Physics for You - June 2013 (Gnv64)

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SOLVED PAPER 2013 r

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EXAMINER'S MIND

THOUGHT PROVOKING PROBLEMS

MCQs for PRACTISE

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Questions for NEET

CHALLENGING PROBLEMS

BRAIN MAP

PHYSICS Vol. XXI

No. 5

May 2013

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edit Q rial Expanding Knowledge and In-depth Research are not contradictory

W

idening knowledge and intensive research are not contradictory but

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

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

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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

Managing Editor

:

Mahabir Singh

why it should be like that.

Editor

:

Anil Ahlawat (BE, MBA)

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

Contents

ideal strings and ideal pulleys one learns more. Each part is a grid. First one masters the grid and then the partitions

JEE Advanced Practice Paper: 2013

6

disappear to become a bigger grid. Still the concepts from electricity, magnetism and other subjects look different. The method of the grid system

Thought Provoking Problems

16

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

(Kinematics)

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.

JEE Main Solved Paper: 2013

21

Expanding knowledge is a different case. The pieces of puzzle are different. Taking the various prices, for example, scattering, fundamental particles,

NCERT Xtract Questions for NEET

29

Einstein's theories and that of de-Broglie, Heisenberg's principle and the advances made by scientists like Bohr are apparently different. To

Target PMTs Practice Questions

37

study them in depth, and seek unity in diversity by removing apparent contradictions from individual chapters and then putting them in shape is a

Brain Map

48

different research. This is on a higher plane.

Examiner's Mind

50

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

Challenging Problems

52

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

For reaching the last stage, one has to prove one's credibility by first-solving shapes are well polished, one cannot put them together. can be solved by going back to the study of the connection of various Exam prep: MCQs for Practise

58

discoveries. Science has no barriers. This has to be fundamental concept. There is also no alternative medicine for Hard Work!

AIIMS Practice Paper: 2013

Anil Ahlawat,

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14 PHYSICS FOR YOU I MAY '13

PRACTICE PAPER 2 Q 1 3

Advanced By : Akhil Tewari, B-Tech, IT-BHU

SECTION Straight Objective Type

(a)

This section contains 6 multiple choice questions numbered

(c)

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)

(b)

J

k

(b)

2 mg (5m + M)

7y[5mg

1 to 6. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. 1.

2V5 mg + M)

(5m

(5m + M)

(d) none of these

3.

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°

4.

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 new particle ? (a) v (b) 2v v (c) —j= (d) None of these

5.

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 ms" 1 ,180° 1 (c) 10 m s" , 90° (d) l O m s " 1 , 60°

X

a^^

r

V2

(c)

(d) none of these 2.

Assuming all the ST surfaces to be 6. A car is travelling at a velocity 10 km h _ 1 on a frictionless find M straight road. The driver of the car throws a the magnitude of _0 10 net acceleration 7777777777777777777777777777777177 of smaller block parcel with a velocity —j= k m h" 1 when car is m with respect to ground passing by a man standing on the side of a road. If

14

PHYSICS FOR YOU I MAY '13

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° (c)

tan

_1

V2

(d)

(a) The minimum time in which he can cross the river is — . v (b) He can reach the point P in time — .

tan (c) He can reach the point P in time V^2

S E C T I O N - II Multiple Correct Answer(s) Type

(d) He cannot reach Pifu>

This section contains 4 multiple choice questions numbered

7.

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) vA = vB (b) vB = u + vA (c)

v.

This section contains 4 multiple choice questions numbered 11 t o 14. Each question contains Statement-1 (Assertion) and

B

H k

VB + U = V A

(d) the two blocks have accelerations of the same magnitude. 8. 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 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. 9.

U2

SECTION - III Assertion Reason type

7 to 10. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE is/are correct.

-

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

10.

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.

14

PHYSICS FOR YOU I MAY '13

Statement-2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. (a)

Statement-1 is True, Statement-2 is True; Statement-2 is a

(b)

Statement-1 is True, Statement-2 is True; Statement-2 is

(c)

Statement-1 is True, Statement-2 is False.

(d)

Statement-1 is False, Statement-2 is True.

correct explanation for Statement-1.

not a correct explanation for Statement-1.

11. Statement-1 : The maximum range along the inclined plane, when thrown downward 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 maximum 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 when 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 when lands on heap of sand.

S E C T I O N - IV Linked Comprehension Type

19. The net velocity at the time when velocity is parallel to the plane is , . ucos0 „ . t/3 5, = 10 m s"- 1

i n 6

a = —r t tan— 2

,

a

Z(j) = tan'

a

a+-cosG 2

6.

12

PHYSICS FOR YOU | MAY '13

(a)

(in 1 s t case)

-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,

= 5>/35_

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 = aa(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 v = —j= m s V3 or

8.

10x2

gcosa = R=

=5ms

R

g cosa 2 0

RB = — m 3v3 16. (a)

10. (a, c, d)

17. (c): at = gsincc = 10 x ~

= 5 m s-2

11. (c) 12. (a): ux = u cosO, ax = 0 vx = ux + ax t = ur = u cosO uy = u sinO ; ay :

/(sin0-gf

= ttsinO -

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,

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.

,

T = — 2

22. (c) : p =

= dt

14. (a): As we know that, F = — At

At

For quarter of a circle,

If Af is more, then F will be less. Av = v\Fl and At = — 2v

10ms~i 15-

( a ) :

2n/2 mv2 nr 23. (b): Velocity of block of mass M\ is

5 ^ 3 m s"

F=

5ms

Time after which velocity vector becomes perpendicular to initial velocity vector is t =

gsin0

10 10sin60°

V\ =

2

VI = V + v' Adding equation (i) and (ii), we get

s

Let Vy be the vertical component of velocity at that instant. Then,

14

V-V'

...(i)

Velocity of block of mass M 2 is

-

PHYSICS FOR YOU I MAY '13 t

V, + VR, n

...(ii)

Thought Provoking Kinematics By : Prof. Rajinder Singh Randhawa* 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 run and catch the train?

2.

A particle is projected from 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

4.

6.

Water is running out of a conical funnel at the rate of a m m 3 s"1. If the radius of the base of the funnel is R m m and the altitude is H m m , find, the rate at w h i c h the water level is d r o p p i n g w h e n it is h m m from the top?

7.

A circular wire frame is fixed in a vertical plane. A s m o o t h wire is slightly stretched b e t w e e n points Px and P 2 . A bead slides f r o m point Pu the highest point of the circle. Determine (a) its velocity v when it arrives at P 2 (b) find the time taken by it?

A particle is projected from 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. 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) Find the speed of the ball with which it varies. (b) After a certain time the sphere reaches a terminal speed, find it?

A small sphere of mass m is released from rest in a large vessel filled with oil where it experiences a resistive force proportional to its speed, i.e., Fd = -kv.

SOLUTIONS 1.

Assume the train is at x = 0 at t = 0, the equation for train is 1

xT =-aTt2

,

1

,

= ±(0.40)f 2

The passenger reached x = 0 at t = f 0 = 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 t

_vp±

or -

aTt2-2vpt+2vpt0=0

2aTvpt0

The roots are real if Vp-2aT vp tQ > 0 vp >2aTt0

= 2x 0.40 x 6 = 4.8 m s" 1 .

Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second FL, Sector-20C, Chandigarh

14

PHYSICS FOR YOU I MAY '13

2.

, 7

(i) Along horizontal direction, h = (Mcos0)f

p/

Since the projectile passes through two points (a, h) and (2a, h), then a and 2a must be roots of Eq. (ii),

™(i)

MCOS(()

a + 2a =

2m sinO cos0

...(iii)

h

arid O

3a tan0 —- = 2a h

...(ii)

0 = Msinij) - gt o r Msin = gt

V 2

2M2 cos 2 0 xh

...(iv)

Divide (iii) by (iv), we get

Q

Along vertical direction, and h = Msinf •

ax2a =

„ 3h tan0 = — 2a

or

.(iii)

5

From Eq. (iv),

Using Eqs. (i) and (ii) in Eq. (iii), we get 1

M

2

^

=

s e c

2

e

^

=

h

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

[

1

+

tan 0 = 2

t a n

2

e ] =

1

h

S ^

h

1 +

9h~ 4 a2

•9/2

h

(ii) Along perpendicular to the plane, or t

0 =i usin(

2

1 r

(b) I

o

2 v

(a) 4p0»o (b) p0v0

(d)

POVO

(yJ/Wo

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/s 2 , the equation of its trajectory is (a) 4y = 2x - 25x2 (b) y = x- 5x2 (d) 4y = 2x-5xz (c) y = 2x - 5x^

14

2B(ol2

0

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 amount of heat, extracted from the source in a single cycle is

i =

2[(15 x 10~ 2 ) 2 + (20 x 10" 2 ) 2 ] 3 / 2

((), = 9.1 x 10~n weber 3.

(d): The 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 shown in the adjacent figure.

(a)

V

IK

Hi—

R 4.

O

(d):

C uooooff^(c)

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 = q 0 (l - e~th) [As q0 = CV] (j = CV(1 „-tlv

R

30. Diameter of a plano-convex lens is 6 cm and thickness at the centre is 3 mm. If speed of light in material of lens is 2 x 10 s m/s, the focal length of the lens is (a) 10 cm (b) 15 cm (c) 20 cm (d) 30 cm

At t =

q = CV(1 - e~x/2x) = CV(1 At t = x

SOLUTIONS

1.

x

q = CV(1 - ~X,X) = CV(

(a)

e'V2)

1-tT1)

At t • 2x, q = CV(1 - e _ 2 t / T ) = CV(1 - e~2)

2.

5.

(b):

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

h

.

B, •B 2

As field d u e to current loop 1 at an axial point •>2 B1=-

2

2

o N

N

2 32

(d +R ) '

Flux linked with smaller loop 2 due to B1 is 2 = B1A2 = "

•

nr

2 (d2+R2)3'2 The coefficient of m u t u a l inductance between the loops is

As the point O lies on broad-side position with respect to both the magnets. Therefore, The net magnetic field at point O is B

net = B j + B2 + B

h

PHYSICS FOR YOU | MAY 13 25

H0

n0

Comparing it with standard equation of SHM,

M2

d2x

An r Ho (MJ + M ) + B = 2 h 4jtr 3 Substituting the given values, we get -

6.

net

dt2 We get

' [1.2 + 1] + 3.6 x 10~5 471 x (10 x 10~ 2 ) 3 ->-7 10" x 2.2+ 3.6x10 -5 10,-3 = 2.2 x 10"4 + 0.36 x 10"4 = 2.56 x 10^ W b / m '

2

c o

471X10

-

2

= - (0 1

Y ^ MV n

1 yP0A2 2TT V MVor (c) : According to Coulomb's law co Frequency, x> = 271

8.

(a)

1

r=_

fill r2

2 ~~ Fr [AT] [AT] _1 3 4 2 1E0J_ 7 7 = [M L" T A ] [MLT" 2 ][L] 2

M

(d):

or co =

MVn

4ne0

7.

= ^ I

9.

FBD of piston at equilibrium

(c):

4

—~w—

2

Po D

Wt

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

2% t; Heat 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 nR For a monoatomic gas, q = —R pv = nRT or T =

v

2

Po o Povo v nR nR ~ 2 Po o Along the path AB, pressure is constant. Hence AQab = nCpAT = nCp(TB 5- TA) A

Mg

(p0 + dP)A

(P0 + dP)A - (-Patm A + Mg) =

M . . . ( i i )

Using (i) and (iii) in (ii), we get 2

26

^ dt2

-

^

x v0

PHYSICS FOR YOU | MAY '13 ?

or

Q,•DA

For monoatomic gas, C p = — R

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 = — V yP0(Ax) dP = ...(hi) (v dV = Ax) Vn

M

2

v

dx dt2

2

2 10 Po2vo P0vo nR nR = y W o .•. The amount of heat extracted from 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

i M : MVn

A

As u = uxi + uyj :. ur = 1 and u„ = 2

i

.•. Voltage across bulb after heater is switched on 120 V .4811 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 M14. (a):

Also x = uxt 1

2

and y = Uyt -—gt :. x = t and s1/ = 2f - 2i x 10 x t 2 = 2f - 5f 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 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 J_ - k I 2 = / j cos 45° =

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

-&)

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 1

JL ll

= 10.61 x 103 Hz = 10.61 kHz.

2L\n Heater

13. (a): As P = Yl R Here, the supply voltage is taken as rated voltage. .'. Resistance of bulb _ 120 V x 120 V ~

60 W

Resistance of heater, R H

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

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

B

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

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

m a = 0.6

U=

31

31

e = J Bmxdx = But 21 _ 5B(ri2 2

—WW 6Q

Bulb

H'

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.

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 ec i 240n + 6 0 0

•'•

U - - .. 2L^pA

Here tension is due to elasticity of wire T = YA

_ L _

. Stress _ TL " As y = = Strain ~ AAL_

1 lYAL 2L' pL i n N/m", p = 7.7 x 10J kg/m J Here, Y = 2.2 x 10" — = 0.01, L = 1.5 m L

Cont. on Page No. 82 PHYSICS FOR YOU | MAY 13

27

FOCUS

Engineering Aspirants 2013 -

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

T

he 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 students will be short listed from 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 (Advanced), 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 may 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 Advanced, 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 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. Cracking t h e JEE (ADVANCED) 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 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 when 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

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PHYSICS FOR YOU | MAY '13 ?

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•

•

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 (a) (c)

3.

4.

sllGm2

(b) •h

f^-i)

>/2G>»

(d)

- 2Gm

2+

2GmA f ^ I

Planet

P_

_

A

5.

\

"A

The t i m e p e r i o d T of the m o o n of p l a n e t mars (mass M m ) is related to its orbital radius jR as (G = Gravitational constant) (a) T2=:

. 1

A planet orbits the sun in an elliptical path as shown in the figure. Let vP and vA be speed of the planet w h e n at perihelion and aphelion respectively. Which of the following relations is correct?

Aphelion

f

J_ 42

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 same 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

P Perihelion

4TC2R3 GMm

2 3 (b) nn2 4k GR M„

In.R3G (d) T 2 = 4rcMmGR3 M„, 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 2 (c) (1.52) x 365 (d) (1.52)3 x 365 2 (c) T =

6.

7.

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 M E and RE is the mass and radius of the earth respectively) GMEm GMptn E a) (b) 4 RE 2 RE GMPm GMEm (d) 16 R £

PHYSICS FOR YOU | MAY '13 2 9

8-

9.

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°. Two uniform solid spheres of equal 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 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

6R

(a) ,

x (c)

4 GM 5 R 3 GM I S I R

5 GM

V

VllT

(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)

~mgR

(b) - m g R E

E

(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 m 2 kg~2] (a) 3.59 x 105 m (b) 3.59 x 106 m 7 (c) 3.59 x 10 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 30

PHYSICS FOR YOU | MAY '13 ?

(a)

(b)

Gm ~T 14. The escape velocity from the surface of the earth is (where RE is the radius of the earth) (c)

-

2

(a) (c)

^

(d) -4V2

(b) JgR^ 2jgRE

(d)

fiK

15. A body weighs 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 u n d e r the gravitational influence of another object. Which of the following quantities is not conserved? (a) Angular momentum (b) Mass of an object (c) Total mechanical energy (d) Linear momentum 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 move in an irregular way because of their small masses and will drift away into outer space. (c) will move around the sun in closed orbits 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 earth towards the sun. 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* 1 0 " m] (a) 2.6 x 104 kg (b) 2.6 x 106 kg 8 (c) 2.6 x 10 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 , , (a) (c)

-

GM r

(b)

GM 2r

(d)

2GM

-

4GM

20. 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 (a) v/

±2

(b)

1 3

(C)

1 J

(d)

1 6

21. In the question number 20, the ratio of the velocity of the satellite at apogee and perigee is 1