleration due to gravity

is independent of the mass of the body but it

depends upon the mass of the earth.

32 PHYSICS FOR YOU | MAY 13 j



(b):

From figure AB = BC = CD = AD = /

AC = BD = 2

Total potential energy of the system of four

particles each of mass m placed at the vertices

A, B, C and D of a square is

Gxm xmU =

Gxmxm^ ( Gxmxm

' AB )+l AC

Gxmxm

BC

Gxmxm

BD

M -

M-

AD

Gxmxm

CD

Gm2Gm

2Gm

2 f i A

Gm2

/V /

+

I ^ J+

/V

+ I

\

Z\

4 Gm2 2Gm2

i f l

Gm

2Gm („ 1

— r ®(d)

(a): Angula r mome ntu m of planet at P,

LP = m pvPr Pwhere mp is the mass of the planet.

Angular momentum of planet at A,

La = m pv Ar AAccording to the law of conservation of angular

momentum

LP = La

mvvPr P = m pv Ar A

'A

U

P

(a): Time period, T =2nR 2nR

3/2

GM m JGM„

R

where the symbols have their meaning as given

in the question.

Squaring both sides, we get

47t 2 R3

T2=-

GM„

6. (b): Accord ing to Kepler's third law

M

T2

R•MS

R6

ixES

where R MS is the mars-sun distance and R ES is

the earth-sun distance.

\ 3 / 2 R

MS

RES

:. Tm = (1.52)3/ 2x 365 days

7. (c) : Initial energy of the satellite is

GM F m£,= —' 4Re

Final energy of the satellite is

E r -

GM E m

Change in energy, AE = Ef - £,

A £ = -GM E m GM E m

\

8R E { 4Re /

GM E m GM E m GM E m

8 R E 4 Re 8Re

8. (b): A geostationary satellite goes around the

earth in west-east direction.

The time period of a geostationary satellite is

24 hours.

The angle between the equatorial plane and the

orbital plane of geostationary satellite is 0°.

9. (c) :

W- -M6 R

Let the projectile be fired with minimum velocity,

v from the surface of sphere of mass M to reach

the surface of sphere of mass 4M. Let N beneutral point at a distance r from the centre of

the sphere of mass M.

At neutral point N,GMm _ G(4M)w

r 2 (6 R-i)2

(6R - r)2 = 4r

2

6R - r = ± 2r or r = 2R or -6R

The point r = -6R does not concern us.

Thus, ON = r = 2R.




It is suffic ient to project the projectil e with a speed

which would enable it to reach N. Thereafter, the

greater gravitational pull of 4M would suffice.

The mechanical energy at the surface of M is

GMm G(4M)m1 2 — mv -2 R 5 R

At the neutral point N, the speed approaches zero.

The mechanical energy at N isGMm G(4M)m _ GMm GMm

2R 4 R 2R R~

According to law of conservation of mechanical

energy,

1 2- mv -2

Ef - E n

GMm 4 GMm GMm GMm

2 2 GMv =

R

R

4 _ 1

5 2

5 R

3 GM

5 R

2 R R

or v =3 GM Y

5 R J

,1/2

10. (b): Mean solar day is the time interval between

two successive noon when sun passes through

zenith point (meridian).

11. (b): Gravitational potential energy at any point

at a distance r from the centre of the earth is

11 =GM E m

where M£ and m be masses of earth and object

respectively.

At the surface of the earth, r = R E

U 1 = ~ GM vm Rr

At a height h from the surface,

r = R E + h = R E + R E = 2R E

U 2 =GM E m

~2Rr

Increase in potential energy is

AU=U 2- U J

GM E m

~ ~ 2 R T '

GMptn

GM E m R?

= 2 mS R E

1 | _ GM E m

2 2RP

GM,

RP2

12. (c) : Time period of satellite,

2k(R e +h) _ 2K (R E +hf' 2

T = -GME

j(R E +h)

J GM e

Squaring both sides, we get

T = -r 2_47i

2(RE +/z)

3

GMr

3 _ GM E T Z

(R E W= A 2471

(.R E +h) =GM F T

Z \l /3

h =GM F T -

4K

,\l/3

4K z -Rv

Here, M £ = 6 x 1024 kg

R e = 6400 km = 6400 x 103 m = 6.4 x 10

6 m

T = 24 h = 24 x 60 x 60 s = 86400 s

G = 6.67 x 10~u N m

2 kg-

2

On substituting the given values, we get

6 .67xl0_ 1 1

x6xl02 4

x(86400)2 ^

4x(3.14)- 6.4 x 10°

= 4.23 x 107 - 6.4 x 106 = 3.59 x 107 m

m Dfr

13. (d): m

From figure,

OA = OB = OC = OD =+ l

lisfi _ i

: 2

VGm \

OA J +

Po te nt ia l at cen tre O du e to gi ven ma ss

configuration is

GOT"| f_Gm) f Gm)

OB J+[ OC J+[ OD J

IN2 I

14. (a): The escape velocity from the surface of the

earth is2 GMC

Rr

GMr

RE

15. (b): Weight of body on the surface of the earth

= mg = 63 N

Acceleration due to gravity at height h is

§Re§h

~(R E+hf

gRl

' R 1x2 2




!

Gravitational force on body at height h is

4 4 4F = mgh = m x -g = - x mg = - x 63 N = 28 N.

16. (d): Linear momentum is not conserved.

17. (d)

18. (c) : Here, M s = 2 x 1030 kg, r = 1.5 * 1011 m

M e = 6 x 1024 kg

Let at a distance x from the earth's centre where

the gravitational forces on the rocket due to sun

and earth become equal and opposite. If m is the

mass of the rocket, then

GM E m

M E =

x2

6x10

_ GMgtn

~ (r-x)2

Mr

( r - x )2

,24 2x10

(r-x)

2

30

r-x

x

1 = 1

2x10,30

1/2

6 x 10

103

rS

24J x l O 6

,1/2

1 . 5 x l O n V 3

V^+103 V3 + 1032.6 x 10° m

19. (d):

Let P is the midpoint of the line joining the centres

of the spheres.

The gravitational potential at point P is

GM GM 2 GM 2GM 4G MV B=—

r/2 r/2

20. (a):Satellite

Perihelion1 Aphelion

Here, r A = 6RE, r P = ?RE

The eccentricity of the orbit is

e =6 R £ - 2 R £

6R£ + 2RE

/ r

21. (b ): According to law of conservation of angular

momentum

Angular momentum at perigee = Angular

momentum at apogee

mvptp = mv Ar A... =

Vp r A 6 R E 3

22. (b): F = 0 for r < R

F k 4 for r > Rr

Hence, option (b) represents the correct graph.

23. (c) : An astrona ut experiences weightlessnessin a space satellite. It is because the astronaut

experiences no gravity.

24. (a):

>x

B E

CRefer figure, ZCBO = 30°

AO= —AE = - x Zsin60°3 3

2 . S I= - x I x — = —j=

3 2 V3

BO = CO = AO = 4=v3

The angle between OC and positive x-axis is

30° and so the angle between OB and negative

x-axis. Then,Force acting on mass 2m at O due to mass m

at A is

F OA~

Gm(2m) 0 6 Gm2 *

( n S y- ]

1l -]

Force acting on mass 2m at O due to mass m

at B is

Gm(2m) a

*F ob = K

r (-1 cos 30° - ; sin 30°)( l / S ) 2




Force acting on mass 2m at O due to mass m

at C is

fo c = G m (

^ c o s 3 0 ° - / s i n 3 0 ° )(//V3)

2

Resultant force on a mass 2m at O is

f R ~F OA+F OB+F OC

6Gm 6Gm ( - i cos 30° - ; sin 30°)

6Gm(i cos 30°- ; sin 30°)

= 0.

25. (b): Total energy of orbiting satellite at a height h is

GM E m=~2(R E +h)

The total energy of the satellite at infinity is

zero.

Energy expended to rocket the satellite out

of the earth's gravitational field is

AE = - E

= 0 -

GM E m

2{R E +h)

GM E m

2 (R E +h)

26. (b): Since the speeds of the stars are negligible

when they are at a distance r, hence the initial

kinetic energy of the system is zero. Therefore,

the initial total energy of the system is

E,=KE+PEGMM \ GM 2

r 'J r

where M represents the mass of each star and ris initial separation between them.

When two stars collide their centres will be at a

distance twice the radius of a star i.e., 2R.

Let v is the speed with which two stars collide.

Then total of the system at the instant of their

collision is given by

E f = 2x ( HGMM

2 R= Mv z GM

' 2 R


energy,

Mv2-GM GM

2 R

or v z=GM

As r>> R

(2R r)GM(r-2R)

2Rr

2 GMrv = or » = ,

2 Rr

27. (a)

28. (c) : All quantit ies var y over an orbit exceptangular momentum and total energy.

GM pm29. (d): Kinetic energy =

2 r

GM cmA - s

Potential energy =

B - r

Total energy

C - p

Orbital velocity = , I M K

GM E m

D - q

30. (a): The gravita tional force is indepe nde nt of the

intervening medium. In other words the force

between two masses remains the same whether

they are in air, vacuum, water or separated by a

brick wall.31. (c) : The angle between the equatorial plane and

the orbital plane of polar satellite is 90°.

32. (c)

33. (b): The escape velocity is ind epend ent of mass

of the body and the direction of projection. It

depends upon the gravitational potential at the

point from where the body is launched. Since this

potential depends slightly on the latitude and

height of the point, the escape velocity depends

slightly on these factors.

34. (a): Let v the speed of the body when it escapes

the gravitational pull of the earth and u be speed of

projection of the body from the earth's surface.


energy,

11 2-mu2

GM E m

Rrmv2- 0

where m and M E be masses of the body and earth

respectively and R E is the radius of the earth.

or v

2 GM E m

RP

2 2V = U-2GM,

' = ^ '2

- v2 =^/(2 ve)

2-v2

i S v =11 .2V3km s"1

35. (c)

3 6 PHYSICS FOR YOU | MAY '13j



i

TARGET

PMTsPRACTICE QUESTIONSA small mass attached

to a string rotates on

a frictionless table

top as shown. If the

tension in the string isincreased by pulling

the string causing

the radius of the circular motion to decrease by a

factor of 2, the kinetic energy of the mass will

(a) remain constant

(b) increase by a factor of 2

(c) increase by a factor of 4

(d) decrease by a factor of 2

In the given network of resistances, the effective

resistance between A and B is

MA/WV4 R

(a) -R3

(b) -R3

(c) 5R (d) 8R

3. Six moles of 0 2 gas is heated from 20°C to 35°C

at constant volume. If specific heat capacityat con sta nt p re ssu re is 8 cal mol"1 K"1 and

R = 8.31 J mo H K-1 . What is the change in internal

energy of the gas?

(a) 180 cal (b) 300 cal

(c) 360 cal (d) 540 cal

[Angular momentum]4. The dimensi ons of

(a) [M3LT~

2A

2]

(c) [ML2A"

2T]

[Magnetic moment]

(b) [MA_1T~X]

(d) [M2L~

3AT

2]

are

5.

Useful forUP-CPMT,

J & K CET,

Karnataka CET,

CMC Vellore,

AMU, AHMS,

WB JEE,

MGMCET

The following diagr am indicates the energy levels

of a certain atom when the system moves from 4E

level to E. A photon of wavelength A., is emitted.

The wavelength of photon produced during its

7transition fro m - £ level to E is K 2 - The ratio -

1-

will be 3

24£

18

(a)

6.

9 4 3 7- (b) 7T (c) - (d) -4 9

w 2 3

Five forces inclined at an angle of 72° to each other

are acting on a particle of mass m placed at origin

of co-ordinates. Four forces are of magnitude

F 1 and one of F2. The resulting acceleration of

particle is

(a)F,-K1 (b) (c) (d)

f 2 - 4 F x

7.

8.

9.

m m m

Two identical piano wires, kept under the same

tension T have a fund am en ta l frequen cy of

600 Hz. The fractional increase in the tension of

one of the wire which will lead to occurence of

6 beats when both the wires oscillate together

would be

(a) 0.02 (b) 0.03 (c) 0.04 (d) 0.01

Of the following, human eyes are most sensitive

to

(a) red ligh t (b) violet light

(c) blue light (d) green light

Let A an d B be the point s respectively above and

below the earth's surface each at a distance equal to

half the radius of the earth. If the acceleration due

to gravity at these points be g A and g B respectively

then g B : g A is

(a) 1 : 1 (b) 9 : 8 (c) 8 : 9 (d) zero




10. A p-n-p transistor having AC current gain of 50 is

used to make an amplifier of a voltage gain of 5.

What will be the power gain of the amplifier?

(a) 125 (b) 178 (c) 250 (d) 354

11. Two point charges +8q and -2cj are located at

x = 0 and x = L respectively. The location of a

point on the x axis at which the net electric field

due to these two point charges is zero is

(a) 4L (b) 8L (c) J (d) 2L

12. The concept of temperature to measure hotness

or coldness of a body is consequence of

(a) Joule's law

(b) first law of thermodynam ics

(c) Newton 's law of cooling

(d) zeroth law of thermodynamics

13. A hollow copper tube of 5 m length has got

external diameter equal to 10 cm and its walls

are 5 mm thick. The specific resistance of copperis 1.7 x 10~

8 Q m. The resistance of the copper

tube is approximately

(a) 5.6 x 10-3 Q (b) 5.6 x 10~

9 £2

(c) 5.6 x 10-5 Q (d) 5.6 x 10"7 £2

14. A particle is dropped from point A at a certain

height from ground. It falls freely and passes

through three points B, C and D with BC = CD.

The time taken by the particle to move from B

to C is 2 s and from C to D is 1 s. The time taken

to move from A to B is

(a) 0.5 s (b) 1.5 s (c) 0.75 s (d) 0.25 s

15. A car is moving on a circular level road of

curvature 300 m. If the coefficient of friction is

0.3 and acceleration due to gravity is 10 m s~2,

the maximum speed of the car can be

(a) 90 km Ir1 (b) 81 km hr

1

(c) 108 km Ir1 (d) 162 km h"

1

16. The gas in a vessel is subjected to a pressure of

20 atmosphere at a temperature 27°C. The pressure

of the gas in the vessel after one half of the gas

is released from the vessel and the temperature

of the remainder is raised by 50°C, is

(a) 8.5 atm (b) 10.8 atm

(c) 11.7 atm (d) 17 atm

17. Same current I is flowing in three infinitely long

wires along positive x, y and z directions. The

magnetic field at a point (0, 0, -a) would be

(a)

(c)

i V .

In a

j V27t a

a - i)

(i-i)

(b)

(d)

M2t za

M

' 2na(i + j+rk)

18. A projectile is fired at an angle of 30° to the

horizontal such that the vertical component of its

initial velocity is 80 m s_1

. Its time to flight is T.

TIts velocity at f = — has a magnitude of nearly

4

(a) 124 ms"1 (b) 134 ms"

1

(c) 144 m s"1 (d) 154 m s"

1

19. Two identical thin rings, each of radius a areplaced coaxially at a distance a apart. Let charges

Qi and Q2 be placed uniformly on the two rings.

The work done in movi ng a charge c\ from the

centre of one ring to that of the other is

(a) zero (b) ^ - ( Q i - Q z )

(c)<7(^2-1)

47ie0fl

<?(V2-1)

47I£,Q2) ( ) 4KE q U (Q1-Q2)

20. A magnet of length 0.1 m and pole strength

10"^ Am is kept in a magnet ic f ield of 30 Wb m~2

at an angle of 30°. The couple acting on it is(a) 7.5 x 10-4 N m (b) 1.5 x 10"4 N m

(c) 4.5 x 10"4 N m (d) 6.0 x 10"

4 N m

21. A point performs simple harmonic oscillation of

period T and the equation of motion is given by

x = asin(o)f + rt/6). After the elapse of what fraction

of time period the velocity of the point will be

equal to half of its maximum velocity?T T T T

(a) - (b) — (c) - (d) i3 12 8 6

22. The angular momentum of an electron in the

hydrogen atom is —. Here, h is Planck's constant .2tc

The kinetic energy of this electron is

(a) 4.35 eV (b) 1.51 eV

(c) 3.4 eV (d) 6.8 eV

23. For the arrangement shown in the figure, the

tension in the string is

[Given : tan"1 (0.8) = 39°]

m = 1 kg

(a) 6 N (b) 6.4 N

(c) 0.4 N (d) zero

24- The voltage time (V-t) gra ph for a triangular wave

having peak Value Vj) is as shown in figure. The

rms value of V is




(a)

Vn Vn(b) y (c) (d)

Ys.

s

25. Four metallic plates each with a surface area of

one side A are placed at a distance d from each

other as shown in figure. The capacitance of the

system is

0» ^ ( 0 ^ („ , ^d d

26. An object is displaced from position vectorA A A A

rx = (2i + 3j)m to r 2 = (4i + 6 ;) m und er a forcer\ A A

F = (3x i + 2y j) N. The work done by this force

is

(a) 63 J (b) 73 J (c) 83 J (d) 93 J

27. If the velocity of the particle is increased three

times, then the percentage decrease in its

de Broglie wavelength will be

(a) 33.3% (b) 66.6%

(c) 99.9% (d) 133.2%

28. The electrical analog of a spr ing cons tant k is

(a) L (b) - (c) C (d)1

L - , C29. If po wer d iss ipate d in the 9 £2 resi stor in the

circuit shown is 36 W, the potential difference

across the 2 Q resistor is

90

-WA-

6 £1-VWV-

a

WA-—IH'v

(a) 4 V (b) 8 V (c) 40 V (d) 2 V

30. A steel ball is dr opped on a har d su rface from a

height of 1 m and rebounds to a height of 64 cm.

The maximum height attained by the ball after

nth

bounce is (in m) (}-\j > 9 m i t

(a) (0.64)2™ IOf

(c) (0.5)2

ai fchmMf,-(d) (0.8)"

31. Four simple harmonic vibrations x t = 8sincof,

x3 = 4sin( cot + rt) an d x2 = 6si n cot h—2

n • I 3n x4 = 2 sin | at + — are sup eri mposed on each

other. The resulting amplitude and its phase

difference with xt are respectively

(a) 20, tan- ' j i j (b) 4 ^ 2 , -

(c) 20, tan"1 (2) (d) 4V2, -4

32. 200 g of a solid ball at 20°C is dropped in an

equal amount of water at 80°C. The resulting

temperature is 60°C. This means that specific

heat of solid is

(a) one-fourth of water (b) one-half of water

(c) twice of water (d) four times of water

33. The length of a potentiometer wir e is I. A cell of

emf e is balanced at a length 1/5 from the positiveend of the wire. If length of the wire is increased

by 1/2, at what distance will the same cell given

a balance point.

(a) —I (b) (c) —I (d) —115 15

w 10 10

34. A uni form thin bar of mass 6m and length 12L is

bent to make a regular hexagon. Its moment of

inertia about an axis passing through the centre of

mass and perpendicular to the plane of hexagon

is

(a) 20mL2 (b) 6ml 2

(c) — ml1 (d) 30mL2

535. A ray of light passes thr ough an equilateral prism

such that angle of incidence is equal to the angle

of emergence and the latter is equal to [ 3

of prism. The angle of deviation is

(a) 45° (b) 39° (c) 20° (d) 30

0 I angle

36. A light rod of length 2 m // // // // // // // // // // /

Brass

suspended from the ceiling

horizonta lly by mean s of

two vertical wires of equal

length. A weight W is hungfrom a light rod as shown

in figure.

The rod hung by means of steel wire of cross-

sectional area A1 = 0.1 cm2 and brass wire of

cross-sectional area A z = 0.2 cm2. To have equal

stress in both wires, —1 _

(a) i; 3+(b) (c) (d)




37. In the circuit shown below, the key K is closed

at t = 0. The current through the battery is

•K

-Troootnr-

Ri

(a)

(b)

(c)

(d)

— at t Rn

0 and

Vat t — 0 and

-WA

V(Ri + R2)

R,R2

VR, R2

at f = <*>

JrI+RI

at t = oo

v(.Ri + R2)

R L + R2

at t = 0 and atf =

VR, R, , v1 — a t t = 0 and — at t = °°

Jrf+Rl R->

38. A force of 200 N is requir ed to push a car of m ass

500 kg slowly at constant speed on a level road.

If a force at 500 N is applied, the acceleration of

the car will be

(a) zero (b) 0.2 m s"2

(c) 0.5 ms"2 (d) 1.0 m s

2

39. A particle of mass 1 x 10~26

kg and cha rg e

1.6 x 10~19

C travelling with a velocity 1.28 x 106 m s"

1

along the positive x-axis enters a region in which

a uni for m electric field £ and a unifo rm magneticfield B are present. If E = -102.4 x 103 k N C"' and

-7 A -9

B = 8 x 10 j Wb m the direction of mot ion of

the particle is

(a) along the posi tive x-axis

(b) along the nega tive x-axis

(c) at 45° to the posi tive x-axis

(d) at 135° to the positive x-axis

40. The power obtained in a reactor using U235

disintegration is 1000 kW. The mass decay of

U235 per hour is

(a) 10 ^g (b) 20 ng (c) 40 ng (d) 1 ng

41. A charge Q is enclosed by a gaussian spherical

surface of radius R. If the radius is doubled, then

the outward electric flux will

(a) increase four times (b) be reduced to half

(c) remain the same (d) be double d

42. A body is projected vertically upw ar ds from the

surface of a planet of radius R with a velocity

equal to the escape velocity for that plane t.

The maximum height attained by the body is

R R(b) T (c)

, . R(a) —

' 2(d) ?3

v ' 5

43. A metal rod of Young's modulus Y and coefficient

of thermal expansion a is held at its two ends such

that its length remains invariant. If its t emper ature

is raised by f°C, the linear stress developed in it is

(a) - f (b) ^ (c) Yat (d)

Y at (Yat)

44. If T denotes the temperature of the gas, the vo lume

thermal expansion coefficient of an ideal gas at

constant pressure is

(a) T (b) T2 (c) | (d) ~

45. Water rises in a capillary tube to a height of

2 cm. In another capillary tube whose radius is

one third of it, how much the water will rise ?

(a) 2 cm (b) 4 cm (c) 6 cm (d) 8 cm

46. Energy required to break one bond in DNA is(a) 10-

10 J (b) 10~

18J (c) 10~

7J (d) 10~

20 J

47. A ray of light strikes a silvered surface inclined to

another one at an angle of 90°. Then the reflected

ray will turn through

(a) 0°

45°

90°

180°

(b)

(c)

(d)

M,

M,

48. An open and a closed pipe have same length.

The ratio of frequencies of their nth overtone is

n + 1(a)

n+1

2n+l(b)

2(w+l)

2n+l(c)

2n+l(d)

2 n

49. An optical fibre communicat ion system works on a

wave leng th of 1.3 jim. The numb er of subscribe rs

it can feed if a channel requires 20 kHz are

(a) 2.3 x 1010

(b) 1.15 x 1010

(c) 1 x 105 (d) 2.3 x 1014

50. A gas is expa nded from volume V 0 to 2V„ under

three different processes as shown in the figure.

Process 1 is isobaric process, process 2 is isothermal

and process 3 is adiabatic. Let A(ij, Aii2 and AU3

be the change in internal energy of the gas in

these three processes. Then

P41

4 0 PHYSICS FOR YOU | MAY'13 ?



(a) Af ij > AU 2 > All 3 (b) ALT! < AU 2 < AU 3(c) AU 2 < ALZj < AU 3 (d) AU 2 < AU 3 < AU,

SOLUTIONS

(c) : Kinetic energy of rotation,

K = H O ^ I ^ ) 2 - 1 * 2

2

1 L

2 I 2 I 2 (mr 2)

(v L = 7(0)

K •

K'

1 L .- j (As is constant dur ing rotation)

• = 4* (r/2)

2

or K' = 4K

Thus kinetic energy of mass increases by a factor

of 4.

(a): MAAA/V . ARC

In the g iven network CDBE is a ba la nc ed

Wheats tone bridge. Thus, the resistance connec ted

across DE beco mes ineffective. The e quival ent

circuit is as shown in figure.

MMMf A R

•-AMAAr•A R

MAAAAr A R

2 R-WWV-

2 R-<WWV-

-WAAr-. 2R

-JWV\ArB

Hence, the equivalent resistance between A and

B is

q 3 3

(d): Here, N = 6, C P = 8 cal mol 1 K 1

R = 8.31 J mol-1 K"1 = 2 cal mol"1 K"1

As C P - C V = R

6.

7.

Cy Cp - R

= 8 cal mol"1 K"1 - 2 cal m o H K-1

= 6 cal mol"1 K 1

Change in internal energy of the gas is

AU = nC v AT = 6 x 6 x 15 = 540 cal

4. (b):[Angular momentum]

[Magnetic moment]

2m

2m ' M "

It AT

5.

= [MA"1 T"

(b): Transition fr om 4 E to E

(4 E- E ) =he

or A,, =he

3 E

Transition from — E to £3 A. 2

he 3hc4 E

:..(i)

— 4E

— £

...(ii)

Divide (i) by (ii), we get

V _ 4

?l2 ~ 9

(a): Acc ordi ng to polyg on law resu ltant of four

forces, each of F 1 acting at 72° is along the fifth

side of polygon taken in opposite order. As F 2 is

acting along this side of polygon, therefore, net

force on the particle = F 2 - F v

F -F.•. Acce lerat ion = — — -

m

(a): Fund ame nta l f requenc y prod uced by a

stretched string is given by

\> = = k\J r (As I an d| i are kep t constant)

dv

v

ldT

2 T

dT „dv „ 6or — = 2 — = 2 x

T V 600

(d)

(b):

• 0.02

Earth's surface

Acceleration due to gravity at point A is

PHYSICS FOR YOU I MAY '13 41



£4 =

f h ]2

f RF )1+ - l +

{ ><E) 2RFV t /

_ 42 V

...(i)

where g is the acceleration due to gravity on the

earth's surface.

Acceleration due to gravity at point B is

1 R

2

...(ii)

E

J k .2Rr

= £2

Divide (ii) by (i), we get

K B=9-

8A 8

10. (c) : The power gain of the amplifierPower gain = Current gain x Voltage gain

= 50 x 5 = 250

11. (d): Let the electric field due to these tw o po int

charges be zero at poin t P at a distance x from

+8 q - 2 q

Ax = 0

Bx = L

M x-

P

-H

L (81) I

1 ( -

2? ) Q

47t£0(L+x)2 4to0 x2

1 (8(7) 1 2 qor ——— = —

4rce0 (L+x)2 47te0 x

2

2 _ 1

L+x x

2x = L + x or x = L

:. AP = AB + BP = L + L = 2L

12. (d): Zero th law of the rmo dyn ami cs leads to theconcept of temperature.

13. (c) : Area of cross-section of the tube is

A = 7t[(5 cm)2 - (4.5 cm)2]

= 4.7571 cm2 = 4.75rt x 10"

4 m

2

Resistance of the tube is

R pi __ 1. 7x10 flmx5m

4.75 x 3.14 x 10 -4 m 2

= 0.56 x 10"4 £2 = 5.6 x 10"5 Q.

1 4 . ( a ) : Let AB = y, BC = CD = h and t AB * t

As per question • ; n

y-\gt 2

y + h = ±g(t+2)2

and y + 2h = ±g(t + 3)2

Solving these three equations, we get

t = 0.5 s J

- D

15. (c) : Here, R = 300 m

p. = 0.3

g - 10 m s~2

The maximum speed of the car is

®max= \t^Rg = a/0-3 X 300 m x 10 m s~2

= 30 m s _ 1 = 30 x — km h" 1 = 108 km h"1

5

16. (c) : According to ideal gas equation

m

PV = — RT M

...(i)

...(ii)

As per question

20 x V = — R x 300 M

p , x V = (jnl2)Rx350

M


P' 175 175 „„— = or P' = x 20 = 11.7 at m20 300 300

17. (a): Point (0, 0, -a) lies on z-axis. Therefore,

magnetic field due to current along z-axis is zero

and due to rest two wires is - - i n mutually2ti a

perpe ndicula r directions along positive indirection

and negative x-direction.

Un / A A

271 a

18. (c) : Vertical component of initial velocity,My = usin30°

u y 80 - ior u = — - — = = 160 m s

sin 30° ( 1/ 2L Q- '(••• Uy =! 80 m s (Given))

Horizontal component of initial velocity,

J 3u = wcos30° =p 160 x — = 80V3m s" 1

2Time of flight,

^ 2u sin 0 2 x 160 x sin 30°T = = = 16 s

X 10

i - — r - : 4s> ni?'4




Let v be the velocity of the projectile at t = —.

Its horizontal and vertical components are given

by

v x = u x = 80 V3 m s"1

v y = M y - gt = 80 - 10 x 4 = 40 m s"1

Its magnitude is given by

v = y l

vx

+ vy =>/( 80V3 f + (40)

2

= 40 7l 2 + l = 40 Vl3 = 144 m s"1

19. (c) : The electrostatic potential at the centre of

the first ring (i.e., at O) with charge Q, is due to

charge Q, itself as well as due to charge Q2 on

the second ring which is given by

Qi

v,= -r r4tc0 a 47te0 a-42

Similarly, the electrostatic potential at the centre

of the second ring (i.e., at O') is given by

V 2 = ——+ —47re0 a-J2 47te0 a

Required work done , W = q( V^ - V 2)

W = q \ — & — — ^

[47ie0 a 4 t i e 0 a\l2 47ce0 a\l2 4ne0 a

471 e0a

4ne0a

42Qx+Q1-Q,-42Q1

4~2

4718,

1 —1t[42(Q1 -Q2) - l (Qj -Q 2 ) ],«v2

qU 2

4nena\f2 (Q1-Q2)-

20. (b): Here, 2/ = 0.1 m, m = 10-4

A m

B = 30 Wb m 2, 6 = 30°, x = ?

x = MBsinO = m(2/)Bsin6

= 10~4 (0.1) x 30sin30° = 1.5 x l ( H N m

. f ( n )in cot + -

I 6 j21. (b): x = asm

V o ,

Velocity = — = — asinfraJ->-i-™4=a(ocos[ <at + y dt dt { & f)

Maximum velocity = am

As per question

ac0 [ K — = fl(OCOS cot + —

2 6

or cos (01 +7C

or cot + — = 60° = —

E

rad6 6

2n n n 2K K Tor cot = = - or — f = - or f = -

6 6 6 T 6 12

3 h22. (b): — = n

In

:. n =3

The kinetic energy of the electron in nth orbit is

n 2n

K, = eV= ^ eV = 1.51 eV13.6 w 13.6

eV = -—3Z 9

23. (d) : If a represent s angle of repose, then

tana = 0.8

a = tan-1

(0.8) = 39°

The given angle of inclination is less than the

angle of repose. So, the 1 kg block has no tendency

to move.

Note that mg sinO is exactly balanced by the force

of friction. So, T = 0

24. (d): = jf for 0 <t<~

16VnV2 =

< v > 0-

T =

4 T/4

dt

r -,T/4

16 V2t 3

j 2 3L J11

WT

0'4

VS

25. (b):

The equivalent circuit of the given network is as

shown in the figure.




1 1

4' '3

The capacitance of each capacitor is

d

Here, the two capacitors are connected in parallel.

Hence, the equivalent capacitance between

A and B is CAB

2 e0 A

26. (c):

h (4-6)

W = j? dr= J (3x2i + 2y])-(dxt + dyj + dzk)

h (2,3)

(4,6)

= J (3x2dx + 2ydy) = [x3 + y2]^)

(2,3)

= [(4)3 + (6)

2 - (2)

3 - (3)

2] = 83 J

27. (b): X = — and A,' = — = - Xmv m3v 3

% decrease in de Broglie wavelength

X - X'• x 100- H x 100

= 11 - ^ |x 100 = 66.6 %

28. (d): The analogy between mechanical and

electrical quantities is as shown in the table.

Mechanical system Electrical system

Mass m Inductance L

Force constant k1

Reciprocal capacitance —

Displacement x Charge q

Velocity v = — y dt

^ , daCurrent i = —-dt

Mechanical energy

1 1 1 2 E = —kx+—mv

2 2

Electromagnetic energy

U = —— +—LI 2

2 C 2

9 P29. (c) : As, P = I Z R or I = J*

V R

Current through 9 Q resistor in figure given

below is

( h +

h )

A

h

9Q-AAAAr

6Q-AAAAr

V

2Q

-J VWV-

(h + h)

h=36 W

\ 9Q= 2 A

As resistors 9 £2 and 6 £2 are connected in parallel,

therefore, potential difference

V A-V b=9I 1=6I 2

or I2 = -Jj = ~x2 A : 3 A6

1 6

Current drawn from the battery

= I1 + 72 = 2 A + 3A = 5 A

Potential difference across 2 £2 resistor

= (5 A) (2 £2) = 10 V

30. (b): When a ball is dro ppe d fro m a height h and

it rebounds to a height hlf then

Here, h = 1 m, h x

10.64 me =

l m

= 64 cm = 0.64 m

0.8

The maximum height attained by the ball after

n[h bounce is

h„ = e2"h = (0.8)2" (1 m) = (0.8)

2" m

31. (d): The resulting ampli tude and cor responding

phase difference can be calculated by vector

method as follows:

A, = 6

/St = 8

IA X = 8 - 4 = 4 and IA y = 6 - 2 = 4

Therefore, resulting ampli tude is and phase

Kdifference with x1 is <f> = —.




t

32. (b): Heat lost by water is

Qi = 200 x swaler x (80 - 60) = 200 x Swaler x 20

Heat gain by solid ball is

Q2 = 200 x Ssol;d x (60 - 20) = 200 x sS0lid x 40

According to principle of calorimetry

Qi = Qi

200 x Swater x 20 = 200 x Ssolid x 40

= 1''solid — 2 ^water

33. (c) : In first case,

Potential gradient, K =

where e0 is the emf of the battery in potentiometer

circuit.

As per question

...(i)ry l EOe = K - = - y - x - = -u-

5 1 5 5In second case,

I 31Length of potentiometer wire = I + — = —

Potential gradient, K' = £

° =-^08

(31 / 2) 3 IIf V is the new balancing length, then

e = K'l' = ^ x l '31

Equating (i) and (ii), we get

...(ii)

^n 2 £n „ „ 3 ,•JL = —— x / or I = — /5 3 1 10

34. (a): Leng th of each side of hexagon = 2 L

Mass of each side = m

Let O be centre of mass of hexagon.

Therefore, perpendicular distance of O from

each sider = Ltan60° =lV3

O

The desired moment of inertia of hexagon about

0 is

1 — 6 [lone side] \2

= 6

= 6

= 6

m(2L)' 2+ mr

12

mL-mi<LV3)2

+ 3 mL1

3= 20 mL

35. (d): i = e = -A

4

where i is the angle of incidence, e is the angle of

emergence and A is the angle of prism.

i = - x 60° = 45°4

When i = e, prism is in min imu m deviation

position,

.-. 8m = 2i - A = 2 x 45° - 60° = 30°

36. (d): Given,

Stress in steel wire = Stress in brass wire

Ti

\ A2

T 2 a2

01

0.2

37. (a): When key K is closed at f = 0, the current

thro ugh induct ance L will increase with time,

resulting induced emf across L, which will opposethe current through the arm having inductance

L. Therefore, the current from battery will flow

through R2. This current is / = - at t = 0. R2

When t = °°, the current will reach to steady state

in the circuit. The presence of L will become

ineffective. Now effective resistance of circuit

R1 + R2

Current through battery,

V(R1+R2)r = -

R1 R2 /(R L + R2)

Net force

R,R2

38. (c): Acceleration =Mass

_ (500-200) N

500 kg

= - m s~2 = 0.6 m s~2

5




39. (a): Here, m = 1 x 10~26

kg

q = 1.6* 10"19 C

1.28 x 106 i m s

_ 1

£ = -102.4 xl03f c N C

_ 1

B = 8x l 0 " 2 / W b m " 2

Force on a charged particle in a uniform electric

and magnetic fields is

F = cjE + q(v x B)

= q(E + vxB)

= (1.6 x 10~19)[(-102.4 x 103 fc)

+ (1.28 x 106 i x 8 x 10~

2 /)]

= (1.6 x 10"19)[(-102.4 x 103 k + 102.4 x 103 k )]

= 0

Acceleration of the particle , a = - • 0

Hence, the particle will move along x-axis.

40. (c) : Acco rdi ng to Eins tein 's mass e ne rgy

relation

E•• mc

cor m = -

Mass decay per second

Am _

~AF~

10'

J_ A E _ I

cz Af ~ c

„-l

1000 x10J W

(3 x 108 m s""

1)2

9x10,16

kgs~

Mass decay per hour

Am

At• x 60 x 60 =

10°

9x10 ,16kgs - 1 (3600 s)

= 4 x 10"8 kg - 40 x 10"

6 g = 40 fig

41. (c) : According to Gauss's law, the total ou twa rd

electric flux linked with gaussian surface

<))£ = — x charge enclosed by sur face.eo

If the radius of the gaussian surface is doubled

the total outward electric flux will remain the

same as charge enclosed by the guassian surface

is unchanged.

42. (d): Escape velocity from planet 's surface is

12 GM

where M and R be the mass and radius of the

planet respectively,

The velocity of projection of the body from the

planet's surface is

1 1 2 GMu = —v„ = — , /

3 3V R

According to the law of conservation of mechanical

energy

Total energy on planet's surface = Total energy

at maximum height h

GMm j_Q | ^ GMm)1 2 —mu +2

1 — m2

\ 2 GM

3 v R

I - i — L .9 R + h

R + h )

GMm GMm

R R + h

or h R

43. (c) : Due to change in temperature t°C, increase

in length,

Al = lat orA/

I= at

Y stress stressstrain AI / /

Al:. Stress = Y x — = Yat

44. (c) : According to an ideal gas equation

PV = nRT

:. PdV = nRdT (P = constant)

dV = |— | dT

or dV = \ Y\dT ...(i)

The volume coefficient of expansion is given by

= dV J ~VdT

dV = yVdT ...(ii)

Equating (i) and (ii) we get

Y =

45. (c) : h =

I

T

2Tcos9

rpg

, 2Tcos6hr = = constant

V i - h2r 2 or h2 =

Subst ituting the given values, we get

h2 = (2.0)(3)

= 6 cm

^ 2 - 1




46. (d): The energy required to break one bond in

DNA is 10"20

J.

47. (d): For mirror M v Zi = 0°

Zr = 0° i.e. the reflected ray would retrace

its path turning through 180°. Mirror M2 has no

effect.

48. (b): Let I be the length of the pip es and v the

speed of sound.

Then frequency of open organ pipe of « th overtone

is

vv' n=(n + 1)

21

and f requency of closed organ pipe of nth

overtone

is

The desired ratio is

•»„ = 2(n + 1)

x>' In +1

49. (b): Optical source frequency,

v _ c _ 3 x 10 m s4

X 1.3 x 10"6m

= 2.3 x 1014

Hz

Number of channels or subscribers

- 2 3 X l

° " =1 .15x10"20 x 103

50. (a): Proc ess 1 is iso bar ic (P = const ant )

expansion.

Hence temperature of gas will increase.

A U 1 = positive

Process 2 is an isothermal process.

A U 2 = 0

Proc ess 3 is an adiab ati c expan sio n. Hence

temperature of gas will fall.

A(J3 = negative

.-. AUj > AU 2 > AU 3.

PHYSICS FOR YOU j MAY '13 47



^ O i i g y l ^ C L ^ ^ I E l e c

t r o s t a t i c s

^

w n

i * i i > i

i n

i i w —

. • • • • > .

• f = z ^ Y .

S u p e r p o s i t i o n o f

H E l e c t r i c C h a r g e s !

. £ = ^ ( b e t w e

e n t w o p a r a l l e l

• F i =

F i 2 +

F i 3 + J r i 4 + - +

J : 1 N

S

c o n d u c t o r s )

• F = J F

1 2

+ f 2 + 2 r F o s 0

[ E l e c t r i c P o t e n t i a l |

—

H e r e , F 1 2 , f 1 3 , F 1 4 f o r c e s e x e r t e

d

• P o t e n t i a l d i f f e r e n c e

E l e c t r i c P o t e n t i a l E n e r g y

o n c h a r g e q 1

b y t h e i n d i v i d u a l

W

\ a r g e s q 2 , q 3 . . . q N r e s p e c t i v e l y .

V = y

, E l e c t r i c p o t e n t i a l e n e r g y o f .

1

• E l e c t r i c

p o t e n t i a l d u e t o

c h a r g e q

a t

s y s t e m o f t w o p o i n t c h a r g e s ,

d i s t a

n c e r f r o m i t .

\

„ n

1

q

U =

n

j A

C o u l o m b ' s L a w

f ~

V = 4 m ~ f

4 n e ° r y i

• F r v a c i u m ^ 1 M ,

• E l e c t r i c p o t e n t i a l a t Y p o i n t d u e t o a

• E l e c t r i c p o t e n t i a l

e n e r g y o f

a

• i - ( v a c u u m ) -

^ r

r

s y s t e m o f N p o i n t c h a r g e s ,

4 J I E Q

f

r

n • c

j i

1

q q

1

1 p c o s 8

( j = — i_

£

. F ( m e d i u m ) = — - j -

1

r 2

4 7 t e 0 a l l p a i r s r j k

, .

0

r

• P o t e n t i a l e n e r g y o f a n e l e c l r i c

d i p o l e i n a u

n i f o r m e l e c t r i c f i e l d ,

U = - p

E ( c o s 6 2 — c o s 9 } )

• I f i n i t i a l l y t h e d i p o l e i s p e r p e n d i -

Q u a n t i s a t i o n o f C h a r g e s

- .

c u l a r t o t h e f i e l d E , 0 , =

9 0 ° a n d

r - — —

n . (

E l e c t r o s t a t i c s )

e 2 = e , t h e n

;

• q

= n

e

4 — E l e c t r i c C h a r g e s

_

• M a s s t r a n s f e r r e d d u r i n g

S M ^ ^ H M ^ ^ ^ ^ .

^ ^ " H B m m a m ^ * ^ ^

= - p

b c o s t ) = - p - h

c h a r g i n g = m e x n

• I f i n i t i a l l y t h e

d i p o l e i s p a r a l l e l t o

t h e f i e l d E v B j = 0 ° a n d 0 2 = 6 , t h e n

U = - p

E (

c o s G - 1 )

r j

E l e c t r i c

F i e l d

k

r / 1 m

\

J

= p E ( 1 - c o s e )

_

c £

= - < 7 o

*

~ j g

r —

E l e c t r i c F i e l d s o f P o i n t

C h a r g e s

E l e c t r i c F l u x a n d G a u s s ' s T h e o r

y

1

1

.

£ = — t

• E l e c t r i c f l u x t h r o u g h a p l a n e s u r f a c e a r e a S h e l d i n a u n i f o r m

e l e c t r i c f i e l d

4 7 t £ o r

,

c_

r c

a

• B y t h e p r i n c i p l e o f s u p e r p o s i t i o n , e l e c t r i c f i e l d

—

c o s o

d u e t o a n u m b e r o f ^ o i n t c h a r

7 e s

d i

l

l

i

f



• F l u x d e n s i t y

• V o l u m e c h a r g e d e n s i t y

d a

,

P = d V

—

A p p l i c a t i o n o f G a u s s ' s L a w

1

• S u r f a c e c h a r g e d e n s i t y

d q

• E l e c t r i c f i e l d o f a l o n g s t r a i g h

t w i r e o f u n i f o r m l i n e a r c h a r g e d

e n s i t y X

C T

~ ^ s

£ = — — —

• L i n e a r c h a r g e d e n s i t y

2 7 1

E 0 r

d a

w h e r e r

i s t h e p e r p e n d i c u l a r d i s t a n c e o f t h e o b s e r v a t i o n p o i n t

f r o m t h

e w i r e .

X = - j j -

^

• E l e c t r i c f i e l d ^ o f a n i n f i n i t e p l a n e s h e e t o f u n i f o r m s u r f a c e c h a r g e d e n s i t y

C T ,

. F o r c e e x e r t e d o n a c h a r g e %

d u e t o a c o n t i n u o u s

^ =

c h a r g e d i s t r i b u t i o n

• E l e c t r i c f i e l d o f t w o p o s i t i v e l y c h a r g e d p a r a l l e l p l a t e s w i t h c h

a r g e d e n s i t i e s c t j a n d

C T 2

F =

i — r

s u c h t h a t c 1 > C T

2 > 0 ,

4 T O

0 J r 2

„

, 1 .

,

, . , „ . , ,

• E l e c t r i c f i e l d

d u e t o c o n t i n u o u s

c h a r g e

E = ± - —

( 0 , +

C T , )

( O u t s i d e t h e p l a t e s )

,

&

2 e 0 K

1

2 J

t

d i s t r i b u t i o n

£ =

( I n s i d e t h e p l a t e s )

i =

• E l e c t r i c f i e l d o f t w o e q u a l l y a

n d o p p o s i t e l y c h a r g e d p a r a l l e l p

l a t e s ,

E = 0 a

( F o r o u t s i d e p o i n t s )

D i p o l e M o m e

n t ,

£ =

^

( F o r i n s i d e P

° i n t s )

D i p o l e F i e l d a n d

T o r q u e

o n a D i p o l e

• E l e c t r i c f i e l d o f a t h i n s p h e r i c

a l s h e l l o f c h a r g e d e n s i t y a a n d r a

d i u s R ,

_

I

a

•

D i p o l e m o m e n t , p = ^ x 2 a

E =

F o r r > R

( O u t s i d e p o i n t s )

• D i p o l e f i e l d a t a n a x i a l p o i n t a

t d i s t a n c e r f r o m

4 7 t e o

r

t h e c e n t r e o f t h e d i p o l e

E = 0

F o r r < R

( I n s i d e p o i n t s )

j 2 ^

E = — — % r

F o r r = R

( A t t h e s u r f a c e )

£ a x i a i = 4 £ , 2_ 2 , 2

4 T O o

R 2

W h e n r » a

°

°

H e r e

q = 4 n R 2 a .

1 2 p

• E l e c t r i c f i e l d o f a s o l i d s p h e r e o f u n i f o r m c

h a r g e d e n s i t y p a n d

r a d i u s R

a x i a i ~ 4 n e o r

3

*

1

q

E =

" ~ 2

F o r r > R ( O u t s i d e p o i n t s )

.

D i p o l e f i e l d a t a n e q u a t o r i a l p o i n t a t d i s t a n c e r

0

r r

f r o m t h e c e n t r e o f t h e d i p o l e i s

E =

F o r r < R

( I n s i d e p o i n t s )

1

p

4 f 0 R

^ e q u a t o r i a l = 4

2

2 3 / 2

£ = — •

4

F o r r = R

( A t t h e s u r f a c e )

w h e n r » «

}

4 t o 0

R 2

r

= J

F

4

3

e q u a t o r i a l

^

H e r e

q = ~ n R p

.

T o r q u e , x = p £ s i n 6 .

, „ „ „



THE LAWS OF CONSERVATION

1.

2.

3.

4.

5.

Conservation of energy and conservation of

mass are the corner stones of classical physics.

Demonstrate this in the formula.

v2-u

2 = las

This formula obeys

(a) conse rvation of mas s

(b) conse rvati on of ene rgy

(c) conse rvati on of kinetic energy

(d) conse rvati on of potentia l energy

A an d B are two identical capacitors.If A has a charge Q and now it is

connected to B which has no initial

charge. Com par e the initial energ y ^ g

with the final energies of A and B.

Choose the correct statement.

(a) The initial ene rgy is equal to the final energi es

of the capacitors.

(b) The initial ene rgy is grea ter than the total final

energy.

How much energy is lost on sharing the charges?

Two capacitors are connected as shown in theproblem (2) to share charges. The laws which are

followed here are

(a) law of conse rvat ion of charge

(b) law of conse rvat ion of mass

(c) law of conse rvat ion of energy

(d) none of these

The essential differences between emission of

X-rays and y-rays are

(a) Produc tion of X-rays such as K w K t > rays are

(b)

spontaneous.

X-rays are mostly by the excitation or ionisationof the inner level electrons.

(c) y-rays are spont aneo usly pr odu ced .

(d) The element is change d to anoth er element in

the production of X-rays.

6. In p+-decay of Na,

(a) elect ron is emit ted (b) po sit ron is emit ted

(c) a negative neu trin o is genera ted

(d) a positive neu trin o is generate d

(e) it is spon tan eous emission.

Neutrinos are neutral particles of negligible mass.

The neutrino is emitted to conserve

(a) mass

(b) charge

(c) energy

(d) none of these

The life-time of neutrinos are very small. Their

mas s is negligible comp are d to even the electrons.

How is it that one has discovered neutrinos

produced by solar emissions when the distance isvery large?

We have seen that in higher physics, there are

many conservation laws. How is it in classical

physics we find that there is conservation of mass

and energy?

SOLUTIONS

(a,b) : (a) is tru e for classical physics. Mass, length

and time do not change due to changes in energy

in kinematics.

(b) v2

-u2

= las ...(i)

Multi plyin g by 1 m in equa tion (i)

1 2 1 2 1 „=> —mv —mil =—mlas

1 1 1

=> Final K.E. - initia l K.E. = Force * distanc e.

The difference in energy is the work done.

Conservation of energy is given in this formula.

(b): Let the initial charg e be Q. As there is

conservation of charges, the final charges are q on

each capacitor = j

r • • , 1 Q2

Initial ene rgy = -

1 q2 1 q

2

The total final energy = — — + — —

=> Total energy =

The total final energy =(Q/2f

C1914 C

5 0 PHYSICS FOR YOU | MAY' 13

}



3. Initial energy = ——

Total final en ergy = — = — I '•' q = ^6 > C 4C I 2 J

1Q2 O

2 1Q

2

Energy lost = — — - — = — —

4. (a,c): (a) Charge is conserved.

Contrary to general feeling, energy is also

conserved. The initial energy before charges are

shared is more than the sum of the final charges.

But the difference in energy is used to pump

the charges from one charged capacitor to the

uncharged one. (c) is true. This is just like water in

a tank flowing to a tub on the ground. The higher

potential energy of water makes it flow to the

ground.

5. (b,c): Norm ally X-rays are pr odu ce d by the

ionisation or excitation to higher empty levels, of

electrons from the inner shell (b).

Y-rays are spontaneously produced (c).

X-rays production does not change the element.

It is the production of y-rays that the element also

changes.

For example, ® Co emits an electron ((3 ) producing

2gNi in excited level. This reac hes the g ro un d

level by the emission of y-rays.

6. (b ,c ,e ) : The P+ emission by 22 Na,2 2

Na —» 2

gNe + e+ +1>

Therefore (b) and (c) are correct.

This is a spontaneous reaction.

Therefore (e) is correct.

7. (c): The neutri no has negligible mass an d no

charge. It conserves energy in the reaction (c). In

the emission \Y by Na, u is emitted and in emission

of e~ by P, antineu tri no is emitte d. The diff erence

between l> and v is in the direct ion of spin.

8. The detection of neut rino s is possible because

their speed is very high, almost a fraction of that

of light.

According to relativity, when the velocity of a

particle is very large, its radius decreases, mass

increases and the time taken increases. It is this

increase of life-time due to high speed that makes

it possible to detect neutrinos from cosmic rays

and solar bursts.

9. The spe ed of the partic les in classical exp er ime nts

is very small and the changes are much smaller

when compared to the experimental error.

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PHYSICS FOR YOU | MAY'13 5<|



CHALLENGING PROBLEMS

JEE AdvancedA machine gun fires a bullet of mass 50 g with a

velocity 1500 m s"1. The man holding it can exert a

ma xim um forc e of 600 N on the gun. Ho w ma ny

bullets can he fire per second at the most?

(a) 8 (b) 7 (c) 4 (d) 2

The heat dissipated in a resistance can be obt ained

by the measurement of resistance, the current and

time. If the maximum error in the measurementof these quantities is 1%, 2% and 1% respectively,

the maximum error in the determination of the

dissipated heat is

(a) 4% (b) 6% (c) — /o3

(d) 2%

3.

4.

A coin placed on a rotating table just slips if it

is placed at a distance 4r from the centre. On

doubling the angular velocity of the table, the

coin will just slip when at a distance from the

centre equal to

(a) 4r (b) 2r (c) r (d) T-

A block of mass 2 kg is free to move along the

x-axis. It is at rest and from t = 0 on wa rd it is

subjected to a time-dependent force F(f) in the

x-direction. The force F(t) varies with t as shown

in the figure. The kinetic energy of the block

after 4.5 s is

F(TY

4 N

04.5 s

3 s • f

(a)

(c)

4.50 J

5.06 J

(b) 7.50 J

(d) 14.06 J

5. A block of mass 0.5 kg is movi ng wit h a spe ed

of 2 m s"1 on a smooth surface. It strikes another

mass of 1 kg and then the y mov e toge ther as a

single body. The energy loss during the collision

is

(a) 0.16 J (b) 1J (c) 0.67 J (d) 0.34 J

6.

7.

8.

9.

Three coherent sonic source emitting sound of

singl e wav el en gth A, are pla ced on the x-axis

at po in ts - x V n,0 , (0,0),

a,VTT The6 J V 6

inte nsi ty rea chi ng a poi nt Ô, j fr om each

source has the same value I0. Then the resultantintensity at this point due to the interference of

the three waves will be

(a) 6J0 (b) 7I0 (c) 410 (d) 5/ 0

The dens ity of wat er at 4°C is 1000 kg m an d

at 100°C it is 958.4 kg rrf 3. The cubic expansivity

of water between these temperatures is

(a) 4.5 x lO-^C"1 (b) 5.4 x lO^CT

1

(c) 4.5 x lO -^ C' 1 (d) 5.4 x lO ^C " 1

On a smooth inclined surface a block of mass

M is attached between two springs. The other

ends of the springs are fixed to firm supports.If each spring has force constant k, the period of

oscillation of the block is

2 k [ ^ J 2 (b)

1/2

(a)MgsinG

a/2

2k

(c) 27i|2M

k(d) 2Ji

2 Mg,1/2

The work done by electric field during the

displa cement of a negatively c harged particle

towards a fixed positively charged particle is 9 J.

As a result the distance between the charges hasbeen decreased by half. What work is done by the

electric field over the first half of this distance?

10.

(a) 3 J (b) 6 J

In the circuit shown,

the cell is ideal, with

emf = 10 V. Each

resistance is of

2 Q. The potential

difference across the

capacitor is

(c) 1.5 J (d ) 9 J

Rr-MAMr-

C = 3 (iF

R

Hwwvv-R

-wm —

10, v

60 PHYSICS FOR YO U | MAY '13



(a) 12 V (b) 10 V

(c) 8 V (d) zero

11. A box weighing 100 N is at rest on a horizon tal

floor. The coefficient of static friction between

the box and the floor is 0.4. What is the smallest

force F exerted eastward and upward at an angle

of 30° with the horizontal that can start the box

in motion?

(a) 27.5 N (b) 37.5 N(c) 14.2 N (d) 45.4 N

12. A particle under goe s unifo rm circular motion.

About which point on the plane of the circle, will

the angular momentum of the particle remain

conserved?

(a) Centre of the circle

(b) On the circumference of the circle.

(c) Inside the circle (d) Outs ide the circle

13. The pressure in an explosion chamber is 345 MPa.

What would be the percent change in volume of a

piece of copper subjected to this pressure?(The bulk modulus for copper is 138 GPa)

(a) 0.1% (b) 0.5% (c) 0.25% (d) 0.2%

14. A unifo rm metre stick of length L and mass M

is hinged at one end, supported in horizontal

direction by a string attached to the other end.

What would be the initial acceleration of the

centre of the stick if the string is cut?

(a) | g (b) / X 3(

c) -!-

(d) 4g

15. A sam pl e of an ideal gas is tak en thr ou gh a

cycle as shown in figure. It absorbs 50 } of heatduring the process AB, no heat during BC, rejects70 J dur ing CA. 40 J of work is done on the gasduring BC. Internal energy of gas at A is 1500 J,the internal energy at C would be

(a) 1590 J £ B

(b) 1620 J

(c) 1540 J

(d) 1570 J

16. A tra nsp are nt cube of 0.21 m edge contains a

small air bubble. Its apparent distance when

viewed through one face of the cube is 0.1 m

and when viewed from the opposite face is

0.04 m. The actual distance of the bubble from

the second face of the cube is

(a) 0.06 m (b) 0.17 m

(c) 0.05 m (d) 0.04 m

17. A pe ndul um bob of mass 80 mg carrying a charge

of 2 x 10~8 C is at rest in uniform horizontal

electric field £ = 20000 V m_t

. Find the tension

in the thread of the pendulum and the angle it

makes with the vertical.

(Take j = 1 0 m s"2)

(a) 2.2 x 10"4 N, 9.1° (b) 4.4 x 10"

4 N, 18.2°

(c) 6.6 x 10"4 N, 20.6° (d) 8.9 x 10"4 N, 26.6°

18. Two concentric coils each of radius equal to

271 cm are placed at righ t angles to each other.

3 A and 4 A are the currents flowing in each coil

respectively. The magnetic induction in Wb m 2

at the centre of the coils will be

(Ho = 4it x 10"7 Wb A

-1 irT

1)

(a) 5 x 10"s

(c) 12 x 10~

(b) 7 x 10"5

(d) 10"5

19. Two coils A and B having turns 300 and 600

respectively are placed near each other. On passing

a current of 3 A in A, the flux linked with A is

1.2 x 10"4 Wb and with B it is 9.0 x 10~

5 Wb. The

mutual inductance of the system is

(a) 4 x 10 H

(c) 2 x 10"5

H

(b) 3 x 10 H

(d) 1.8 x 10~2

H

20. The rope shown at an instant is carrying a wave

travelling towards right, created by a source

vibrating at a frequency u. Consider the following

statements.

(a) The speed of the wave is 4u x ab.

(b) The phase difference between b and e is(c) Both are correct (d) Both are wrong

3JI

21. The length of a sonometer wire AB is 110 cm.

Where should the two bridges be placed from A to

divide the wire in 3 segments whose fundamental

frequencies are in the ratio of 1 : 2 : 3?

(a) 3 0 cm and 9 0 cm (b) 4 0 cm and 8 0 cm

(c) 6 0 cm and 9 0 cm (d) 3 0 cm and 6 0 cm

22. A solenoid of 0.4 m length with 500 turns carries

a current of 3 A. A coil of 10 turns and of radius

0.01 m carries a curent of 0.4 A. The torque required

to hold the coil with its axis at right angles to that

of solenoid in the middle part of it, is

(a) 6TT2 x 10 "

7 N M ( b ) 3TT

2 X 10"

7 N M

(c) 9n2 x 10"

7 N m (d) 12ti x 10 N m

23. A meta l disc of rad ius R rotates with an angular

velocity CO about an axis perpendicular to its plane

passing through its centre in a magnetic field of

induction B acting perpendicular to the plane of

the disc. The induced emf between the rim and

the axis of the disc is




24.

25.

26.

29.

30.

(a) BnR2

(c) BnR2<x>

(b)

(d)

2Bn2 R

2

CO

1 9 — BR (o2

The resonant frequency of a series LCR circuit, which

comprises an inductance of 200 jxH, a capacitance

of 5 x 10"4 JJ.F and resistance of 10 Q is

(a) 402 kH z (b) 452 kHz(c) 504 kH z (d) 552 kH z

If the bind ing ene rgy of the electron in a h ydr oge n

atom is 13.6 eV, the energy required to remove the

electron from the first excited state of Li++ is

(a) 122.4 eV (b) 30.6 eV

(c) 13.6 eV (d) 3.4 eV

In Young's double slit experiment the wavelength

of l ight X = 4 x 10 7 m and separatio n bet wee n

the slits is 0.1 mm. If the fringe width is 4 mm,

then the separation between the slits and screen

will be

(a) 100 mm (b) 1 m

(c) 10 cm (d) 10 A

27. A point particle of mass 0.1 kg is executing SHM

of amplitude of 0.1 m. When the particle passes

through the mean position, its kinetic energy is

18 x 10 3 J. The equ at ion of moti on of this pa rtic le

when the initial phase of oscillation is 45° can be

given by

(a) O.lcosl 6 f+ —

(c) 0.4 sin f +

(b) 0.1 sin |6f + -

28.

(d) 0.2s in|2 f + -

A perso n throw s vertically n balls per second wit h

the same velocity. He throws a ball whenever the

previous one is at its highest point. The height

to which the ball rise is

(b) 2gn (c) (d) 2gn2

(a) ^n 2 n

z

Two blocks of mass es 10 kg and 4 kg are connec ted

by a spring of negligible mass and placed on a

frictionless horizontal surface. An impulse gives

a velocity of 14 m s~1 to the heavier block in thedirection on the lighter block. The velocity of the

centre of mass is

(a) 30 m s

(c) 10 m s~(b)

(d)

20 m s"

5 m s"1

An ideal gas heat engine operates in a cycle

between 227°C and 127°C. It absorbs 6.0 * 104 cal

at the higher temperature. How much work per

cycle is this engine capable of performing?

(a) 1.2 x 10 cal

(c) 12 x 104 J

(b) 1.2 x 104 J

(d) 1.2 x 104 cal

SOLUTIONS

1. (a): Mass of the bul le t is m = 50 g = 0.05 kg

Initial velocity of the bullet is u = 0 m s~J

Final velocity of the bullet is v = 1500 m s -1

Force exerted = change in momentum of one

bullet x no. of bullets fired per second

600 = m(v -u)xn

600 = 0 .0 5(1 500 -0) xn

600 _ 600

0.05x 1500 " 75= 8

2. (b): Heat dissipated in a resistance is given by

H = I 2 Rt

H ~ I +

R +

t

For maximum percentage error,

x 100 = 2( — x l oo l + — x 100 + — x 100 H { i J R t

= 2 x 2% + 1% + 1% = 6%

3. (c) : The coin placed on the rotating table slips

when mrar is greater than or equal to the static

force of friction \img.

Mathematically,2 W?

mr or > \xmg or r >co

jigThe coin just slips when r = — r

CO

For radii r, and r 2 , let an gu la r velo cit ies be CO]

and (02 respectively.

'l to;

- f -T - 11^2(0 J 4

Hence, r, = 4r, when a), = co and for r 2, co2 = 2co

\ 2r JL

4 rr 2 = r

(c) : From the figure the slope of the force-time,

curve is obta ined as There fore, the equ ati on

of force is

F = \ —f + 4 N I 3 J =

where area under F-t curve gives the change in

momentum

Ap = jF(t)dt

4.5

A p = J

o-—f

2 +4t

I 3 J 3

Ap = m(v - u) = mv

Ap >4 .5kgms _ 1

m 2 kg

= 4.5 kg m s 1

( v « = 0)

-l

5 4 PHYSICS FOR YOU I MAY'13



The kinetic of the block after 4.5 s is

K = ~mv2 = | x2x (2 .2 5)

2 = 5.06J

5. (c): Ac co rd in g to law of co ns er vat io n of

momentum,

OTjM + m2 x 0 = (m1 + m2)v

0. 5x2 2 iv = = — m s

1 + 0.5 3Therefore, loss in kinetic energy is given by

1 1 AK = — mjW 2 ~~(m1 + m2)v

2

= i ( 0 . 5 x 22) - i ( l + 0 .5) f |

8 .

(b ) :

1 2= 1 — = - = 0.67 J

3 3

WiT/6 D

AB = ^J~AD2 + BD2 =X = AC

So sound reaching from B and C will be in same

phase.

(2A)2 + (A)2

+ 2 x 2 A x A x c o s | |

= A2 + 2 A2 = A\p7

••• I = 7I ()

(c) : As Y = V (1 + yAT) where AT is the difference

in tempera tur e, y is the coefficien t of vol ume

expansion

V'-V 1or = y

V AT

or y = •AT

( p - p ' ) x P

(PP') a t, (p-p ' ) J _

p' AT

= 1000-958.4 ^

958.4x96, , . /\

(a ): It is a system of two

springs in parallel. The

restoring force on the

block is due to springs

and not due to gravity

pul l . Therefore s lope

is irrelevant. Here the

effective spring constant

= k + k = 2 k

Thus, time period, T = 2n.

(a): Here, u = Q U l . u .4ne0r

II, - LL =

QH)

47ten

Q H )

4jter

Qq

4tc e0r = 9 ..(i)

When negative charge travels first half of distance,

i.e., r/4, potential energy of the system

u = Q H ) Q<? x4

3 4jre0(3r/4) 4rt:e0r 3

Work done = U 1-U 3

_ Q H ) ,4jxe0r 4 ne0r

4x —

3

Q<7 1 9 o T

= ^ X 3 = 3 = 3 J ( U s i n g ( 1 ) ) .

10. (c) : A fully charged capacitor draws no current.

Therefore, no current flows in arm GHF. So the

resistance, R of arm HF is ineffective.

The equivalent resistance of the resistors in

circuit is

R-vm-

C = 3nF— I K - n H

RWWW~

D

10, V

RA/WM

Req = {R+R)xR + R=(2 + 2)x2

+ 2 A(R + R) + R

VTotal cur rent, I =

(2 + 2)-

10 V

Req (10/ 3) Q.--3 A

In parallel circuit, the curren t divides in the inverse

ratio of resistance, so current in arm ABGD = 1 A

and current in arm AD = 2 A.

Potential difference between G and D

= V G-V D=1 Ax2Cl = 2V

Potential difference between D and F

= VD-VF = 3A x2 C 2 = 6V

••• ^g-^f =(V g-V d ) + (V d -V F )=2 + 6 = 8 V

11. (b):N FsinO

/////An;/}//;/

y

->Fcos8

W

PHYSICS FOR YO U I MAY'13 5 5



Consider the forces in the x-direction and

apply the conditions for equilibrium, noting

/ equa ls its max im um value to star t motion .

2 F x = 0, => Fc os O- / = 0

FcosO = /

Fcos30° = / = ii.N = 0.4N ...(i) 1

Now apply the conditions for equilibrium to the

forces in {/-direction

2 Fy = 0 => N + FsinB - W = 0N + Fsin30° - 100 = 0

N = 100 - Fsin30°

Substituting this in equation (i)

Fcos30° = 0.4N

= 0.4(100 - Fsin30°)

or 0.866F + 0.2F = 40

F = 37.5 N

12. (a): Here,III = mvr

16. (a ): Here refractive index, |i =real distance

apparent distance

or(0 .21 - x )

(0.21-x)

O p = mv

The direction of L (about the centre) is

perpendicular to the plane containing the circular

path. Both magnitude and direction of the angular

momentum of the particle moving in a circular

path about its centre O is constant.

13. (c) : The bulk modulus is defined as B = —A P

AV IV

where the minus sign is inserted because AV is

negative when AP is positive.

A Vx 100 A 100 = ^ x 1 0 0 = 0.25%

V B 138 x 10

14. (c) : Here, angular acceleration

L

x M8 2 3 ga = - =

I ML2 2 L

Mg

Now, acceleration at C ar = ra = —c 2

15. (a) : AW^tp = 0 as V = constant

.-. AQab = AU AB = 50 J

U,4 = 1500 J (Given)

U B = (1500 + 50) J = 1550 J

A W BC = - A U BC = - 4 0 J (Given)

AU BC = 40 J

Uc = (1550 + 40) J = 1590 J

l l2 L

0.1 0.04

or 0.21 x 0 . 0 4 - i x 0.04 0.21 m

= x x 0.1

On solving, we get x = 0.06 m :

17. (d): Tension T is the result of forces mg an d qE.

mg = Tc os0

qE = TsinO

,. tanO = -2—mg

_ 2 x l 0 ~ 8 x 20000

8 0 x l 0- 6

xl0

_ 1

_

20 = 26.6°

and T = \[{mg)2 + (qE)2

llllllllljlilllll

18.

= 7(8 x 10"4)2 + [(4 x 10"4)2] - 8.9 x 10-4 N

Held at(a ): The magnet ic field at the centre due to circular

coil

R - ô1! . R2 r

As both the coils are perpendicular to each other,

hence B, is per pendic ula r to B2.

B = JB21+ B

22 =

4K x 10- 7

2 x 2tc x 10""2

+ 4

= 5 x 10-5

Wb rcf 2

19. (d): Here, e j = N,

i i dI v

and e = M— 1-1 1 dt

dt

d<h dl N—=

dt

or M = N„

Jji

M =

dl„

600x9x10"= 1.8 x 10"' H

20. (c) : Speed of the wave = vX - u(4ab)

= 4u x ab

ab--V

5 6 PHYSICS FOR YOU ! MAY'13



Path difference between b and i

2K

31

:. Phase difference = — x Path differenceA-

_2K3X_3K

~ X 4 ~ 2

21. (c) : Fundamental frequency

1I) oc -

1

Given : u2 : t>3 = 1 : 2 : 3

1 1 1K '2 '3

= 1:2:3

1 1 1or h-l2-l3 l • 2 • 3

or Zj : Z2 : / 3 = 6 : 3 : 2

L = —x 110 = 60 cm11

L =— x 110 = 30 cm2 n

2and L = —x 110 = 20 cm

11

60 cm 30 cm ^ ^ 20 cm

t t x22. (a): B for solenoid = u0n/ = 471 x 10

7 x — x 3

0.4Magnetic moment of the coil,

M = IAN

M = 0.4 x n x (0.01)2 x 10

' = 4n x 0.0001

.-. x = MB sin 90°

= 4K x 10~ 7 x — x 3 x 4ti x 0.0001

0.4

= 6ti2 x io~

7 N m

23. (d)

24. (c) : Given : L = 200 ^ H = 200 * 10~

6

HC = 5 x 10"

4 |xF = 5 x 10~

10 F

R = 10 £2

Resonant frequency of a series LCR circuit is

1 1

2n4lC 271 (200 x l0"6)x (5 xlO"

10)

504 x 103 Hz = 504 kHz

25. (b): Here, £ H = 13.6 eV, n = 2

£„(Li++

) =

E,( Li++ ) =13.6 x(3)

2

(2)

26. (b): Fringe width, (3 =

(3rf

= 30.6 eV

XD

d

D -

4xl0~

3

xO.lxlO"

3

1 m4x l 0"

7

27. (b): Here, A = 0.1 m, m = 0.1 kg,

X = 18 x 10"3 J, d) = —

4Kinetic energy at mean position is

K = -m(02 A z

2

or coJ 2 K )

V 2 =

mA )

2 x l 8 x l 0 "3

O.lx(O.l)2

1/2

= 6

Equation of SHM is

y = Asin(cot + <f>> = O.lsinl 61 +

28. (c) : Number of balls thrown per second = n

1Time interval between two balls thrown = — s

In this time it reaches highest point

v = u + at

0 = u-g or u = -

v2 - m2 = 2as or 02 -1 - 2 x (~g)h

h =2 n A

29. (c) : i'cm =n]v1 + m2v2 _1 0x l4 + 4x0

ttu +m, 10 + 4:10ms"

30. (b): The efficiency of heat engine,

Work done

Heat taken

W T 2or — = 1 — -

Q T,

Given, Tj = 227°C = 227 + 273 = 500 K

T 2 = 127°C = 127 + 273 = 400 K

Q = 6.0 x 104 cal

W

6.0 x 104

400

500

W500-400

500x6.0xl0 4

= —- x 6.0 x 104 = 1.2 x 10

4 J

500




• _r"M a - I

For Practisel . A 40.0 kg boy is standing on a plank of mass

160 kg. The plank original ly at rest, is free to slide

on a smooth fr ozen lake. The boy walks along the

plank at a constant speed of 1.5 m s_1

relative to

the plank. The speed of the boy relative to the

ice surface is

(a) 1.8 m s-

1

(c) 1.2 ms-1 (b) 1.6 ms"

1

(d) 1.5 m s"1

2. A radioactive nucleus of mass M emits a photon

of frequency u and the nucleus recoils. The recoil

energy will be

h2 x>2

(a) Mc2 - hv> (b)

(c) zero

2Mc

(d) hv

3.

4.

A drunkard walking in a narrow lane takes 8 steps

forward and 6 steps backward, followed again by

8 steps forward and 6 steps backward and so on.

Each step is 1 m long and r equir es 1 s. Det ermine

how long the drunkard takes to fall in a pit

18 m away from the start,

(a) 18 s (b) 126 s (c) 78 s (d) 62 s

Which of the following statements is true?

(a) Sound waves cannot interfere.

(b) Only light waves may interfere.

(c) The de Broglie waves associated with moving

particles can interfere.

(d) The Bragg formula for crystal structure isan example of the corpuscular nature of

electromagnetic radiation.

5. In an L-R circuit, the value of L is0.4

H and the

value of R is 30 Q. If in the circuit, an alternating

emf of 200 V at 50 cycl e/s is connec te d, the

impedance and current of the circuit will be

(a) 11.4 £2, 17.5 A (b) 30.7 Q, 6,5 A

(c) 40.4 a , 5 A (d) 50 £2, 4 A

6.

7.

8.

9.

The following physical quantity has a ratio of 103

between its SI units and CGS units

(a) Universal gravi tational constan t

(b) Boltzmann's constant

(c) Planck's constant

(d) Young's modulus of elasticity

Which of the following letters do not suf fer lateral

inversion?

(a) HGA (b) HOX

(c) VET (d) YUL

By using only two resistance coils singly, in

series, or in parallel, one should be able to obtain

resistances of 3, 4, 12 and 16 ohm. The separate

resistances of the coils in ohm are

(a) 3 and 4 (b) 4 and 12

(c) 12 and 16 (d) 16 and 3

When a solid sphere rolls without slipping down

an inclined plane making an angle 0 with thehorizontal, the acceleration of its centre of mass

is a. If the same sphere slides without friction its

acceleration a' is

(a)5

(b) (c)7-a5

5(d)

10. If unit vectors A and B are inclined at an angle

0, then \A-B\ is

(a) 2 si n-2

(b), 02 cos-

2

(c) 2 tan

0

(d) tan0(<fl

11- The distance between two movi ng particles at any

time is a. If v be their relative velocity and vt and

v2 be the component s of v along and perpendicular

to a. The time when they are closest to each other is

aw,(b)(a) ^

v

zl„2

(C)

av(d)

av




12. An infinite number of charges each equal to

0.2 uC are arranged in a line at distances 1, 2, 4,

8 .... metre from a fixed point. The potential at

the fixed point is

(a) 1800 V (b) 2000 V (c) 3600 V (d) 2250 V

13. Let x = xmcos(cof + 0). At t = 0, x = xm. If time

period is T, what is the time taken to reach

(a) (b) | (c) = T (d) -

14. A small bob attached to a string of length I is

suspended from a rigid support and rotates with

uni form speed along a circle in a horizontal plane.

Let 0 be the angle made by the string with the

vertical. Then the length of a simple pendulum

having the same period is

I I(a) ^ (b ) ZsinG (C) ^ ( d) /COS0

15- In space two point charges are placed as shown

in figure. The electric field line at point P may

be in direction

© @N

->E

S

(a) sou th (b) sout hwes t

(c) nor theast (d) west

16- A ball A is thrown up vertically with a speed u

and at the same instant another ball B is released

from a height h. At time f, the speed of A relative

to B is

(a) u (b) 2u (c) u-gt (d) Ju2-gt

17- A coil in the shape of an equilateral triangle of

side 0.02 m is suspen ded from its vertex such that

it is hanging in a vertical plane between the pole

pieces of permane nt magnet p roduc ing a unifor m

field of 5 x 10~2 T. If a curren t of 0.1 A is pas sed

through the coil, what is the couple acting?

(a) 5V3 x 10~7 N m (b) 5V 3x lO ~

1 0Nm

V3(c) — xl O

- 7 N m (d) ^

18- A body of mass 1 kg is th rown upw ar d wi th a

velocity 20 m s" It momentar ily comes to rest

after attaining a height of 18 m. How m uch energy

is lost due to air friction?

(Take g = 10 m s^2)

(a) 30 J (b) 40 J (c) 10 J (d) 20 J

5 , x l 0 " 7 N m

19. When a lens of refract ive index is placed in a

liquid of refractive index \x2/ the lens looks to be

disappeared only if

(a)

(c) Hi = \i2

(b) H =

5(d) = - h 2

20. A planet revolves around the sun in an elliptical

orbit of eccentricity e. If T is the time period of the

planet, then the time spent by the planet between

the ends of the minor axis and major axis close

to the sun is

(b) f ^ - 1(a)

(c)

2e

h2K

(d) Te

2n

21. The depletion layer of a p-n junction

(a) is of constant wid th irrespect ive of the bias(b) acts like an insulating zone under reverse bias

(c) has a widt h that increases with an increase

in forward bias

(d) is dep let ed of ions

22. A body is rolling down an inclined plane. If

kinetic energy of rotation is 40% of kinetic energy

in translatory state, then the body is a

(a) ring (b) cylinder

(c) hol low ball (d) solid ball

23. Two liquid drops of equal radii are falling through

air with the terminal velocity v. If these two dropscoalesce to form a single drop , its terminal velocity

will be

(a) y]2v (b) 2v (c) (d) ^f lv

24. The decreasing order of wavelength of infrared,microwave, ultraviolet and gamma rays is

(a) micro wave, inf rar ed, ultrav iolet, g amma

rays

(b) ga m m a r a ys , u l t r a v i o l e t , i n f r a r e d ,

microwaves

(c) mic row ave s , gam ma rays , in f r a r ed ,

ultraviolet(d) infrared, microwave, ultraviolet, gamma rays

25. Two blocks of masses 1 kg

and 2 kg are connected by

a metal wire going over a

smooth pulley as shown

in figure. The breaking

stress of the metal is

2 x iO9 N rrr2.

What should be the minimum radius of the wire

used if it is not to break?

imiuitmuiimiimm

1i k g n

• 2kg




(Take g = 10 m s~2)

(a) 4.6 x 10"5 m (b) 4.6 x 10~

6 m

(c) 2.5 x 10"6 m (d) 2.5 x 10"

5 m

26. The equivalent capacitance between A an d B is

29.

30.

H H H HH H BO~

c(a) ± (b) \ c (c) j C (d) C

3 3

4

27. When a plas tic thin film of refract ive index 1.45 is

placed in the path of one of the interfering waves

then the central fringe is displaced through width

of five fringes. The thickness of the film, if the

wavelength of light is 5890 A, will be

(a) 6.54 x 10"4 cm (b) 6.54 x 10~

3 cm

(c) 6.54 x 10"5 cm (d) 6.54 x 10~2 cm

28. A solid sphere of mass 10 kg is placed over two

smooth inclined planes as shown in figure. Normal

reaction at 1 and 2 will be

(Take g = 10 m s~2)

(a) 50>/3 N, 50 N

(c) 50 N, 50>/3 N

(b) 50 N, 50 N

(d) 60 N, 40 N

The atmospheric pressure on the earth's surfaceis P in MKS units. A table of area 2 m 2 is tilted

at 45° to the horizontal. The force on the table

due to the atmosphere is (in newton) ;

(a) IP (b) 72P (c) 2^ 2P (d)

Two spheres of radii r1 and r2 have densities p,

and p2 and specific heats Sj and s2 respectively.

If they are heated to the same temperature, the

ratio of their rates of cooling will be

h P2 S 2 r

2 P2 S

1

h Pi S

1 r

l P lS

2

rlPl

Sl ( d )

r2Pl

S2

r 2 P2 S

2 h P2 S

1

31. A ball of mass m is moving towards a batsman

with a speed v. The batsman strikes the ball and

deflects it by an angle 0 without changing its

speed. The impulse imparted to the ball is

(a) mvsinO (b) mvcosQ

(a)

(c)

(c) 2musin0

(d) 2mwcos^

32. A particle is executing simple harmonic motion

with an amplitude A and time period T. The

displacement of the particle after 2 T period from

its initial position is

(a) A (b) AA (c) 8A (d) zero

33. A thin disc having radius r and charge q distributed

uniformly over the disc is rotated u rotations per

second about its axis. The magnetic field at thecentre of the disc is

3^0(a) M ( b ) W2 r r

(0 M ( d )4 r 4 r

34. In the arrangement of resistances shown below,

the effective resistance between points A and B is

(a) 23.5 Q. (b) 38 Q (c) 19 £2 (d) 25 Q

35. A radioactive element X converts into another

stable element Y. Half life of X is 2 h. Initially

only X is present. After time f, the ratio of atoms

of X and Y is found to be 1 : 4, then t in hours is

(a) 2 (b) 4

(c) between 4 and 6 (d) 6

36. Let Mj and n2 moles of two different ideal gases

be mixed. If ratio of specific heats of the two

gases are yx an d y2 respectively, then the ratio of

specific heats y of the mixture is given through

the relation

(a) («i + n2)y = Mj y2 + «2 Yi

(»l+" 2) wi , "2

(b)Y - l Y i "l Y2~l

(C) ("i+ n

2) = J ! i _ +

Y+l Yi- 1 Y2- 1

(d) (n , + n2)(Y - 1) = M(Yi + 1) + "2(Y2 + 1)

37. The following configuration of gates is equivalent to

(a) NAND

(c) OR

(b) XOR

(d) AND

6 0 PHYSICS FOR YOU | MAY '13



38. At a given temperature, velocity of sound in

oxygen and in hydrogen has the ratio

(a) 4 : 1 (b) 1 : 4 (c) 1 : 1 (d) 2 : i

39. In the Davisson and Germer experiment, the

velocity of electrons emitted from the electron

gun can be increased by

(a) increasing the potential difference between

the anode and filament

(b) increasing the filament current

(c) decreas ing the filament current

(d) decreas ing the potential difference between

the anode and filament

40. Magnetic flux through a stationary loop with a

resistance R varies during the time interval x as

(|) = at(z - t) where a is a constant. The amount

of heat generated in the loop during the time

interval x is

(a)

(c)

a2T3

6 R

a2 x3

3 R

(b)

(d)

a2x3

4 R

a2 x

3

2R

41. First overtone frequency of a closed organ pipe is

equal to the first overtone frequency of an open

organ pipe. Further nth harmonic of closed organ

pipe is also equal to the mth harmonic of open

pipe, where n and m are

(a) 5, 4 (b) 7, 5 (c) 9, 6 (d) 7, 3

42. A vernier calipers has 1 mm m arks on the main

scale. It has 20 equal divisions on the vernier

scale which match with 16 main scale divisions.

For this vernier calipers, the least count is

(a) 0.02 mm (b) 0.05 mm

(c) 0.1 mm (d) 0.2 mm

43. The centre of mass of a non-uniform rod of

Kx2

length L whose mass per uni t length X = ,

where K is a constant and x is the distance from

one end is

- (d) \2 3

44. Two containers of equal volume contain thesame gas at the pressure Pj and P2 and absolute

temperature Tj and T2 respectively. On joining

the vessels, the gas reaches a common pressure

P and a common temperature T. The ratio P/T

is equal to

, , 3 L(a) T

(b) k (C)

(a)

(c) ±v > 2

t xT 2

P1T 2 + P2T t

t xT 2

(b)

(d)

T 1+T 2

Pih ~ P2Tl

T{T 2

45. A point P moves in

counter-clockwise

direction on a circular

path as shown in the

figure. The movement

of P is such that it

sweeps out a length

s = t 3 + 5; where s is

in metres and t is inseconds.

The radius of the path is 20 m. The acceleration

of P when t = 2 s is nearly

(a) 12 m s"2 (b) 7.2 m s"2

(c) 14 m s- 2 (d) 13 ms- 2

46. In a hydrogen atom, the magnetic field at the

centre of the atom produced by an electron in

the nth orbit is proportional to

1 „ 1 1 . .. 1(a) (b) (c) (d)

47. A force F acting on a body depends on its

displacement S as F S~1/3. The power delivered

by F will depend on displacement as

(a) S 2' 3 (b) S1/3 (c) S° (d) S"5/3

48. A bat tery of interna l resistance 4 £2 is connected

to the network of resistances as shown. In order

that the maximum power can be delivered to the

network, the value of R in Q should be

(a) | (b) 2 (c) | (d) 18

49. A tiny spherical oil drop carrying a net charge

q is balanced in still air with a vertical uniform

electric field of strength x 105 V rrT1. When

the field is switched off, the drop is observed to

fall with terminal velocity 2 x 10-3 m s_1.

Given, g = 9.8 m s - 2 , viscosity of the air

= 1.8 x 10 -5 N s m - 2 and the density of oil

= 900 kg m-3, the magnitude of q is

(a) 1.6 x 10~19 C (b) 3.2 x 10-19 C

(c) 4.8 x 10-19 C (d) 8.0 x 10"19 C

50. Internal energy of n, moles of hydrogen at

temperature T is equal to the internal energy of

«2 moles of helium at temperature IT. Then the

ratio nt /n 2 is

<d> I<«» I (b) < o f

74 PHYSICS FOR YOU | MAY '13 ;



1 .

SOLUTIONS

(c) : The system is not subjected to any external

force and hence conservation of momentum can

be used. Let mb and mp represent the masses of

the boy and the plank respectively. Let vbi , v pi and

vbp be the velocity of the boy with respect to ice,

that of the plank with respect to ice and that of

the boy with respect to the plank respectively.Then,

mbvbi + m pv pi = 0 ...(i)

vbi = vh + v pi ...(ii)

or v pi = vbi - vbp

Substituting the value of v pi in Eq. (i), we get

mbvbi + m p(vbl- vbp) = 0

vbi{mb + m) = m vb or vbi =mu + m„


160 kg x 1.5 m s

(40+ 160) kg

-l240 _i „ „ _i

: m s 1 = 1.2 m s '

200

2. (b): Momentum of emitted photon,

_ hvPphot on

Let v the speed of recoil nucleus.

According to law of conservation of momentum,

Pnucleus — V photon

hv hv:. Mv = — or v = •

c Mc

The recoil energy of the nucleus

= -Mv2 =-M

hv

Mc

h2v2

2 Mc2

3. (c) : When the drunkard walks 8 steps forward

and 6 steps backward, the displacement in the

first 14 steps = 8 m - 6 m = 2 m

Time taken for first 14 steps = 14 s

Time taken by drunkard to cover first 10 m of

jou rn ey

14= — x 10 = 70 s

2If the drunkard takes 8 steps more, he will fall

into the pit, so the time taken by the last 8 steps

= 8 s

Total time taken = 7 0 s + 8 s = 7 8 s

4. (c) : Both so un d and light wave s exhi bit t hephenomenon of interference.

The Bragg formula for crystal structure is an

example of the wave nature of electromagnetic

radiation.

5. (d): Her e, L = H,R = 30£2'OA{ n

•• 200 V, u = 50 Hz

The inductive reactance isM-ms

X L = (aL=2nvL • 271 x 50 x0.4

40 £2

The impedance of the circuit is

z = ^R2

+ xl = 7 ( 3 0 H)2

+ (40 Q.)2

= 50 Q

Current in the circuit is

I„V 200 V

= 4 A

6.

Z 50 Q.

(a ): The value of univer sal gravitati onal c onstant

in CGS system is 6.67 x 10~8 dyne cm2 g~2and in

SI system is 6.67 x 10"11 N m2 kg"2.

Their corresponding ratio is

G in CGS unit 6.67 x 10"°

G in SI unit 1-11- = 1 0

7.

8.

6.67 x 10"

(b): The letters do not suffer lateral inversion areA, H, 1, M, O, T, U, V, W, X, Y

(b): The least value of two resistances R, and R2

can be obtained when connected in parallel. The

maximum value of two resistances is obtained

when connected in series. In question, the least

resistance is 3 Q and maximum resistance is

16 £2. So

3 = • R1R2

Rj + R2

an d 16 = Rj + R2

Using (ii) in (i), we get R, R,

3 =_ 2

16RjR 2 = 48

...(i)

...(ii)

..(iii)

9.

Solving (ii) and (iii), we get

Ri = 4 Q. and R2 = 12 £2

or Rj = 12 £2 and R2 = 4 £2

(c) : Acce lerat ion of the solid sphere, whe n it rolls

without slipping down an inclined plane is

rsinGa = —

1 I

MR

...(i)

For a solid sphere , / = - MR 2

gsinO 5 .a = -—— = - f s i n G

1 + — 7

5Acceleration of the same sphere, when it slides

without friction down an same inclined plane is

a' = ^-sinO ...(ii)


a ' 7 , 7— = - or a =-aa 5 5

6 2 PHYSICS FOR YO U I MAY' 13



10. (a): \ A-B \ = (A - B) • (A - B)A A A A A A A A

= A-A -A-B-B-A + B-B

= 1 - A-B-A-B + 1 = 2 - 2 cos6 = 2[1 - cos0]

= 2 1 - 1 + 2 sin2 - = 4 sin -

a a aor \ A - B \ - 2 s i n -

211. (a): The situation is as shown in the figure.

ê \

90°

2

AB = aFrom figure,

7 7 9 V-iV= + v2 or tan 0 = —i

1

v2

COS0 =

and sin0 =

_ V 2 ..V 2

yjv 1 + v\ V

_ ^

Jv l + v\V

Minimum distance between A and B isaV' ,

Smi n = BC = AB cos0 = —^

vThe time when they are closest to each other is

t = '- AC AB sin0 av-

v v

12. (c) : The given charge configuration is as shown

in the figure.

0.2 nC0.2 nC 0.2 nC 0.2 hC

= 0 l m 2m 4m 8m

The potential due to the charge configuration at

the fixed point (x = 0) is

1V47ie0l 1

0.2 0.2 0.2 0.2 0.2+ + + + K,2 4 8 16

= 9 x 109 x 0.2 x 10~

6

: 1.8 X 10

1 1 1 1 1- + - + - + - + —

1 2 4 8 16

1

' 2

= 3.6 x 103 V = 3600 V

13. (d): Given, x = xmcos(o)f + <j>)

When t = 0, x - xm ,

xm= xmcos(co x 0 + 0)

or cosij) = 1 = cosO0 or 0 = 0°

X - X mCOS(Ot

X XWhen x = — , then — = x cos cot

2 2 m

, 1 7t Kor coscot = - = cos — or cot = -2 3 3

t = -k _ nT

3co 3 x 2rc

T

6

. 2it(••• ® = y )

14. (d): When a small bob

of mass m attached to

a string of length I is

suspended from a rigid

support and rotates with

uniform speed along

a circle in a hor izonta l ,pl an e as sh ow n in

adjacent figure. Such

arrangement is known as

conical pendulum.

Time period of a conical pendulum is

/ cosOT = 2n

Time period of a simple pendulum of length L is

T = 2n -

For having the same time period as that of conical

pendulum, then

L = /cos0

6115. (c)

16. (a): Refer figure.

At time t,

Velocity of A, v A = u - gt (upwards)

Velocity of B, v B = gt (downwards)

= -gt (upwards)

Relative velocity of A w.r.t. B is f

V AB = v A-V B = {u- gt) - (-gt) = u " A

17. (a): Area of equilateral triangle of side / is

A = i x base x height

. 1 • l S S i 2

A = - x / x =2 2 4

Here, I = 0.02 m

, . A = ^ X(°-°2m)2 = S x l O - i m 2

PHYSICS FOR YOU | MAY 3 5 3



Couple acting on the coil is

T = MBsinG

Here, I = 0.1 A, A = J3 x 10"4 m

2,

B = 5 x 10"2 T, 9 = 90°

x = 0.1 x Vi x 10"4 x 5 x 10~

2 sin90°

= x 10"7 N m

18. (d): Initial ene rgy is

£. = I m w2 = I x 1 kg x (20 m s"1)2 = 200 J

Final energy is

Ef = mgh = 1 kg x 10 m s"2 x 18 m = 180 J

Therefore, loss of energy due to air friction is

£lo s, = £, - E7 = 200 j - 180 ] = 20 ]

19. (c) : If re fract ive in dex of len s (X, is equal to

refractive index of liquid ji2, then lens behaves

as a plane glass plate and becomes invisible in

the medium.

20. (d):

i O

As areal velocity of a planet around the sun is

constant. Therefore, the desired time is

f AB~area ABS

area of ellipsex time period

If a = semi-major axis and b = semi-minor axisof ellipse, then

area of ellipse = nab

Area ABS = ~ (area of ellipse)

- Area of triangle ASO

1 1= - x nab - - (ea) x (b)

n(ab) 1eab

n abxT = T

1 __e_

4 2ti

21. (b) '

22. (d): Rotational kinetic energy is

\2

K = ~ / c o 2 = ^ M K 2 ^ j (v I = MK2andi> = Rco)

= -Mv2

2

Translational kinetic energy is

K r =-Mv2

T 2As per question,

K R = 40% K T

or

-Mv2

2

K 2

R2

KA

X

40

100

= 40%~Mw2

2

2

5

K 2For solid sphere, — = —

R2 5

Hence, the body is a solid ball.

23. (c) : Let R be the radius of big dr op fo rme d and

r be radius of each small drop. Then

-7iR3 = 2x-7 tr

3 or R = 2

1 /3r

3 3

Terminal velocity °=r2

\ 2

V

V

Ror v --Ifiv

24. (a): The decreasing orde r of wavel ength of the

given electromagnetic waves is as follows:

^•Microwave În fra red ^ ^-Ultraviolet ^G am ma rays

25. (a): Here , m, = 1 kg, m2 = 2 kg

Breaking stress = 2 x 109 N m

2

The tension in the string is

_2m x m2

_ 2 x l k g x 2 k g x l 0 m s -

~ (1 + 2) kg

If r is the minimum radius, then

40

Breaking stress = -2-

40N

nr

r 2 = -40

3 x 7C x 2 x 10or r

3nx 10"

r = 0.46 x 10-4 m = 4.6 x 10"5 m

26. (b): The equivalent circuit of the given network

is as shown in the figure.

4 C C C R

HHHh

Hh

In the upper arm three capacitors are connected

in series. So, their equivalent capacitance is

6 4 PHYSICS FOR YO U | MAY 131



1 1 1 1 3 ^ C — =—+—+—=— or Cc= —C s C C C C s 3

Hence, the equivalent capacitance between

A an d B is

C 4Ab £> 3 3

27. (a): As x = A

Here, x = 5(3

X = 5890 A = 5890 x lO"10

m, n = 1.45

(3 (0.45) t5(3 = -

t = •

5890 x10

5 x 5890 x 10

i-io

- 10

0.45

= 6.54 x 10"4 cm

28. (a): Nx

= 6.54 x 10~6m

W= 100 N

All three forces W,-N 1 an d N 2 will pass through

centre.

According to Lami's theorem

N, N, 100

sin (90° + 30°) sin (90° + 60°) sin 90°

100sin(90° + 60°) = 1 0 ( ) c o s 3 0 O = 5 0 ^ N

1 sin 90°

N 2 =100 sin(90° + 60°)

100cos60°= 50 Nsin 90°

Note that plane 2 is at 60° with horizontal, hence

normal to it i.e., N 2 will make an angle 30° with

horizontal.

29. (a)

30. (a): As the temperatures of both spheres are

equal, therefore energy emitted per unit second

per area by both the spheres is the same.

Qi = Q2

(4jtr2)Af (4jcr2)Af

mjSjATj m2s2 AT 2r 2(At) r 2( At)

L ^ P X A J , P 7 C R 23 P 2 | S 2 A T 2

r2(Af) r2(Af)

rlPi%

AT,

AT,

At1 - r 2 P2 S 2

A T 2

~~AT

At _r lP2S 2

A T2

AfPi s i

31. (d):

psin

O

-Bat

The ball hits the bat along AO and goes along OB,

after striking the bat. Rectangular componentsQ

of v along AO a r e w c o s - a l o n g AC and

9 ^csin- along AD.

The rectangular components of v along OB are

9 9i'cos - along OD and wsin- along OE.

Velocity component along the bat before and

after collision are the same. Therefore, there is no

change in velocity along the bat. However, velocity

component perpendicular to the bat is reversed

in direction. Therefore,

Impulse imparted = change in momentum

0) , 9, -mvcos- =2m»cos—

2 ( 2 J 2

32. (d): The particle completes one oscillation in time

T. Therefore, in time 2 T, it will complete two

oscillations and will reach to its starting point,

i.e., initial position. Therefore, the displacement

is zero.

33. (b): Cons ider a hypothetical ring of radius x and

thickness dx of a disc as shown in figure.

= mv cos-

Charge on the ring, dcj = J_nr 2 x (2nxdx)

Current due to rotation of charge on ring is

dqd l J j r = _dq_ = x)d q = vq2xdx




Magnetic field at the centre O due to current of

ring element is

d B = = P-Q l^xdx = uqrf.t

2x r-(2x)

Total magnetic field due to current of whole disc is

v2 J r 2 r

34. (c):

For the given network, the straight line AFB is

the line of symmetry. If a cell is connected across

A and B, the points D, F an d H will be at the

same potentials. Hence no current will flow in

arms DF an d FH. The equivalent circuit will be

as shown in figure. D

The resistance between C and E of bridge CDEF is2 7 x 5 4

= 18 a

35. (c) : Let N0 be the number of atoms of X at time

t = 0. Then at f = 4 h (two half lives)

N x = -

WY

N,o and y 4

Nv

and at t = 6 h ( three half lives)

N x = :N„ and Ny = -7N n

N x __

" N r 7

1 1 1The given ratio - lies between - and - .& 4 3 7Therefore, f lies between 4 h and 6 h.

36. (b)

37. (b): Out pu t of Gt = (A + B)

Output of G2 = A-B

Output of G3 is

Y = (A + B)-(Alt)=(A + B)-(A + B)

= A-A + A-B + B-A + B-B=A-B + A-B •

It is the Boolean of XOR gate.

Hence, the given configurat ion of gate is equivalent

to XOR gate.

38. (b): Speed of sound in gas is

v =7RT

M

where the symbols have their usual meaning.

Since both oxygen and hydrogen are diatomic

gases and at same temperature

"o,

H,

MH .2_ _

M r

27 + 54Resistance of the network of upper portion

between A and B is

Ri = 10 + jR' + 10 = 10 + 18 + 10 = 38 £2

Similarly, resistance of the network of lower

portion between A and B = 38 Q.

Therefore, effective resistance between A and B is

38 + 38

39. (a) : As eV = - m v 2 or2

Thus velocity of electron emitted from electron

gun can be increased by increasing the potential

difference between anode and filament in Davisson

and Germer experiment.

40. (c) : Magnetic flux through the stationary loop is

<J> = at(x - t)

Induced emf,

dt dt2at] = (2at - ax)

The amount of heat generated in the loop during

a small time interval dt is

dQ =(2at - ax) z

dt R R

Hence, the total heat generated is

6 6 PHYSICS FOR YOU I MAY '131



I

a"S

(2 at - atf

~Rdt = - J (4 a

212

+ fl2x

2 - 4 a

2 it) dt

o

-a2t 3 +a2 x2t--a2it 2

3 2 "3 R

41. (c) : First overtone frequency (i.e. 3rd

harmonic)

of a closed pipe of length lc is

= 3

/ \V

v4 hjwhere v is the speed of sound in air

First overtone frequency (i.e. 2nd

harmonic) of an

open pipe of length /„ is

= 2

/ \V

Given, 3

/ \ ( \

V _ o V

l^J — z.SJ

Now, n

3

4 ...(i) /

Vf i l»o)

= 2'lc \ 6 3 9

Jo J 4=2 6(Using (i»

nor —

m

Thus, n = 9 and m = 6.

42. (d) : 20 VSD = 16 MSD

1 VSD = — MSD = -MSD20 5

VC = 1 MSD - 1 VSD

- H )MSD = - MSD

5

= — x 1 mm = 02 mm5

43. (a): x = 0dx

-Hx-L

Mass of a small element of length dx of the rod at

a distance x from the one end of the rod isKx2

dm = Adx = dx L

The centre of mass of the rod is

L L

j xdm JKxc

dx

X,CM_ 0

L- L.

\dm JKx'dx

3V /

3 L

' 4

44. (c) : According to ideal gas equation

P.V^n.RT, ...(i)

an d P2V = n2 RT 2 ...(ii)

As the number of moles remain conserved,

n = nl + n2

P(2V) _ Pt V | P2V

RT RTi RT 2

P

T >1 _ 1~p1t 2+p2t 1'

T iT 2_

~ 2 L T1T2 J45. (c)

Here,

s = t 3 + 5; r ^ 20 m

Velocity, v = — = 3t 2 J dt

When, t = 2 s,

v = 3 x 22 = 12 m s"1

dvTangential acceleration, at = — = 61

dt

When, t = 2 s, at = 6 x 2 = 12 m s"2

Centripetal acceleration,

122 _ _2

a = — = = 7.2 m s 2

c

r 20Effective acceleration,

a = yjaf + a2 = Jl22 + 7.22 = 14 m s"2

46. (d): Electric current due to electron motion in

nth orbit is given by

In =

For hydrogen atom

Frequency of electron in n[h orbit is

u =

47cen

A 2 44?r e m

Radius of «th

orbit is

4llZrM2h

2

r = — «n a 2 2 An e m

( - 2 54K e m

4KE,0 J n3h3(Using (i))

...(i)

...(ii)

...(iii)

The magnetic field at the centre of the current

carrying coil is given by




B =2R

Hence, the magnetic field produced at the centre of

the atom due to electron motion in nth orbit is

2 r 'B,

By, ~

47ien

' M *4*

7™

2 (Using (ii) and (iii))

n h

B„

47. (c) : A s F « S "1/3, therefore,

acceleration a « S~1/3

_dv _dv dS _ dv

dt dS dt dSdv „ _ i / 3

dS

Integrating both sides, we get

v2 °c S2'3 or v oe S1'3

As P = F»

p oc S-1/3

S1/ 3

or P °c S°

i.e. Power is independent of S.

48. (b):

— v \ ^ w —

R

WWA 1

;6R |

i4R i ^

The given circuit is that of a Wheatstone

bridge.

The circuit is a balanced one since,

resistance across AC _ resistance across CB

resistance across AD resistance across BD

Thus, no current will flow across 6R of the side

CD. The given circuit will now be equivalent to

R 2R

—WWW VWIMr-|

2 R-VWWV-

4R-JVWWV-

4 £2,

3R-VvWW-

6 R-wmr-

4£2,i

—WWW-2R

h—

For maximum power,

net external resistance = total internal resistance

or 2R = 4 or R = 2 £2

49. (d): Here,

„ 8l7tx 10

5 V m

-1,i> = 2 x 10"

3 m s"

1,

T) — 1.8 x 10"5 N s m-

2; p = 900 kg mr

3

Whe n dr op is balanced in still air unde r the effect

of electric field, then

4 3:Kr PS

4 aqE = - nr pg or

3 E...(i)

When the electric field is swi tched off, let the

drop falls with terminal velocity v, then

~|l/22r z(p - a)g

v = — or r =9r|

1 4q = E X 3 n p g

9vr\

9r\v

2(p -0 )g

3/2

2(p-o)g

8171 x i o3

4x - x jt x 900 x

1

9 x l . 8 x l 0 "5x 2 x l 0N-3

3/2

2 x 900 x <

On solving we get, q = 8x 10"19

C

50. (c) : Internal energy of n moles of an ideal gas

at temperature T is given by

U = ^ nRT ( / = degrees of freedom)

U He = J n2 RT 2

As U-H , = <i,H e

••• finiTj =f 2n2T 2

or ^ = ^n2

As hydrogen is a diatomic gas and helium is a

monoatomic gas

.•. / i = 5 and f 2 = 3

Here, T, = T, T2 = 2T

n1 _( '3Y2TN l 6

5 A T




F^Fllj^^CZ'TI CZeE!

fences

1.

2.

3.

4.

5 .

A potential difference V = 100 ± 5 V, when applied

across a resistance, gives a current I = 10 ± 0.2 A.

What is the percentage error in R?

(a) 2% (b) 5%

(c) 7% (d) 8%

A car covers the first one-third of a distance x

at a speed of 10 km h -1 , the second one-third ata speed of 20 km IT1

and the last one-third at a

speed of 60 km h_1

. Find the average speed of

the car over the entire distance x.

(a) 10 km h_1

(b) 12 km h"1

(c) 18 km h_1

(d) 20 km f r1

A simple pendulum performs simple harmonic

motion about x = 0 with an amp litude a and time

period T. The speed of the pendulum at x = a/2

will be

(a)

(c)

TW

Y

7CflV3

(b)

(d)

3K 2a

T

7tflV3

2T

A parallel plate condenser has a uniform electric

field E (V/m) in the space between the plates. If

the distance between the plates is d (m) and area

of each plate is A (m2) the energy (joules) stored

in the condenser is

(a) E 2 Ad/e0

(c) e0 EAd

(b) - e 0 £

(d) -ea E 2 Ad

An explosion blows a rock into three parts. Two

parts go off at right angles to each other. These

two are 1 kg first par t mov ing with a velocity

of 12 m s_1

and 2 kg second part moving with a

velocity of 8 m s_1

. If the third part flies with a

velocity of 4 m s_1

, its mass would be

(a) 3 kg (b) 5 kg

(c) 7 kg (d) 12 kg

A parallel monochromat ic beam of light is incident

norm ally on a na rrow slit. A diffract ion pattern is

formed on a screen placed perpendicular to the

direction of incident beam. At the first maximum

of the diffraction pattern, the phase differencebetween the rays coming from the edges of the

slit is

(b) 52

(d) 2K

(a) 0

(c) it

A uniform wire of

resistance 36 ohm is

bent in the form of

a circle. The effective

resistance across the

points A and B is

(a) 5 £2 (b) 15 Q

(c) 7.2 Q (d) 30 Q,

A 50 kg mass is travelling at a speed of 2 m s"1.

Another 60 kg mass is travelling at a speed of

12 m s_1

in the same direction, strikes the first

mass. After the collision the 50 kg mass is

travelling with a speed of 4 m s_1

. The coefficient

of restitution of the collision is

(a)

(c)

19

30

20

11

(b)

(d)

30

19

11

20

By sucking through a straw, a student can reduce

the pressure in his lungs to 750 mm of Hg (density

= 13.6 g cm-3). Using the straw, he can drink water

from a glass upto maximum depth of

(a) 10 cm (b) 75 cm

(c) 13.6 cm (d) 1.36 cm




10. A charged particle moves through a magnetic field

in a direction perpendicular to it. Then the

(a) speed of the particl e remains unchan ged

(b) direction of the particle remains unchanged

(c) acceleration remains uncha nge d

(d) velocity remains unchanged

11. An eng ine ha s an efficiency of 1/6. When the

temperature of sink is reduced by 62°C, itsefficiency is doubled. Temperatures of the source

is

(a) 37°C (b) 62°C

(c) 99°C (d) 124°C

12. Two solenoids of equal num ber of turns h ave

their lengths and the radii in the same ratio

1 : 2. The ratio of their self-inductances will be

(a) 1 : 2 (b) 2 : 1

(c) 1 : 1 (d) 1 : 4

13. A transistor-oscillator usi ng a resonant circuitwith an inductor L (of negligible resistance) and

a capacitor C in series produce oscillations of

frequency x>. If L is doubled and C is changed to

4 C, the frequency will be

(a) u/2 (b) u/4

(c) 8 u (d) d /2 V2

14. At wha t temp era ture will the speed of sound in

hydrogen be the same as in oxygen at 100° C?

Molar masses of oxygen and hydrogen are in the

ratio 16 : 1.

(a) - 249.7°C (b) - 239.7°C

(c) 259.7°C (d) 269.7°C

15. The frequency of a light wave in a materia l is

2*1014

Hz and wavelength is 5000 A . The refractive

index of material will be

(a) 1.50 (b) 3.00

(c) 1.33 (d) 1.40

16. Liquid oxygen at 50 K is heated to 300 K at constant

pressure of 1 atm. The rate of heating is constant.

Which one of the following graphs represents the

variation of temperature with time?

(a)

(c)

y(b)

Time Time

(d)

Time Time

17. A ball rolls wi tho ut sl ipp ing . The rad iu s of

gyration of the ball about an axis passing through

its centre of mass is K. If radius of the ball be R,

then the fraction of total energy associated with

its rotation will be

(a) (b)

21.

22.

(c)

R2

K 2

R2

K 2 + R2(d) R

18. A simple pendul um has a time period T 1 when

on earth's surface and T2 when taken to a height

2 R above the earth's surface where R is the radius

of earth. The value of TJT2 is

1

9(a)

(c) V3

(b) iv ' 3

19. The circuitNOR

(d) 9

NAND NOT

is equivalent to

(a) NOR gate (b) OR gate

(c) AND gate (d) NA ND gate

20. A bar magnet is oscillating in the Earth's magneti c

field with a period T. What happens to its period

and motion if its mass is quadrupled?(a) motion remains simple harmon ic with time

period = T/2

(b) motion remains S.H.M. with time period

= 2 T

(c) mo tio n rem ain s S.H.M. wit h time pe rio d

= 4T

(d) motion remains S.H.M. and period remains

nearly constant

A particle is projected from the ground with an

initial speed of v at an angle 0 with horizontal. The

average velocity of the particle between its pointof projection and highest point of trajectory is

(a) - J l + 2cos20 (b) - J l + cos

20

2V

(c) + 3cos20 (d) wcosO

For inelastic collision, between two spherical rigid

bodies

(a) the total kinetic energy is conserved

(b) the total poten tial energy is conserved




(c) the linear mo ment um is not conserved

(d) the linear mome ntum is conserved.

23. A point source emits sound equally in all directions

in a non-absorbing medium. Two points P and

Q are at distances of 2 m and 3 m respectively

from the source. The ratio of the intensities of

the waves at P and Q is

(a) 9 : 4 (b) 2 : 3

(c) 3 : 2 (d) 4 : 9

A A A24. If a vector 2i + 3; + 8k is perpendicular to the

A A Avector 4; - 4 i + a A:, then the value of a is

(a) - 1 (b) \

(c) (d) 1

25. A resistor and a capacitor are connec ted in series

with an ac source . If the pote ntia l dr op acrossthe capacitor is 5 V and that across resistor is

12 V, then applied voltage is

(a) 13 V (b) 17 V (c) 5 V (d) 12 V

26. A radio transmit ter radi ated 1 kW power at a

wavelength 198.6 m. How many photons does

it emit per second?

(a) 1010 (b) 1020

(c) 1030 (d) 1040

27. Total angu lar mome nt um of a rota ting b ody

remains constant, if the net torque acting on thebody is

(a) zero (b) max imu m

(c) min imu m (d) uni ty

28. In the fusion reaction 2

H + \ H > \ He + g n,

the masses of deuteron, helium and neutron

expressed in amu are 2.015, 3.017 and 1.009

respectively. If 1 kg of deut eri um und erg oes

complete fusion, find the amount of total energy

released.

1 amu = 931.5 MeV/c2

.(a) 9 x 1013 J (b) 6 x 1010 J

(c) 9 x 1010 J (d) 6 x 1013 J

29. From the top of a tower, a par ticl e is thr ow n

vertically downwards with a velocity of 10 m s_1

.

The ratio of distances covered by it in the 3 rd and

2nd seconds of the motion is

(Take g = 10 m s~2)

(a) 5 : 7 (b) 7 : 5

(c) 3 : 6 (d) 6 : 3

30. The essentia] distincti on bet wee n X-rays and

y-rays is that

(a) y-rays have smaller wavelength that X-rays

(b) y-rays emanate from nucleus while X-rays

emanate from outer part of the atom

(c) y-rays have greater ionizing power than

X-rays

(d) y-rays are more penetrating than X-rays

31. Given figur es show the arr ang eme nts of two

lenses. The radii of curvature of all the curved

surfaces are same. The ratio of the equivalent

focal length of combinations P, Q and R is

(P) (Q) (R)

(a) 1 : 1 : 1 (b) 1 : 1 : - 1(c) 2 : 1 : 1 (d) 2 : 1 • 2

32. Whe n bot h the listener and source are moving

towards each other, then which of the following

is true regarding frequency and wavelength of

wave observed by the observer?

(a) More frequency, less wavelength

(b) Mor e frequency, mor e wavelength

(c) Less frequency, less wavelength

(d) More frequency, constant wavelength

33. A block of steel of size 5 cm x 5 cm x 5 cm isweighed in water. If the relative density of steel

is 7, its apparent weight is

(a) 4 * 4 * 4 * 6 ^ (b) 5 * 5 x 5 * 9 j

(c) 4 * 4 * 4 * 7g (d) 5 * 5 * 5 * 6 g

34. An electro n is accelerated un de r a pote ntia l

difference of 182 V. The maximum velocity of

electron will be

(a) 5.65 x 106 m s"

1 (b) 4 * 10

6 m s"

1

(c) 8 x 106 m s"

1 (d) 16 x 10"« m s"

1

(Charge of electron is 1.6 x 10~19

C and its mass

is 9.1 x 10~31 kg)

35. A force F, of 500 N is requ ired to pu sh a car

of mass 1000 kg slowly at constant speed on a

levelled road. If a force F2 of 1000 N is applied,

the acceleration of the car will be

(a) zero (b) 1.5 m s~2

(c) 1.0 m s"2 (d) 0.5 m s~

2

36. In the circuit sho wn in the figure, the potential

difference across the 4.5 pF capacitor is




37.

4.5 |aF

H h -

3 (jF

6 (iF

12 V

(a) - V (b) 4 V

(c) 6 V (d) 8 V

A uniform cylinder has a radius R and length L.

If the moment of inertia of this cylinder about an

axis passing through its centre and normal to its

circular face is equal to the moment of inertia of

the same cylinder about an axis passing its centre

and perpendicular to its length, then

(a) L = R (b) L = SR

R(c) L =

s(d) L =1*

38. A conducting wire of cross-sectional area 1 cm2

has 3 x 1023

charge carriers per metre3. If wire

carries a current 24 mA, then drift velocity of

carriers is

(a) 5 x 10"2 m s-

1 (b) 0.5 m s"

1

(c) 5 x 10"3 m s^ (d) 5 x 10"

6 m s"

39. An engin e has an eff icie ncy of - . When the6

temperature of sink is reduced by 62°C, its

efficiency is doubled. Temperature of the source

is

(a) 124°C (b) 37°C

(c) 62°C (d) 99°C

40. The magneti c field B at a distance r from a long

straight wire carrying current varies with distance

r as shown in the figure.

(a) B

(c)

Direct ions : In the fol lowi ng quest ions (41-60), a

statement of assertion (A) is foll owe d by a statement

of reason (R). Mark the correct choice as :

(a) If both assertion and reason are true and reason

is the correct explanation of assertion.

(b) If both assertion and reason are true but reason

is not the correct explanation of assertion.

(c) If assertion is true but reason is false.(d) If both assertion and reason are false.

41. Assertion : The light year and wavelength consist

of dimensions of length.

Reason: Both light year and wavelength represent

distances.

42. Assertion : The tyres of an aircraf t are m ade

slightly conducting.

Reason : Frictional charges develop ed dur ing

take off and landing.

43. Assertion : A body falling freely may do so withconstant velocity.

Reason: The body falls freely, whe n acceleration of

a body is equal to acceleration due to gravity.

44. Assertion : Surface energy of an oil drop is same

whether placed on glass or water surface.

Reason : Surface energy is dependent only on

the properties of oil.

45. Assertion : In case of metallic wires the I-V

characteristics are linear as long as the current

flowing through the wires is small.

Reason : Joule heat ing is directly proportional tothe square of the current.

46. Assertion : In circular motion, work done by

centripetal force is zero.

Reason : In circular motion centripetal force is

perpendicular to the displacement.

47. Assertion : Diamagnetic materials can exhibit

magnetism.

Reason : Diamagnetic materials have permanent

magnetic moment.

48. Assertion : A shell at rest, explodes. The centre ofmass of fragments moves along a straight path.

Reason : In explosion the linear momentum of

the system remains always conserved.

49. Assertion : Alth oug h the same current flo ws

through the line wires and the filament of the

bulb, the filament gets heated up but not the line

wires.

Reason : Line wires are ma de of copper a nd the

filaments are made of tungsten.

7 4 PHYSICS FOR YOU | MAY '13 ;



50. Assertion : A rocket moves forw ard by pus hin g

surrounding air backward.

Reason : It derives the necessary thrust to move

forward according to Newton's third law of

motion.

51. Assertion : The bob of a simpl e pe nd ul um is a

ball full of water, if a fine hole is made in the

bottom of the ball, the time period first increasesand then decreases.

Reason : As water f lows out of the bob the wei ght

of bob decreases.

52. Assertion : A bar ma gne t is dr op pe d in a long

hollow copper tube. The acceleration of the magnet

can be zero.

Reason : The cop per tu be has neg l ig ib l e

resistance.

53. Assertion : Total energy is conserved in moving

a satellite to higher orbit.Reason : Sum of change in PE and KE is same

in magnitude and opposite in nature.

54. Assertion: The ratio of the energies of the hyd rog en

9atom in its first to second excit ed stat e is - .

4

Reason : Energy of an electron in the nth orbit is

inversely proportional to n2.

55. Assertion : Work done in lifting a bucket full of

water from a well by means of a rope tied to the

bucket is negative.

Reason : Work done by gravitational force is

positive when a bucket full of water is lifted

upwards.

56. Assertion : No free charge carriers are available

in depletion layer.

Reason : Thickness of depletion layer is fixed in

all semiconductor devices.

57. Assertion : A ladder is more apt to slip, when

you are high up on it than when you just begin

to climb.

Reason : At the high up on a ladder, the torque islarge and on climbing up the torque is small.

58. Assertion : A domestic electrical appliance,

working on a three pin will continue working

even if the top pin is removed.

Reason : The third pin is used only as a safety

device.

59. Assertion : When observer moves away from

the source, the frequency of sound appears to

increase.

Reason : The appare nt freq uency of sou nd does

not depend on whether the observer is moving

towards source or away from the source.

60. Assertion : The propa gat ion of radio wav es is

termed as sky wave propagation.

Reason : All radio wav es are called sky w aves .

SOLUTION

1. (c) : Potential difference is V = 100 ± 5 V

Current is I = 10 ± 0.2 A.

Percentage error in voltage is obtained as

AV 5— = — x 100% = 5%V 100

Percentage error in current is obtained as

^ = 0 ^ x l 0 0 % = 2o /o

I 10

Using, R = y , the percen tage erro r in resista nce

is calculated as R = 5% + 2% = 7%.

2. (c) : For first one-third of distance

xDista nce cover ed = — km

3

speed = 10 km h - 1 .

The time taken for the journey,

t - E l l h = ± rh10 30

For the next one-third of distance

xDista nce cover ed = — km.

3

Speed = 20 km h"1

The time taken for travel is

x/3f-, =

20 60

For the last one-third of distance :x

Distance covered = — km.3

Speed is 60 km h 1

The time taken for travel is

60 180

total distanceAverage Speed =

total time

x

X X X

30 60 180

180x

10*= 18 km h -l




(c) : For simple harmonic motion,

v = (o\la2 - x2

When x = — v = £0. |a2- —

2 V 4

As (0 =2n In S

v = aT 2

v = -Ky/3a

T T 2 T

(d): Capacitance of a parallel plate condenseris

e nAC = —

d

Potential difference across the plates is

V = Ed

Energy stored in the condenser is

...(i)

...(ii)

u = I C v2

= i2 2

e0 A(Ed)2

= - e 0 E2 Ad

(Using (i) and (ii))

(b): Here, initial mome ntum of rock is zero, sinceno external force exists. Hence mom ent um mus t

remain conserved i.e.

P\+P2+P3 = 0' P3=-(P\+P2)As two parts right angle to each other

\p3 \=\p1 + p2 \ = s[tf2 2+ P2

p3 = N / ( lx l 2 )2+( 2x8)

2 = 20

m : T = 5k g.

(d): The phase difference (<)>) between the wavelets

from the top edge and the bottom edge of the slit

2 Kis <(> = — (d sinO) where d is the slit width.

X

The first minima of the diffraction pattern occurs

Xat sinO = —

„ , 2K (* X ) „

(a):

Let R be the resistance of total length of circular

wire, R = 36 Q.

The resistance of smaller arc AB, Rj = R/6 = 36/6

= 6 Q.. The resistance of bigger arc AB, R2 = 5R/6

= 5 x 36/6 = 30 Q.

8.

Now J?! a nd R2 are in parallel in between points A

and B. So, effective resistance between A and B

6 + 30

(a): Acco rdin g to conse rvat ion of an gular

momentum,

m,Mj + m2u2 = m^-y + m2v2

50 x 2 + 60 x 12 = 50 x 4 + 60 x p2

62= — m s

2 6

Coefficient of restitution,

W, -M,i.e., e =

4- (62/6)

2 - 1 2

IS! 19

30

9. (c): Pressu re differ ence betw een lungs and

atmosphere

= 760 mm - 750 mm = 10 mm = 1 cm of Hg.If one can draw from a depth of I cm of water,

then pressure difference

= 1 x 13.6 x 980 = I x 1 x 980

or I = 13.6 cm of water.

10. (a): If a moving charged par ticle is subjected to

a perpendicular uniform magnetic field, then

according to F = ijwBsinO, it will experience a

max imum force which will provide the centripetal

force to particle and it will describe a circular

path with uniform speed.

11. (c) : Efficiency of an engine, T) = 1 — —

where T, is the temperature of the source and T2

is the temperature of the sink.

1 T

6 t;or, -±. = - ...(i)

When the temperature of the sink is decreased

by 62°C efficiency becomes double.

Since, the temperature of the source remain

unchanged

... 2 x l = l ~ ^ >6 T,

or,1 , (T 2- 62)

(i)

or —T 2- 62

2 or, 27; =3T 2 —l86T,

or 27] = 3 37 —186 [using (i)]

74 PHYSICS FOR YOU | MAY'13;



1 - 22

T, =1 86 or, ZL2

186

or, T, = 372 K = 99°C.2 „ 2

12. (a): Self-inductance, L = ^

or

h hl2 r 2,

1

L2 2

- I M !

13. (d): Frequency of LC oscillation

1/2

271 >/LC

According to the problem,

1

2 L x 4 C

L x C= 2A /2

_ _ t)

2 2V2 2V2'

14. (a): Speed of sound in an ideal gas,

tyRTV ~\ M

here v

h2 ~ v

0 2

...(i)

(Using(i))

V ^ H l \yo2R T

o2

M.

and YH2 = YO2

M r

M,

Mr J 2 /

0 o 2 ) :

('.' Both are diatomic)

(100 + 273)

15. (b): H =

= 23.31 K = -249.7°C

velocity of light in vacuum (c)velocity of light in medium (v)

v = vX = 2 x 1014

* 5000 x 10-10

In the medium, v = 108 m s"

1

3x10s „c

H = - =v 108

16. (a) Temperature of liquid oxygen will first increase

in the same phase. Then, the liquid oxygen will

change to gaseous phase during which t emperature

will remain constant. After that temperature of

oxygen in gaseous state will increase. Hence opt ion

(a) represen ts corresponding temper ature- time

graph.

17. (c) : Total kinetic energy = K.E. of translation

+ K.E. of rotation

= -M»2 + 4 /c o2

2 2

K.E. of rotation

Total K.E.

\MV 2 \

- f )

IMK 2 42 R

2

^Mi;2

2

K 2

R2 K 2

i K21 H r

R2

K 2 + Ri K21 H r

R2

18. (b): Time period, T = 271,

Ti = 2n Jb T

2=2n Ji

Here, a' =

i.e., ? = —« 9

T2 =27t

R A

T 2 3

R2 _ §R2

(R + ft)2 (R + 2 R)2(•/ h = 2R)

- L = 2 7 t J - = 3 T 1

r /9 1

19. (a)

20. (b): Initial mass of the magnet m, = m and final

mass of the magnet m2 = 4m.

TThe time period, T = 2n ,

:2n.

MB

Imk 2

'MB

or T°c \fm

Tj _ _ yfm _ 1

T2 Jnu, \Tim 2

or T 2 = 2Tj = 2 T

74 PHYSICS FOR YOU | MAY '13;



21. (a): Average velocity, vav = ^

Here, H = maximum height

v2 sin 29

R = range =g

and T = time of flight

2v sinG

T12

v2 sin

2 9

2*

-(i)

H

• » a v = | N / l + 3cos2e

22. (d)

23. (a): From a point source, energy spre ads over

the surface of sphere of radius r.

PEnergy Power

Intensity, 1 = ^ — = — ,Time x Area Area 4ji r

or / <

U

/ \2

Ju UJ9

4

24. (c) : Let a = 2i + 3i + 8k_ A A A

b = 4; - 4 / + afcA A A

= - 4i + 4/ + ak

According to the question, alb

or a • b = 0

or (2f + 3; + 8fc )-( -4f + 4; + afc) = 0

or - 8 + 12 + 8a = 0

1a = —

2

25. (a): Let the applied voltage be V volt

Here, V R = 12 V, Vc = 5 V

V = JV 2 + V 2 =J(12)2

+ (5)2

= ^144 + 25^ M

— 1 1 —

= Vl69r

26. (c) : Here, number of photons emitted per second,

H )1000 x 198.6

E hv he

6.6 x 10"34

x 3 x 108

:1030

.

27. (a): Torque t = Rate of ch an ge of a ng ul ar

momentum (L)

or i = (constan t L)dt

or x = 0

28. (a): Am = 2(2.015) - (3.017 + 1.009) = 0.004 am u

.-. Energy released = (0.004 x 931.5) MeV

= 3.726 MeV

Energy released per deuteron

= = 1.863 MeV2

Number of deuterons in 1 kg

= 6

-°2 X 1026

=3.01x 10*2

Energy released per kg of deuterium

fusion

= (3.01 x 1026 x 1.863) MeV

= 5.6 x 1026

MeV

= 9 x 1013 J

29. (b): Distance travelled in 3rd second

S3 = 10 + y (2 x 3 - 1) = 35 m

Distance travelled in 2nd

second,

S2 = 10 + y (2 x 2 - 1) = 25 m

SO 7=> — = -

S2 5

30. (b): The wav el engt h of the y-rays is shor ter.

However the main distinguishing feature is thenature of emission.

31. (a): In given diagra m P, Q and R lenses are in

contact.

For P combination of lenses

J _ _ I I = 1

F P'f + f ~ f

or FD =

1 1 1— = —(- — for combination of lenses

F A h

/Similarly for Q a nd R combinations

F Q={ ^D F R = |

Therefore, F P : FQ : F R is equal to 1 : 1 : 1

32. (a) 33. (d) 34. (c)

35. (d) 36. (d) 37. (b)

38. (c) 39. (d) 40. (c)

41. (a) 42. (a)




43. (b): Wh en th e bo dy is fal lin g freely, on ly

gravitational force is acting on it in the vertical

downwar d direction. Because of this acceleration,

the velocity of the body increases and will

be maximum when it touches ground. When

downward accelerating force is balanced by

upward retarding force, the body falls with a

constan t velocity. This constan t velocity is kn own

as terminal velocity.

44. (d): If a mat eri al is in contact with ano the r

material, the surface energy depends on the

interaction of molecules of the materials. If the

molecules of the materials attract each other,

surface energy is reduced and when they repel

each other, the surface energy is increased. Thus

the surface energy depends on both the materials,

so both the assertion and reason are false.

45. (a): Heat produced per unit time by Joule

effect = PR. This increases the temperature ofthe conductor and thus

the resistance. For Ohm's

law to be valid, the

temperature must remain

constant. The increase in

temperature makes the

current versus vol tage

graph non-linear.

46. (a): In circular motion centripetal force is

perpendicular to the displacement, i.e., 0 = 90°.

Work done W = Fscos0

= Fscos 90°

= 0

Hence, work done by centripetal force is zero.

47. (d): There is no inherent magnetic moment in

atoms or molecules of diamagneti c materials. This

characteristic, magnetic mom ent of the atoms is the

bases of classification of materi als as diama gnet ic

paramagnetic and ferromagnetic. Diamagnetic

materials do not show magnetism.48. (d): According to the law of conservation of linear

momentum, the centre of mass after explosion

remains at rest. While other parts of the shell move

in all directions making total linear momentum

of the system to zero.

49. (b): The resistance of filament is extremely h igh

whereas the resistance of line wires is very low.

Therefore, for the same current the filament gets

heated up a lot more than the line wires.

V

The reason that filaments are made of tungsten

and line wires of copper is also true. But this does

not explain the high value of resistance of filament

wires. High resistivity and high temperature

co-efficient of tungsten alone cannot account for

the extremely high value. The filament wires

are extremely thin which makes their resistance

very high. The reason is incomplete and does not

explain the assertion.

50. (a): Acc ord ing to Newt on 's thi r d law of

mo t io n, ther e is an equ al and opp osi te

reaction to every action. In the same manner

the burning fuel in rocket ejects gases with

a speed backward due to this, the rocket

moves forward.

51. (b): When ball is completely filled with water,

the centre of gravity of the pendulum is at the

centre of the ball when water starts flowing out,the centre of gravity sh ifts below, thus increasing

the length of pendulum, and increasing time

period. When ball is more than half empty the

centre of gravity again rises up so length of

pendulum decreases and t ime period also

decreases.

52. (a): When a bar magnet is dropped in a long

hollow copper tube initially the velocity of the

magnet increases because of the acceleration

due to gravity. The magnetic field of a bar

magne t is non un iform . When the magnet movesdownwards with increasing speed, the non-

uniform magnetic field causes eddy currents to

flow in the copper tube. As the copper tube has

negligible resistance large eddy currents are set-

up in the tube which oppose the motion of the

bar magnet. This opposing force on the magnet

increases with increasing velocity of the magnet.

When the opposing force becomes equal to the

gravitational force the net acceleration becomes

zero. The magnet attains a constant terminal

velocity.53. (c) : Total energy is given by

E=-GMm

2 R

where R is the radius of orbit.

54. (a) 55. (d)

57. (a) 58. (a)

60. (c)

56.

59.

(c)

(d)




Contd. from Page No. 27


D = -2x1.5'

10J

2.2 x l On

x 0.01

7.7 x 10J

Hz = 178.2 Hz3 V7

17. (a): Loss of energy is max imu m when collision

is inelastic.

Maximum ener gy loss _ 1 u2

f =mM

2 ( M + m)

(M + m)

Hence, Statement-I is false, Statement-II is true.

18. (a): q

M- dx-H

Consider a small element of length dx at a distance

x from O.

Charge on the element, dQ = j-dx

Potential at O due to the element is

1 dQ 1 QdV = -

4jte ,o 4;te0 Lx-dx

Potential at O due to the rod is

2 LV: • J f l r . J

Q

471 e0 Lxdx

1 ^ l n x £ =

Q l n 2

47 te 0 L

19. (d): Le t k be the sp ri ng

constant of spring and it

gets extended by length x0

in equilibrium position.

In equilibrium,

kx o + f B = Mg

4 j t e 0 L

MgaL AI

XN=-

M g f c L A

2 M

20. (a): When the screen is placed perpendicular to

the line joining the sources, the fringes will be

concentric circles.

21. (d):

/ / / / 7 / 7 / / / / / / / / / / / / T 7 A > / / / / /

According to law of conservation at point of

contact,

mr 2 co0 = mvr + mr co

2 I v= mvr + mr —

mr co0 = mvr + mvr

mr z(i)0 = 2mvr

v-J M S).2

22. (d): The amplitude of a damped oscillator at a

given instant of time t is given by

A = A0e~mm

where A0 is its amplitude in the absence of

damping, b is the damping constant.

As per question

After 5 s (i.e. t = 5 s) its amplitude becomes

0.9 A0 = A0e' H5)l2m = A0e~5b,2m

0.9 = e~5bl2m ...(i)

After 10 more second (i.e. t = 15 s), its amp litude

becomes

oA0 = A0e-6(15)/2m

= A0e~15bl2m

a = (e-5W2m

)3 = (0.9)

3 (Using (i))

= 0.729

23. (a)

24. (b): Energy of the satellite on the surface of the

planet isGMm j GMm

E; = KE + PE = 0 - R R

If v is the velocity of the satellite at a distance 2R

from the surface of the planet, then total energy

of the satellite is

GMm1 2 E f =- mv +

' 2

1

= -m2

(R + 2R))

\2G M

(R + 2R)

1 GMm GMm

GMm

3 R

GMm

2 3 R 3 R 6 R

Minimum energy required to launch the

satellite is

GMm ( GMm \ AE = E f - E t =

GMm

6 R +

6 R

GMm

R

R )

5 GMm

6 R




25. (a): In a hydrogen like atom, when an electron

makes an transition from an energy level with n

to n - 1, the frequency of emitted radiation is

v = RcZ

(» - i r

= RcZ 2n2-(n-lf

(n2)(n-1)2

As n > > 1

RcZ 2 In IRcZ 2:. u = — = —

or a) oc -

RcZ 2 (2w-l)

n 2 ( n - l ) 2

26. (b): The situation is as shown in the figure.. . , *<? % 1

When a particle of mass m and charge q01 = ^ j

placed is at the origin is given a small displacement

along the y-axis, then the situation is shown in

the figure.

y

Fsin8

jfcose•Fcosf)''

» *FsinO0T:«

y•.B

a " a

By symmetry, the components of forces on the

particle of charge q0 due to charges at A and B

along .x-axis will cancel each other where along

y-axis will add up.

.-. The net force acting on the particle is

i w o y= 2Fcos0 = 2-

471 e n

y

2

+«

2 h2+*2)

2 <6) y471 e0 (y2+a2

)2V(y2 + i

2)

*•' 0 = 1 (Given)

471 e0 (y2+f l2)3/2

As y < < a

p =neti 2 y

4tc e0 a3or F„

27. (d): According to Newton's law of cooling the

option (d) represents the correct graph.

28. (c): For po te nt ia l to be ma de zero , af ter

connection

120Q = 200C2

6 C, = 10C2

3 Q = 5C2

29. (b): The I-V characteristics of a LED is similar

to that of a Si junction diode. But the threshold

voltages are much higher and slightly different

for each colour.

Hence, the option (b) represents the correct

graph.

30. (d): According to lens maker's formula

Ri R*,

As the lens is plano-convex R1 = R,R2 = oo

" /

or / = -

R

R

(l i-1)

As speed of light in the medium of lens is

2 x 108 m/s

...(ii)_ c _ 3 x 10

a m/s _ 3

^ v 2 x10

s

m/s 2

If r is the radius and t is the thickness of lens

(at the centre), the radius of curvature R of its

curved surface in accordance with figure will

be given by

R2 = r 2 + {R - tf

R2 = r 2 + R2 + t 2- 2Rt

2 Ri = r 2 + t 2

2

R = -2t

(v r » t )

Here, r = 3 cm, t = 3 mm = 0.3 cm

(3 cm )2

R--2 x 0.3 cm

- = 15 cm

On substituting the values of ji and R from Eqs.

(ii) and (iii) in (i), we get

, 15 cm f = = 30 cmJ (1-5-1)

• ! • • ! • • ! •




Clear Your Concepts ADVT.

ROLLING MOTION

Most of the students who are serious about JEE preparationwere not confident about the characteristics of pure rolling

motion, charge distribution on spherical shells and plates

etc, But in our Saxena Success Point, it is game play for

our students.

Pure rolling as a pure rotation: We know that a rigid

body is said to be in pure rotation about an axis only

when the angular velocity (co) of every point of that

body is same about any reference point on that axis.

Therefore to consider the pure rolling motion as a pure

rotation about an axis which is instantly passing throu gh

the point of contact and perpendicular to plane of the

rolling body, we have to prove that the co of va riouspoints of the body is same w.r.t. point of contact and it

is also equal to co of the rolling body.

Consider a ball of mass M and radius R is roll ing

without slipping on a horizontal floor. The linear

velocity of the centre of mass C, is vcm to the right.

The angular velocity of the ball rotating clockwise

ab ou t C is co. For rol l in g wi th ou t sl ip pi ng ,

vcm = R(ji. Let us consider a general point A, on the ball.

The angle made by the radius CA with the vertical

is 0 as shown in the figure. The linear velocity of A

relative to the floor is the resultant of a horizontal

velocity. vcm , and a tangential velocity of magnitudevcm. The angle between these two combining velocities

is 0 as well. It follow from the paralle logram law that

the magnitude of v A is 2vcm cos (0/2) and its direction

bisects the said angle. Since ZCAD = ZFAG = (0/2) a nd

AF1 CA. We conclude that v A is directed perpendicu lar

to the line DA.

c / F8/ 7 /

Now, in triangle ABD, ZBAD =

simple trigonometry :

90° and ZADB = 0/2. By

cos ZADB

DA 0

or DA = DB cos ZADBÎR cos - DB 2Therefore, the angular velocity of A about D is :

VA 2 V ™ C ° 4 Van ft.

D A 2* co s 9

2

"cm

R= —CO

R

The proof will be so much easier for the points lying

on the vertical through D. The angular velocity of the

highest point, B, about D is :

v.B 2i>„,- = <0

DB 2 R

The angular velocity of C about the same point is :

CODC

cm

R

:. From the above analysis we can say that the ball

is in pure rotation about an axis passing through D

instantly.

ELECTROSTATICS

Dear Students the following two questions should serve as

food for thought. Happy problem solving!

1. In the fig. shown, a point charge

-q is at the centre, A is earthed,

B and C are given -2q and +3qrespectively. A and C are connected

with a conducting wire without

touching B. Find the charges

appear on all the 6 surfaces of the

system

In the system of plates shown in the fig., plates A & D

are connected, C & E also connected by a conducting

wire and B is earthed. Find the charges that appear

on all the faces of the system of plates.

Q2

Qi

2.

_L

ML

B

rf/4 d/ 8

D

ftd! 16,

EU

The above content has been contributed by : Saxena Success Point, P2 Complex, Brig. Hoshiar Singh Road,

Sarojini Nagar, New Del hi 110023 Contact No.: 011-24673656, 8447405006,

For any further clarification feel free to mail your queries to [email protected].






*0TEST Y©UR

^ PHYSICS APTITUDEThis aptitude test is the ultimate challenge for a smart student. The questions push you to your limits, stretch

your abilities, and of course, tease you.

This aptitude test is also scored, which means at the end we will tell you "how smart you are!" i.e. you are

genius, expert, progressive expert, average or

1. A wo oden block of mas s 1 kg is attached to the

hook of a spring balance. The spring balance is

then raised with an acceleration of 9.8 m s~2. The

apparent weight of the block is

(a) 1 kg wt (b) 2 kg wt

(c) 3 kg wt (d) 4 kg wt

2. Two homogene ous spheres A and B of masses

m and 2m having radii 2a an d a respectively are

placed in touch. The distance of centre of mass

from first sphere is

(a) a (b) 2a

(c) 3a (d) 4a

3. A flywheel rotates with a unif orm a ngula r

acceleration. Its angular velocity increases from20n rad s"1 to 4071 ra d s"1 in 10 s. How ma ny

rotations did it make in this period?

(a) 80 (b) 100

(c) 120 (d) 150

4. The bob of a simp le pen du lu m is of mas s 10 g.

It is suspended with a thread of 1 m. If we hold

the bob so as to stretch the string horizontally and

release it, what will be the tension at the lowest

position? (Take g = 10 m s~2)

(a) zero (b) 0.1 N

(c) 0.3 N (d) 1.0 N

5. A current of 7 A flow s th rou gh the circuit as sho wn

in the figure. The potential differe nce across point

B and C is

B4 £2—WWW^

7 A

A

2Q-A/WW

10 Q•NMh-

5Q-VWVW

7 A

(a)

(c)

16 V

10 V

(b) 8 V

(d) 5 V

6. The tor que requir ed to hol d a small circular coilof 10 turns, 2 x 10"4 m2 area and carrying 0.5 A

current in the middle of a long solenoid of 103

turns per m carrying 3 A current, with its axis

perpendicular to the axis of the solenoid, is

(a) 12ti x 10~7 N m (b) 67 tx l0 " 7 N m

(c) 4ti x 10"7 N m (d) 27 ix l0 - 7 N m

7. A dy namo dissipate s 20 W, wh en it supp lies a

current of 4 A through it. If the terminal potential

difference is 220 V, the emf pro duc ed is

(a) 220 V (b) 225 V

(c) 215 V (d) 300 V

8. 200 V ac source is fed to series IC R circuit hav ing

X L = 50 £2, X c = 50 Q and R = 25 Q. Potential drop

across the inductor is

(a) 100 V (b) 200 V

(c) 400 V (d) 10 V

9. The vari atio n of velocity of a parti cle mo ving

along a straight line is shown in figure. The

distanc e trans vere d by the particle in 4 second is

(a) 60 m

(c) 55 m

(b) 25 m

(d) 30 m

74 PHYSICS FOR YOU | MAY '13;



10. A rubber rope of length 8 m is hung from the

ceiling of a room. What is the increase in length

of the rope due to its own weight? (Given Young's

mod ulu s of elasticity of rubber = 5 * 106 N m~

2

and density of rubber = 1,5 » 103 kg m~

3 and

g = 10 m s"2)

(a) 1.5 mm (b) 6 mm

(c) 24 mm (d) 96 mm

11. A car sounding its horn at 480 Hz moves towards

a high wall at a speed of 20 m s"1. If the speed of

sound is 340 m s-1 , the frequency of the reflected

sound heard by the man sitting in the car will be

nearest to

(a) 480 Hz (b) 510 Hz

(c) 540 Hz (d) 570 Hz

12. An a-particle of mass 6.4 x 10~27 kg and charge

3.2 x 10~19 C is situated in a uniform electric field

of 1.6 x 105 V m

_1. The velocity of the particle at

the end of 2 x 10~2 m path when it starts from rest

is

(a) 2V3xl05 m s"

1 (b) 8 x 105

m s"1

(c) 16 x 105 m (d) 4\ /2 xl 0

5m s

13. The natura l boron of atomic weight 10.81 is found

to have two isotopes B10

and B11

. The ratio of

abundance of isotopes in natural boron should

be(a) 11 : 10 (b) 81 : 19

(c) 10 :1 1 (d) 19 :8 1

14. The dimens ions of Planck's constant and angular

momentum are respectively

(a) [ML2T

-1] and [MLT"

1]

(b) [ML2T_1] and [ML2T_1]

(c) [MLT"1] and [ML

2!"

1]

(d) [MLT"1] and [ML

2T~

2]

15. A body dro ppe d fr om the top of a tower covers adistance 7h in the last second of its journey where

h is the distance covered in the first second. How

much time does it take to reach the ground?

(a) 3 s (b) 4 s

(c) 5 s (d) 6 s

16. A body of mass M kg is on the top point of a

smooth hemisphere of radius 5 m. It is released

to slide down the surface of the hemisphere. It

leaves the surface when its velocity 5 m s_1

. At this

instant the angle made by the radius vector of the

body with the vertical is

(Take g = 10 m s~2)

(a) 30° (b) 45°

(c) 60° (d) 90°

17. A toy gun uses a spring of force constant k. When

charged before being triggered in the upward

direction, the spring is compressed by a distance x. If the mass of shot is m, on being triggered it

will go up to a height of

kr 2 r 2

(a) (b) JL_ mg kmg

(c) - ^ L (d) 2 mg mg

18. The escape velocity f rom the earth is 11.2 km s_1

.

The escape velocity from a planet having twice

the radius and the same mean density as the earthis

(a) 22.4 km s-1 (b) 11.2 km s"

1

(c) 5.6 km s"1 (d) 15.8 km s"

1

19. A steel ring of radius r and cross-sectional area

A is fitted onto a wooden disc of radius R(R > r).

If Young's modulus be Y, then the force with the

steel ring is expande d is

R(a) AY- (b) AY

R-r

(c)Y (R-r)

(d) "777

Xl

AR

20. A rod 70 cm long is clamped from middle. The

velocity of sound in the material of the rod is

3500 m s_1. The frequency of fundamental note

produced by it is

(a) 700 Hz (b) 1250 Hz

(c) 2500 Hz (d) 3500 Hz

21. A source of sound emits a sound of frequency600 Hz and is rotating in a circle of radius 4 m at

a linear speed of 40 m s_1

. What is the lowest and

highest frequency heard by an observer at long

distance away at rest with respect to the centre of

circle?

(a) 545 Hz and 687 Hz (b) 683 Hz and 535 Hz

(c) 535 Hz and 683 Hz (d) 687 Hz and 545 Hz

22. 200 cc of a gas is compressed to 100 cc at the

atmospheric pressure 106 dyne cm-2. If the change is




sudden , what is the final pressure? (Given y = 1.4)

(a) 1.6 x 106 dyne cnr

2 (b) 2.0 * 10

6 dyne cnr

2

(c) 2.6 x 106 dyne cnr

2 (d) 3.0 * 10

6 dyne cnr

2

23. Starting with the same initial conditions, an

ideal gas expands from volume Vj to V 2 in three

different ways. The work done by the gas is W( if

the process is purely isothermal, W2 if the processis purely isobaric, and W3 if the process is purely

adiabatic, then

(a) W r >W 2 >W 3 (b) W 2 >W 3 >W,

(c) W 3>W2>W i (d) W 2 >W,>W 3

24. The specific heat of water in cal g"1 is s = 0.5f

2,

where t is the temperature on the Celsius scale. If

the tempera ture of 10 g of water is raised through

10°C, what is the amount of heat required?

(a) 60 cal (b) 200 cal

(c) 0.6 kcal (d) 2kca l

25. A transistor connected at common-emitt er mode

contains load resistance of 5 k£2 and an input

resistance of 1 k£L If the input peak voltage is

5 mV and the current gain is 50, find the voltage

gain.

(a) 250 (b) 500

(c) 125 (d) 50

26. The surface of the metal is illuminated wi th the

light of 400 nm. The kinetic energy of the ejectedphotoelectrons was found to be 1.68 eV. The work

function of metal is

(a) 1.42 eV (b) 1.51 eV

(c) 1.68 eV (d) 3.0 eV

27. A parallel beam of light of wavelength

3141.59 A is incident on a small aperture. After

passing through the aperture, the beam is no

longer parallel but diverges at 1° to the incident

direction. What is the diameter of the aperture?

(a) 180 nm (b) 18 |um

(c) 1.8 m (d) 0.18 m

28. An object is placed 30 cm to the left of a diverging

lens whose focal length is of magnitude 20 cm.

Which one of the following correctly states the

nature and position of the virtual image formed?

Nature of image Distance from lens

(a) inverted, enlarge 60 cm to the right

(b) erect, diminished 12 cm to the left

(c) inverted, enlarged

(d) erect, diminished

60 cm to the left

12 cm to the right

29. In an ac generator, a coil wit h N turns, all of the

same area A and total resistance R, rotates with

frequency oo in a magnetic field B. The maximum

value of emf generated in the coil is

(a) NAB (b) NABR(c) NAB(o (d) NABRu)

30. The magnetic flux linked with the coil varies

with time as <)> = 312 + 4f + 9. The magni tude of the

induced emf at 2 s is

(a) 9 V (b) 16 V

(c) 3 V (d) 4 V

31. The total current suppli ed to the circuit by the

battery as shown figure is

(a) 1A

(c) 4 A

(b) 6 A

(d) 2 A

32. If two electric bulbs, each designed to operate wi th

a power of 500 W in 220 V line, are put in series in

a 110 V line, what will be the power generated byeach bulb?

(a) 31.25 W (b) 21.25 W

(c) 11.25 W (d) 41.25 W

33. A particle moves in a straight line with a constant

acceleration. It changes its velocity from 10 m s_1 to

20 m s_1

while passing thr ough a distance 135 m in

t second. The value of f is

(a) 12 (b) 9

(c) 10 (d) 1.8

34. A block of mass m is resting on a smooth

horizontal surface. One end of a uniform rope of

mass — is fixed to the block, which is pul led

in the horizontal direction by applying a force F

at the other end. The tension in the middle of the

rope is

(a) (b) - F7




(C) (d) 7

-F

35. A partic le is acted upon by a force F which varies

with position x as shown in figure. If the particle

at x = 0 has kinetic energy of 25 J, then the kinetic

energy of the particle at x = 16 m is

14 16 x(m)-+

(a) 45 J

(c) 70 J

(b) 30 J

(d) 20 J

36. A child is standing with fo lded hands at the centre

of a platform rotating about its central axis. The

kinetic energy of the system is K. The child now

stretches his arms so that the moment of inertia

of the system doubled. The kinetic energy of the

system now is

(a) 2K (b) |

(c) | (d) 4K

37. Two solid spherical planets of equal radi i R havingmasses 4M and 9M have their centres separated

by a distance 6R. A projectile of mass m is sent

from the planet of mass 4M towards the heavier

planet. What is the distance r of the point from

the lighter planet, where the gravitational force

on the projectile is zero?

(a) 1.4R (b) 1.8R

(c) 1.5R (d) 2.4R

38. The terminal speed of a sphere of gold

(density = 19.5 kg m~

3

) is 0.2 m s

_1

in a viscousliquid (density = 1.5 kg nr3). Then the terminal

speed of a sphere of silver (density 10.5 kg m-3

) of

the same size in the same liquid is

(a) 0.1 ms "1 (b) 0.4 ms "

1

(c) 0.2 ms "1 (d) 0.3 ms"

1

39. Two capacitors A and B are connected in series

with a battery as shown in figure. When the switch

S is closed and the two capacitors get charged

fully, then

2 (iF 3fiF

(a) the poten tial di fference across the plates of A is 4 V and across the plates of B is 6 V.

(b) the potentia l difference across the plates of

A is 6 V and across the plates of B is 4 V.

(c) the ratio of electrical energies stored in A and

B is 2 : 3.

(d) the ratio of charges on A and B is 3 : 2.

40. A copper wire of length 1 m and radius 1 mm

is joined in series with an iron wire of length

2 m and radius 3 mm and a current is passed

through the wires. The ratio of current density in

the copper and iron wires is

(a) 2 : 3 (b) 6 :1

(c) 9 : 1 (d) 18 :1

41. A horizontal long straight wire placed east-west

carries a current 3.6 A. What is the distance of

neutral point from the wire if the horizontal

component of the earth's magnetic field from

south to north is B H = 3.6 * 10~5 T.

(a) 1 x 10~

2

m (b) 2 * 10"

2

m(c) 1.5 x 10"2 m (d) 3 x 10-2 m

42. The mutual inductance of an induction coil is

5 H. In the primary coil, the current reduces from

5 A to zero in 10"2 s. What is the induced emf in

the secondary coil?

(a) 2500 V (b) 25000 V

(c) 2510 V (d) zero

43. A coil of resistance 200 £2 and self inductance

1.0 H has been connected to an ac source of

200frequency —— Hz. The phase difference between

voltage and current is

(a) 30° (b) 63°

(c) 45° (d) 75°

44. If an electron in n = 3 orbit of hyd rogen atom

ju mp s down to n = 2 orbit, the amount of energy

released and the wavelength of radiation emitted




(a) 0.85 6V, 6566 A (b) 1.89 eV, 1240 A

(c) 1.89 eV, 6566 A (d) 1.5 eV, 6566 A

45. A photon of energy 4 eV is incident of a metal

surfac e who se wor k funct ion is 2 eV. The min imu m

reverse potential to be applied for stopping the

emission of electrons is

(a) 2 V (b) 4 V

(c) 6 V (d) 8 V

46. In a carbon monoxide molecule, the carbon and

the oxygen atoms are separated by a distance

1.12 x 10"10 m. The distance of the centre of mass

from the carbon atom is

(a) 0.48 x 10"10

m (b) 0.51 x 10"10

m

(c) 0.56 x 10"10 m (d) 0.64 x 10~10 m

47. A block of mass M is pulled up by a uniform

string of mass-M tied to it, by applying a force F at

the other free end of the string. The tension at the

midpoint of the string is

(a) F (b) 0.5F

(c) 0.75F (d) IF

48. The work don e in increasing the size of

a rectangular soap film with dimensions

8 cm x 3.75 cm to 10 cm x 6 cm is 2 x 10"4 J. The

surface tension of the film in N nr 1 is

(a) 1.65 xlO"2 (b) 3.3 xl0 - 2

(c) 6.6xl0- 2 (d) 8.2 5xl 0- 2

49. Two capillaries of length L and 2L and of radii R and 2 R are connected in series. The net rate of

flow of fluid through them will be (Given rate of

the flow through single capillary X = nPR4 /8r\L)

(a) | X (b) f x

(c) ^ X (d) ^ X7 5

50. A 2 kg copp er block is hea ted to 500°C and then

it is placed on a large block of ice at 0°C. If the

specific heat capacity of copper is 400 J kg -1 °C_1

and la tent heat of fus ion of water is 3.5 x 10s J kg-1,

the amount of ice that can melt is

(a) | kg (b) | kg

(c) | kg (d) | kg

V Check Your Score! If your score is J •

> 90% you are a Genius

9 0 - 8 1 you are an Expert

80-71 you are a Progressive Expert7 0 - 6 1 you are Average

<60 you Need to work hard!

(o) OS (B) '6* (q) "8* (o) i t (P)

(B) •Sfr 0) •w (q) 'Efr (B) (q) •Ifr(o) w (q) '6£ (B) '8E (P) ZE (q) '9£(B) •SE (P) K (q) EE (B) Z£ (o) "IE

(q) •0£ (o) '6Z (q) '8Z (q) LZ (B) '92(B) S2 (P) •frZ (P) £2 (3) ZZ (o) '12

(o) "02 (q) '61 (B) 81 (0) Li (o) '91

(q) 'SI (q) "N (P) El (p) Zl 0) 'U

(P) 01 (o) '6 (o) 8 (q) L (B) 9

(o) 'S (o) "fr (P) ' £ (q) Z (q) l

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Physics for You - June 2013 (Gnv64)