Physics for You - June 2013 (Gnv64)

MCQs for PRACTISE

ICERT/ft

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Questions for

NEET

THOUGHT PROVOKING

PROBLEMS

EXAMINER'S

MIND

CHALLENGING

PROBLEMS

BRAIN MAP

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JEE Main

r.'iMi'.'i 11 w. i n n a m m m

_{SOLVED PAPER 2013}

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PHYSICS APTITUDE

MCQs for PRACTISE

Questions for

NEET

THOUGHT PROVOKING

PROBLEMS

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MIND

CHALLENGING

PROBLEMS

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PHYSICS

Vol. XXI No. 5 May 2013 Corporate Office:

Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (HR),Tel: 0124-4951200 Regd. Office

406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029.

e - m a i 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 Aptitude 86

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Editor: Anil Ahlawat

Copyright© MTG Learning Media (P) Ltd. All rights reserved. Reproduction in any form is prohibited.

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Q

rial

Expanding Knowledge and In-depth

Research are not contradictory

W

idening 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 finally 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 partitions disappear 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!

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PRACTICE PAPER 2 Q 1 3

Advanced

By : Akhil Tewari, B-Tech, IT-BHU S E C T I O N

Straight Objective Type

This section contains 6 multiple choice questions numbered 1 to 6. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct.

1. Velocity of a particle moving along a straight line

a

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

(V

a and b are positive constants. The approximate

graph of acceleration versus position is

(a) > X (b)

J

k a^^

r

(c) 2. (d) none 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) none 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 n e w 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 moving due 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 n e w particle ?

(a) v (b) 2v

v

(c) —j= (d) None 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 and its direction.

(a) 10 m s"1,120° (b) 10 m s "1, 1 8 0 ° (c) 10 m s"1, 90° (d) l O m s "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= k m h"1 w h e n car is passing by a m a n 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 m i n i m u m time in which he can cross the

(c) t a n_ 1V 2 (d) t a n

S E C T I O N - II

Multiple Correct Answer(s) Type This section contains 4 multiple choice questions numbered 7 to 10. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE is/are correct.

7.

8.

9.

In the figure, the pulley

P moves to the right

with a constant speed u. The d o w n w a r d speed of A is Vfy and the speed of B to the right is Vg. Then, (a) vA = vB vB = u + vA VB + U = VA B H k (b) (c) (d) 10.

the two blocks have accelerations of the same magnitude.

Two particles A and B start simultaneously f r o m 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 paths must intersect at some point (b) They must collide at some point (c) They will collide only if a\U-[ = a2u2

(d) If Mi > u2 and 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 and aimed towards 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 either (a), (b) or (c), depending 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 w i d t h d, flowing with a speed u. The point opposite him across the river is P.

river is — .

v

(b) He can reach the point P in time — . (c) He can reach the point P in time

V^2 - U2

(d) He cannot reach Pifu> v. SECTION - III Assertion Reason type

This section contains 4 multiple choice questions numbered 11 t o 14. Each question contains Statement-1 (Assertion) and Statement-2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of w h i c h ONLY ONE is correct.

(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(b) Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1.

(c) Statement-1 is True, Statement-2 is False. (d) Statement-1 is False, Statement-2 is True.

11. Statement-1 : The m a x i m u m range along the

inclined plane, w h e n thrown d o w n w a r d is greater than that w h e n thrown u p w a r d along the same inclined plane with constant velocity.

Statement-2 : The m a x i m u m range along inclined

plane is independent of angle of inclination.

12. Statement-1 : A particle is projected 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 horizontal direction

remains same but velocity along vertical direction 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 mass 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 : Normal reaction always acts

perpendicular to the contact surface.

14. Statement-1: In high jump, it hurts less w h e n an

athlete lands on a heap of sand.

Statement-2 : Because of greater distance and

hence greater time over which the motion of an athlete is stopped, the athlete experience less force w h e n lands on heap of sand.

S E C T I O N - IV

Linked Comprehension Type

This section contains 3 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct.

P1 5_ 1 7 : Paragraph for Question Nos. 15 to 17

When a particle is projected at some angle with the horizontal, the path of the particle is parabolic in nature. In the process the horizontal velocity remains 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° with the horizontal then

15. The radius of curvature of path of particle at the instant when the velocity vector of the particle becomes perpendicular to initial velocity vector

(a) (c) 20 3 ^ 3 4 0 3 ^ 3 10 (b)

WT

m (d) 80 3 ^ 3

16. The magnitude of acceleration of particle 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 m s "2 (b) 20 m s -2

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

P18-20 : Paragraph for Question Nos. 18 to 20 A particle is projected with

velocity u at an a n g l e 0 with an inclined plane of inclination 0 < 45° w i t h the horizontal.

18. The time taken when velocity of projectile

becomes parallel to the plane (a) (c) wsinO 2wtan0 (b) (d) !(COt0 M t a n 0 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. Radius of curvature of the path of projectile w h e n velocity is parallel to the plane

(a) (c) w2cos20 U COS0 (b) (d) M2C O S22 0 g c o s30 ..2 g c o s30

P21-23 : Paragraph for Question Nos. 21 to 23 An object at rest remains at rest and an object in motion will continue its motion with a constant velocity unless it experiences a net external force. But the magnitude of force given by Newton'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, w h e n 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 Newton's laws, concept of centre of mass is included.

21. Pulley and strings are massless. The force acting on the block of mass m

(a) 2F (b) F

(d) 4F

I

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

22. A particle of mass m moves along a circle of radius

R. The m o d u l u s 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) V 2 w 2\fln k r (b) ( d ) 2 V 2 1 mv mv

23. If velocity of movable pulley is v and velocities of block of masses M] and M2 are v j and V2 then the correct relation between them

1 0 PHYSICS FOR YOU | MAY'13

M 777777777777777777777

£

(a) v = i>i + (c) V = 2{VX + V2) ( n M l U [ j M 2 a + l>i (b) l> = — 2 ( d ) o = S O L U T I O N S 1. (a)

2. (a): Free body diagram for m yr N

r

2a ... (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

_-N

1

- > r On solving, a = 2mg (M + 5m) Net acceleration of m, am = = ^ a = jSmg_ m (5m + M ) 3. (a): t a n a = t a n a = BsinO A + Bcos6 sin0 1 + cosG ...(i) t a ri n 6 tan— = — 2 , a a a + - c o s G 2 a sin9 tan— = 2 (2 + cos0) . . . ( i i ) 2 t a n a t a n a = 1 - t a n 2a sin0 sin0 2 + cosG 1 + cos0 _ sin2 9 (2 + cos0) On solving, Cos0 = — / . 0 = 1 2 0 ° 2

4. (c): Refer the given below. H n m # — mvi+ mvj = 2 m(v'xi + v'yj) 2 A A A A or — (vi + vj) = v'x i+ v' j v , v or vY = or v.. = -2 y -2 2

"T2

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

Vrm = (Vrx ~ vm)i+vryi V

Case (i): tan90° = — or vrx = 5 m s_ 1

vrx~5

Case (ii): vrm = (5i - 1 0 / ) + vry) (y vm = 10/)

tan60° = or vrv = -5S 5 - 1 0 r y vr = 5i - 5>/3 5, = 10 m s" - 1 Z(j) = tan' 6 . (a) - 5 ^ 3 or d> = 120°

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 = vA

••• vB = u + vA Also z>b~va or flg = a

a-8. (a, c, d)

9. (a, b, c): Superimpose an upward 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 downward acceleration.

10. (a, c, d) 11. (c) 12. (a): ux = u cosO, ax = 0 vx = ux + ax t = ur = u cosO uy = u sinO ; ay --:/ ( s i n 0 - g f = ttsinO - 2usin0 vy = -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 = — A t 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 2

t =

gsin0 10sin60°

_s

Let Vy be the vertical component of velocity at that instant. Then, = 5>/3-5_ 1 0 x 2 10 v = —j= m s V3 ! = 5 m s or g c o s a = R = R g cosa B 2 0 R = — m 3v3 16. (a) -2 17. (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)

- n V, + VR,

1 4 PHYSICS FOR YOU I MAY '13

Thought

Provoking

Kinematics

1. A train starts from a station with a constant

acceleration of 0.40 m s'2. A passenger arrives at a 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 r u n and catch the train?

2. A particle is projected f r o m a point 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 particle strike the plane: (i) horizontal (ii) at right angle?

3 A particle is projected f r o m a point on the level ground and its height is h w h e n at horizontal distances a and 2a from its point of projection. Find the velocity of projection.

4. A bullet is fired into a viscous liquid with a

velocity v0. The retarding force is proportional to square of velocity, so that the acceleration becomes

a = -kv2.

(a) Derive an expression for the distance travelled in the liquid.

(b) What is the distance travelled in the liquid when velocity is reduced to — and the corresponding time ?

5. A small sphere of mass m is released f r o m rest in a large vessel filled with oil w h e r e it experiences a resistive force proportional to its speed, i.e.,

Fd = -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 r u n n i n g out of a conical f u n n e l at the rate of a m m3 s"1. If the radius of the base of the f u n n e l is R m m and the altitude is H m m , find, the rate at w h i c h the w a t e r level is d r o p p i n g w h e n it is h m m f r o m the top?

7. A circular wire f r a m e is fixed in a vertical plane. A s m o o t h w i r e is slightly s t r e t c h e d b e t w e e n points Px a n d P2. A b e a d slides f r o m point Pu the highest point of the circle. Determine (a) its velocity v w h e n it arrives at P2 (b) find the time taken by it?

1.

SOLUTIONS

A s s u m e the train is at x = 0 at t = 0, the equation for train is

1 , 1 ,

xT =-aTt2 = ±(0.40)f2

The passenger reached x = 0 at t = f0 = 6 s, so his coordinate at time tis xP = v,,(t -10).

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

~aTt2=vp(t-t0) or aTt2-2vpt+2vpt0=0

or t _vp± - 2aTvpt0

The roots are real if Vp-2aT vp tQ > 0

vp >2aTt0 = 2x 0.40 x 6 = 4.8 m s"1.

R a n d h a w a Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second FL, Sector-20C, Chandigarh

2. (i) Along horizontal direction, h = (Mcos0)f p

_/

,

7

MCOS(() h O _Q

Along vertical direction, 0 = Msinij) - gt o r Msin<|> = gt

and h = Msin<j>f •

V

2 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 - ~(^cos45°)f2 or t i 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 - 1 or - =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 is

J2.

•••(i)

y = xtanO - - —

2 m2 cos20

With origin at the point of projection,

gx2 - 2u2x 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 cos2 0 xh ...(iii) ...(iv) Divide (iii) by (iv), we get

3 a tan0 „ 3 h — - = or tan0 = —

2a h 2 a

From Eq. (iv),

M 2=^s e c 2e = ^ [ 1 + t a n2e ] =S ^ h h 1 h 1 + 9h~ 4 a2 h •9/2 4. Acceleration of bullet, ~ = - k v2 at dv r jt or — --kdt v On integrating, we get, " J_{• dv} F ...(i) v0 0 k dt or -kt 1 1 _-kt Vn V dx 1 dt or v = — = — or dx = — dt . 1 1 kt + kt + v, _o v. jdx=j o o kt + dt 1, —— or x = - In f k vn o kt + ...(ii) ...(iii) . . ( i v )

We calculate time t, when velocity is reduced to

From Eq. (ii)

vn₀ v. kvn Put t in Eq. (iv), we get

1 , „ 0.693

x = - n 2 = . k k

5. (a) Force acting on sphere = mg - kv where k is constant. Acceleration of ball, — = g- — v dt m or -dv mg m -dt ...(i) ...(ii) 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 . . . ( i i i ) m In I k or In t r : ?• td -s v 1 r 3 3 3 ? f i f s v 1 r 3 3 3 ? f i f 3 3 3 mg -v mg k ,-TS. l - e

- t

1

) !

- ( i v ) o r YH

where, x = — is called time constant.

k

(b) When the sphere reaches terminal speed, the acceleration of the sphere becomes zero. Then

mg mg = kvt or vt = —1-. V

V 1

I - 7 H

V I

At 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 t r h. T/ 1 (hRY, dV nR2h2 dh V = -71 — « or — = 3 {H

J

dt h2 dt dh a3H2

(a) v2 = v2 + lax = 02 + 2(^cos9)(2Rcos6) •Of = 2 ^Rg cos Q

t j ± j 4 R g c o s e = 2Ir a gcos6 ]j g

2 0 PHYSICS FOR YOU | MAY 13

«

SOLVED PAPER 2 Q

1 3

Main

1. The anode voltage of a photocell is kept fixed. The

wavelength X of the light falling on the cathode 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 bigger loop. 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 weber (b) 9.1 x 10"11 weber (c) 6 x 10"n weber (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. N o w switch S. is closed, S2 kept open, (q is charge on the capacitor and T = RC is

capacitive time constant). Which of the following statement is correct? V R

\

Si c _JS 2 5. L (a) At f = 1 q = CV( 1 - e"1)

(b) Work done by the battery 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 magnets of length 1 cm each

have magnetic moments 1.20 Am2 and 1.00 Am2

respectively. They are placed on a horizontal table 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 joining their centres is close to

( H o r i z o n t a l c o m p o n e n t of e a r t h ' s m a g n e t i c 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. Of

the four choices given after the Statements, choose the one that best describes the two Statements.

Statement-I : H i g h e r the range, greater is the

resistance of ammeter.

Statement-II: To increase the range of ammeter,

additional shunt needs to be used across it. (a) Statement-I is false, Statement-II is true. (b) S t a t e m e n t - I is t r u e , S t a t e m e n t - I I is t r u e ,

Statement-II is the correct e x p l a n a t i o n of Statement-I.

(c) S t a t e m e n t - I is t r u e , S t a t e m e n t - I I is t r u e , Statement-II is not the correct explanation of Statement-I.

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

7. An ideal gas enclosed in a vertical cylindrical

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 V0 and its pressure 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 A2y (b) 2K V0M (c) < d > 2K V MV, _o

8. Let [GQ] denote the dimensional formula of the

permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then

(a) [£0] = [ M1 L2 r1 A] (b) [e0] = [M"1 L"3 T2 A] (c) [ G0] = [M"1 L"3 T4 A2] (d) [GQ] = [M"1 L2 T"1 A'2] 2Pn ii i _{> 1} r I o 2 v0 T h e a b o v e p-v d i a g r a m r e p r e s e n t s t h e thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The a m o u n t of heat, extracted f r o m the source in a single cycle is (a) 4p0»o

(b) p0v0

<o i f

POVO

( d ) ( y J / W o

10. A projectile is given an initial velocity of

(f+2/)m/s, where i is along the ground and j is along the vertical. If g = 10 m/s2, the equation of its trajectory is

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

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

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) IQ ( C ) I0/2 (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 parallel with a load resistance 100 kilo ohm. Find the m a x i m u m modulated frequency which could be detected by it.

(a) 5.31 k H z (b) 10.62 M H z (c) 10.62 k H z (d) 5.31 M H 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, w h e n a 240 W heater is switched on in parallel to the bulb?

(a) 10.04 Volt (b) zero Volt (c) 2.9 Volt (d) 13.3 Volt

14. A metallic rod of length '/' is tied to a string of length 21 and m a d e 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 electromagnetic 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 s o n o m e t e r w i r e of l e n g t h 1.5 m is m a d e of steel. The tension in it p r o d u c e s an elastic strain of 1%. What is the f u n d a m e n t a l frequency of steel if d e n s i t y a n d elasticity of steel are 7.7 x 103 kg/m3 and 2.2 * 10" N/m2 respectively? (a) 770 Hz (b) 188.5 H z

(c) 178.2 H z (d) 200.5 H z

17. This question 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 with stationary point particle

of mass M. If the m a x i m u m energy loss possible is given as / [ - m y * then / = m

M + m

Statement-II: Maximum energy loss occurs w h e n

the particles get stuck together as a result of the collision.

(a) Statement-I is false, Statement-II is true. (b) S t a t e m e n t - I is t r u e , S t a t e m e n t - I I is t r u e ,

S t a t e m e n t - I I is a c o r r e c t e x p l a n a t i o n of Statement-I.

(c) S t a t e m e n t - I is t r u e , S t a t e m e n t - I I is t r u e , Statement-II is not a correct explanation of statement-I.

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

18. A charge Q is uniformly distributed over a long

rod AB of length L as s h o w n in the figure. The electric potential at the point O lying at a distance L f r o m the end A is AWfflBZMZSZmK o (a) (c) Q l n 2 471 e0L 3 Q 4K e0L (b) (d) Q l n 2 8n e0L Q 4 j t e0L In 2

19. A uniform cylinder of length L and mass M having cross-sectional area A is s u s p e n d e d , w i t h its length vertical, f r o m 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 w h e n 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

s e p a r a t e d by a s m a l l d i s t a n c e ' d ' as s h o w n . The fringes obtained on the screen will be

(a) concentric circles (b) (c) straight lines

21. A hoop of radius 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. W h a t will be the velocity of the centre of the h o o p w h e n it ceases to slip?

26.

(a) m0 (b) r CON "0 (c) (d) r ( 0o 4 3 2

22. The a m p l i t u d e of a d a m p e d oscillator decreases to 0.9 t i m e s its original m a g n i t u d e in 5 s. In another 10 s it will decrease to a times its original m a g n i t u d e w h e r e 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 s u r f a c e energy, so t h a t its t e m p e r a t u r e remains unchanged. What should be the minimum r a d i u s of the d r o p for this to be possible? The surface tension is T, density of liquid is p and L is its latent heat of vaporization.

^ (b) Bk

p L T

£ (d) £

VP L PL

24. What is the m i n i m u m energy required to launch a satellite of mass m f r o m 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 hydrogen like atom electron makes transition f r o m an energy level with q u a n t u m n u m b e r n to another with q u a n t u m n u m b e r (n - 1). If n > > 1, the frequency of radiation emitted is proportional to

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 charge

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 h e a t e d to t e m p e r a t u r e 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)

t

(C) 8t

O

(d) e0

o

28. Two capacitors C, and C2 are charged to 120 V and 200 V respectively. It is f o u n d that by connecting them together the potential on each one can be m a d e zero. Then

(a) 9C1 = 4C2 (b) 5CI = 3C2

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

29. The I-V characteristic of an LED is

(a)

(c) R

30. D i a m e t e r of a p l a n o - c o n v e x lens is 6 cm a n d thickness at the centre is 3 m m . 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 d u e to current loop 1 at an axial point •>2

B1 =

-2 (d2+R2)3'2

Flux linked with smaller loop 2 d u e to B1 is <t>2 = B1A2 = " • nr 2 (d2+R2)3'2 The coefficient of m u t u a l i n d u c t a n c e b e t w e e n the loops is M _ $2 _ M2™ "2 = T ~ _{2 (d2 + R2)3'2}

Flux linked with bigger loop 1 is = = TTK

2(d + R ) '

Substituting the given values, we get

4TI x I P '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 / 2

4>i =

((), = 9.1 x 10~n weber

3. (d): T h e g r a p h b e t w e e n angle of deviation(S) and angle of incidence (i) for a triangular prism is as s h o w n

in the adjacent figure. O

4. ( d ) : V Hi— R

I K

C uooooff^-L

As switch S, is closed and switch S2 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 q} 0 = CV] At t = q = CV(1 - e~x/2x) = CV(1 - e'V2) At t = x q = CV(1 -2x, ~X,X) = CV( 1-tT1) At t • 5. q = CV(1 - e_ 2 t / T) = CV(1 - e~2)

(c) : The situation is as s h o w n in the figure. N bh . 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 Bn e t = B j + B2 + Bh

H0 n0 M2

An r = Ho

4jtr3

(MJ + M2) + Bh

Substituting the given values, we get - - 4 7 1 X 1 0 ' [1.2 + 1] + 3.6 x 10~5 n e t 471 x (10 x 10~2)3 ->-7 1 0 " 10 ,-3 x 2.2+ 3 . 6 x 1 0 -5 = 2.2 x 10"4 + 0.36 x 10"4 = 2.56 x 10^ W b / m ' 6. (a) 7. (d): M FBD of piston at equilibrium PgA Mg P3TMA + Mg = PqA ...(i) FBD of piston w h e n piston is p u s h e d d o w n a distance x Mg (p0 + dP)A (P0 + dP)A - (-Patm A + Mg) = M . . . ( i i ) As the system is completely isolated f r o m its surrounding therefore the change is adiabatic. For an adiabatic process

PVT = constant VdP + V^PdV = 0 yPdV or dP = — dP = V yP0(Ax) Vn (v dV = Ax) ...(hi) Using (i) and (iii) in (ii), we get

M ^ - ^ x or dt2 v0 d2x dt2 i M : MVn

Comparing it with standard equation of SHM,

d2x dt2 We get 2 = - (0 1 c o 2 = ^ I MVn or co = Y ^ MVn Frequency, x> = co 1 yP0A2 271 2TT

V

MV_o r

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

r =_ 1 fill 4ne0 r2 ~~ Fr2 [AT] [AT] 1E0J_ 7 7 [MLT"2][L]2 9. (c): 4 2Po Wt = [M_ 1L"3T4A2] —~ w— D t; 2%

H e a t is extracted f r o m the source in p a t h DA and AB.

Along path DA, volume is constant. Hence,

AQda = nCvAT = nCv(TA - TD) According to ideal gas equation

pv pv = nRT or T =

nR

For a monoatomic gas, q = —R

v 2

2Povo Povo

nR

AQ, •DA

nR ~ 2 Povo Along the path AB, pressure is constant. Hence

AQab = nCpAT = nCp(TB - TA) 5

For monoatomic gas, Cp = — R

2Po2vo 2P0vo

nR nR

10 = y W o .•. The a m o u n t of heat extracted f r o m the source in a single cycle is A Q = A QDA + A QAB 3 10 13 = 2povo+ y ? W o = y W o 10. (c) : Given: u=i + 2j A A As u = uxi + uyj :. ur = 1 and u„ = 2

i Also x = uxt 1 2 and y = Uyt -—gt :. x = t and 1/ = 2f - i x 10 x t_{s 2} 2 = 2f - 5f2 Equation of trajectory is y = 2x - 5x2.

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

1 2

N o w this light will pass t h r o u g h the second polaroid 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 m a x i m u m f r e q u e n c y w h i c h can be detected is 1 "0 = 2nmax where, x = CR

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

R = 100 kilo ohm = 100 * 103 ohm

ma = 0.6 1 U = 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 bulb _ 120 V x 120 V B ~ 60 W Resistance of heater, RH — W W 6Q Bulb H ' 120 V = 240 Q 120 V x 120 V = 60 Q 240 W

Voltage across bulb before heater is switched on,

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

As bulb and heater are connected in parallel. Their equivalent resistance is

(240Q)(60n) = 4 8 n eci 2 4 0 n + 6 0 0

.•. Voltage across bulb after heater is switched on

120 V . 4 8 1 1

2 48 Q + 6 Q

Decrease in the voltage across the bulb is

\V = V1 - V2 = 10.41 V = 10.04 V

CO 21 ... I

M-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

E0 = B0c

E0 = (20 x 10""9 T) (3 x 108 m s"1) = 6 V/m 16. (c) : Fundamental frequency of vibration of wire

is

JL ll

2L\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 _ . Stress As y = = Strain _ TL " ~ AAL_ 1 lYAL 2L' pL i n Here, Y = 2.2 x 10" N/m", p = 7.7 x 10J kg/mJ — = 0.01, L = 1.5 m

L Cont. on Page No. 82

FOCUS

Engineering Aspirants 2013

-Are you Prepared for JEE Advanced?

(Here are the few Tips to get your optimum)

T

he JEE Advanced (formerly k n o w n 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 students will be short listed f r o m JEE Main 2013 to appear for the JEE Advanced 2013 on June 2nd, 2013. A serious aspirant ideally must have completed the syllabus by now.

Schedule of JEE ( A d v a n c e d ) , 2 0 1 3

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 contributes about 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 m a y be given in particular on the following topics:

Mathematics : Quadratic Equations & Expressions, Complex Numbers, Probability, Vectors & 3D Geometry, Matrices in Algebra; Circle, Parabola, Hyperbola in Coordinate Geometry; Functions, Limits, Continuity and Differentiability, Application of Derivatives, Definite Integral in Calculus. 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 A d v a n c e d , 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. According to the general trends, Mechanics and Electricity and Magnetism are the most important topics in terms of the n u m b e r 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 f r o m 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.

Cracking t h e JEE (ADVANCED) 2 0 1 3

"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 u p o n 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 w h e n you take a break. 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 2 8 PHYSICS FOR YOU | MAY '13 ?

• • •

NCERTXtract

1 • • / / • • •

uestions forNEET

GRAVITATION

l . Which of the following statements is correct?

(a) Acceleration due to gravity increases with increasing altitude.

(b) Acceleration due to gravity increases with increasing depth.

(c) Acceleration due to gravity increases with 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 2GmA f ^ . 1 3. 4.

The earth is an approximate 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 everywhere but not directed towards the centre.

(c) will be s a m e e v e r y w h e r e in m a g n i t u d e 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 and vA be speed of the p l a n e t w h e n at p e r i h e l i o n and a p h e l i o n respectively. Which of the following relations is correct? Planet P Perihelion >A Aphelion 5. P _ _ fA \ "A The t i m e p e r i o d T of t h e m o o n of p l a n e t

mars (mass Mm) is related to its orbital radius

jR as (G = Gravitational constant) 4TC2R3 nn2 4k2GR3 (a) (c) T2=: T2= GMm In.R3G M„, (b) M„ (d) T2 = 4rcMmGR3 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 r a d i u s 2RE about the earth. Energy required to transfer it to a circular orbit of radius 4RE is (where ME and RE is the mass and radius of the earth respectively) GMptn (b) GMEm a) E 2 RE (b) 4 RE GMPm (d) GMEm GMPm (d) 16 R£

8- Which of the following statements is true? (a) A geostationary satellite goes a r o u n d the

earth in east-west direction.

(b) A geostationary satellite goes a r o u n d the

earth in west-east direction.

(c) The time-period of a geostationary satellite

is 48 hours.

(d) The angle between the equatorial plane and the orbital plane of geostationary satellite is 90°.

9. Two uniform solid spheres of equal radii R, b u t mass M and 4M have a centre to centre separation 6R, as shown in figure. A projectile of mass m is projected from the surface of the sphere 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

V l l T

, x 3 GM ( c ) I S I R (d) 5 GM 3 R

10. The time interval between two successive noon

when sun passes through zenith point (meridian) is known as

(a) sidereal day (b) mean solar day (c) solar year (d) lunar month

11- An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth. The gain in its potential energy is (where RE is the radius of the earth)

(a) m§RE

( c ) ~ m g RE

(b) - m g RE

(d) | mgRE

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, RE = 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 106 m

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

A system of four 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) -4V2 Gm

~T

14. The escape velocity from the surface of the earth is

(where RE is the radius of the earth)

(a) (b) JgR^

(c) 2 j g RE (d) fiK

15. A body weighs 63 N on the surface of the earth.

What is the gravitational force on it d u e 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 m o t i o n of an object u n d e r the

gravitational influence of another object. Which of the following quantities is not conserved? (a) Angular m o m e n t u m

(b) Mass of an object (c) Total mechanical energy (d) Linear m o m e n t u m

17. In our solar system, the inter-planetary region has chunks of matter (much smaller in size compared to planets) called asteroids. They

(a) will not move around the sun since they have very small masses compared to sun. (b) will m o v e in an irregular w a y because of

their small masses and will drift away into outer space.

(c) will move a r o u n d the sun in closed orbits b u t not obey Kepler's laws.

(d) will move in orbits like planets and obey Kepler's laws.

18- A rocket is fired from the earth towards the sun. At w h a t 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* 1 0 " 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 spheres each of mass M and radius 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 RE is the radius 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 is

1 6 (a) ± v / 2 (b) <b) 1 v ' 3 (d)

The gravitational force between a hollow spherical shell (of radius R and u n i f o r m density) and a point mass is F.

Which of the following g r a p h s 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 a s t r o n a u t experiences weightlessness in a space satellite. It is because

(a) the gravitational force is small at that location in space.

(b) the gravitational force is large at that location in space.

(c) the astronaut experiences no gravity. (d) the gravitational force is infinitely large at

that location in space. Three masses each of m a s s m a r e p l a c e d at the vertices of an e q u i l a t e r a l t r i a n g l e ABC of side I as shown in f i g u r e . T h e f o r c e acting on a m a s s 2m placed at the centroid O of the triangle is *-x (a) zero (c) -6Gm2 r (b) (d) 6Gm2 o Gm2 *

25. A satellite of mass m orbits the earth at a height h above the surface of the earth. H o w much energy must be expended to rocket the satellite out of earth's gravitational influence?

(where ME and RE be mass and radius of the earth respectively) (a) (c) GMEm 4 (RE+h) GMEm (b) (d) GMEm 2(R£+/i) 2 GMEm 26. (.RE+h) ' ' (RE+h)

Two stars each of mass M are approaching each other for a head-on collision. When they are at a distance r, their speeds are negligible. The radius of each star is R(r >> R). The speed which they collide is (a) (c) (b) (d) 2 GM R

27. Different points in earth are at slightly different distances f r o m the s u n and hence experience different forces d u e to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the centre of mass causing translation and a net torque at the centre of mass causing rotation around an axis through the centre of mass. For the e a r t h - s u n system ( ap pr o ximatin g the earth as a uniform density sphere)

(a) the torque is zero.

(b) the torque causes the earth to spin.

(c) the rigid body result is not applicable since the earth is not even approximately a rigid body.

(d) the torque causes the earth to move around the sun.

28. _{A comet orbits the sun in a highly elliptical orbit.} Which of the following quantities remains constant throughout its orbit?

(i) Linear speed (ii) Angular speed (iii) Angular m o m e n t u m (iv) Kinetic energy (v) Potential energy (vi) Total energy

(a) (i), (ii), (iii) (b) (c) (iii) and (vi) (d)

(iii), (iv), (v) (ii), (iii) and (vi)

29. Match the following.

For a satellite in circular orbit,

33.

Column I Column II

(A) Kinetic energy (P) GMEm

2 r (B) Potential energy _(q) GME V r (C) Total energy (r) GMEM r (D) O r b i t a l velocity (s) GMEm 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 gravitational force is d e p e n d e n t of the intervening medium.

The g r a v i t a t i o n a l force is a c o n s e r v a t i v e force.

The gravitational force is a central force, (d) The gravitational force obeys the inverse

square law.

31. What is the angle between the equatorial plane

and 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 towards M. (c) m will move towards 2M. (d) m will have oscillatory motion.

The escape velocity of a b o d y f r o m the earth depend on

(i) the mass of the body.

(ii) the location from where it is projected. (iii) the direction of projection.

(iv) the height of the location f r o m where the body is launched.

(a) (i) and (ii) (b) (ii) and (iv) (c) (i) and (iii) (d) (iii) and (iv)

34. The escape speed of a body on the earth's surface is 11.2 km s"1. A body is projected with twice of this speed. The speed of the body when 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.2V2 km s_ 1 (d) -4=- km s_ 1

V2

35. As observed from earth, the sun appears to move

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 between earth and mercury is not inverse square law. (c) not be true because the major gravitational

force on mercury is d u e to sun.

(d) not be true because mercury is influenced by forces other than gravitational forces.

SOLUTIONS

1. (c) : Acceleration d u e to gravity at a altitude h

above the earth's surface is SH=I

2h

Rr

...(i) where g is the acceleration d u e to gravity on the earth's surface and RE is the radius of the earth. Eq. (i) shows that acceleration d u e to gravity decreases with increasing altitude.

Acceleration d u e to gravity at a depth d below the earth's surface is

1 i -REJ

...(ii) Eq. (ii) shows that acceleration d u e to gravity decreases with increasing depth.

Acceleration d u e to gravity at latitude A.

gx = g - RE®2 COS2X ...(iii)

where co is the angular speed of rotation of the earth.

Eq. (iii) shows that acceleration d u e to gravity increases with increasing latitude.

Acceleration d u e to gravity of body of m a s s m is placed on the earth's surface is

...(iv) gm£

Rl

Eq. (iv) shows that acceleration d u e to gravity is i n d e p e n d e n t of the mass of the b o d y b u t it d e p e n d s u p o n the mass of the earth.

3 2 PHYSICS FOR YOU | MAY 13

(b):

From figure

AB = BC = CD = AD = /

AC = BD = 2

Total potential energy of the system of f o u r particles each of mass m placed at the vertices

A, B, C and D of a square is Gxm xm U = Gxmxm^ ( Gxmxm ' AB )+l AC Gxmxm BC Gxmxm BD

M

-

M-AD Gxmxm CD Gm2 Gm2 Gm2 f i A Gm2 / V / +

I ^ J

+ / V + I \ Z\ 4 Gm2 2Gm2 i f l Gm 2Gm („ 1

— r ®

(d)

(a): Angular momentum of planet at P,

LP = mpvPrP

where mp is the mass of the planet. Angular momentum of planet at A,

La = mpvArA

According to the law of conservation of angular momentum

LP = La

mvvPrP = mpvArA

'A UP

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

3 / 2

GMm JGM„

R

where the symbols have their meaning as given in the question.

Squaring both sides, we get

47t2 R3 T2=

-GM„

6. (b): According to Kepler's third law M

T2

R _•MS

R6 ixES

where RMS is the mars-sun distance and RES is

the earth-sun distance. \ 3 / 2

R _MS R _ES

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

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

GMFm

£,= —

' 4Re

Final energy of the satellite is

E r -GMEm Change in energy, AE = Ef - £, A £ = -GMEm GMEm \ 8RE

{

_{4Re /} GMEm GMEm GMEm 8 RE 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- -M

6 R

Let the projectile be fired with minimum velocity,

v from the surface of sphere of mass M to reach

the surface of s p h e r e of mass 4M. Let N be neutral point at a distance r from the centre of the sphere of mass M.

At neutral point N,

GMm _ G(4M)w r2 (6 R-i)2

(6R - r)2 = 4r2

6R - r = ± 2r or r = 2R or -6R

The point r = -6R does not concern us. Thus, ON = r = 2R.

It is sufficient to project the projectile 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)m

1 2

— mv

-2 R 5 R

At the neutral point N, the speed approaches zero. The mechanical energy at N is

GMm G(4M)m _ GMm GMm

2R 4 R 2R R~ According to law of conservation of mechanical energy, 1 2 - mv -2 Ef - En GMm 4 GMm GMm GMm 2 2 GM v = 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 w h e n 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 = GMEm

where M£ and m be masses of earth and object

respectively.

At the surface of the earth, r = RE U1 =~

GMvm

Rr

At a height h from the surface,

r = RE + h = RE + RE = 2RE

U2 = GM

Em

~2Rr

Increase in potential energy is

AU=U2- UJ GMEm ~ ~ 2 RT ' GMptn GMEm R? = 2mSRE 1 | _ GMEm 2 2RP GM, RP2

12. (c) : Time period of satellite,

2k(Re +h) _ 2K(RE+hf'2

T =

-GME

j(RE+h)

JGMe

Squaring both sides, we get

T = -r 2_ 4 7 i2( RE+ / z )3 GMr 3 _ GMETZ (REW= A 2 471 (.RE+h) = GMFTZ \l/3 h = GMFT -4K ,\l/3 4Kz -Rv Here, M£ = 6 x 1024 kg Re = 6400 km = 6400 x 103 m = 6.4 x 106 m T = 24 h = 24 x 60 x 60 s = 86400 s G = 6.67 x 10~u N m2 kg-2

On substituting the given values, we get 6 . 6 7 x l 0_ 1 1x 6 x l 02 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 = + ll isfi _ i : 2 V Gm \ OA J + P o t e n t i a l a t c e n t r e O d u e t o g i v e n m a s s 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 is 2 GMC

Rr

GMr

R E

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

= mg = 63 N

Acceleration d u e to gravity at height h is

§Re

§h~(RE+hf

gRl

' R 1x2 2

!

Gravitational force on b o d y at height h is 4 4 4 F = mgh = m x -g = - x mg = - x 63 N = 28 N .

16. (d): Linear m o m e n t u m is not conserved.

17. (d)

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

Me = 6 x 1024 kg

Let at a distance x f r o m the earth's centre w h e r e the gravitational forces on the rocket d u e to sun and earth become equal and opposite. If m is the mass of the rocket, then

GMEm ME = x2 6 x 1 0 _ GMgtn ~ (r-x)2 Mr ( r - x )2 ,24 2 x 1 0 (r-x)2 30 r-x x

1 =

1 2 x 1 0 ,30 1/2 6 x 10 103 rS 24 J x l O6 ,1/2 1 . 5 x l OnV 3 V^+103 V3 + 103 2.6 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 4GM VB=— r/2 r/2 20. (a): Satellite Perihelion1 Aphelion Here, rA = 6RE, rP = ?RE The eccentricity of the orbit is

e = 6 R£- 2 R£

6R£ + 2RE

/ r

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

m o m e n t u m

Angular m o m e n t u m at perigee = Angular m o m e n t u m at apogee mvptp = mvArA ... = Vp rA 6 RE 3 22. (b): F = 0 for r < R F k 4 for r > R r

Hence, option (b) represents the correct graph.

23. (c) : An a s t r o n a u t experiences weightlessness

in a space satellite. It is because the astronaut experiences no gravity.

24. (a):

>x

B E C

Refer figure, ZCBO = 30°

AO= —AE = - x Zsin60°

3 3 2 . S I = - x I x — = —j= 3 2 V3 BO = CO = AO = 4= v 3

The a n g l e b e t w e e n OC a n d positive x-axis is 30° and so the angle between OB and negative x-axis. Then,

Force acting on m a s s 2m at O d u e to mass m at A is FOA~ Gm(2m) 0 6 Gm2 * ( n S y - ] 1l -]

Force acting on m a s s 2m at O d u e to mass m at B is

Gm(2m) a *

Fob = K r (-1 cos 30° - ; sin 30°) ( l / S )2