1. A force acts on aparticle and displaces it through

. The value of x for zerowork is

1) 0.5

2) −2

3) +23) +2

4) 6Vikasana - CET 2012

)

Ans - 3 (2)Soln :-

Zero = (6×2)+(2×-3)+(-3×x)∴ x = 2

Vikasana - CET 2012

2. Two bodies with K.E. in the ratio 4 : 1 are2. Two bodies with K.E. in the ratio 4 : 1 aremoving with same linear momenta. Theratio of their masses isratio of their masses is

1) 1 : 42) 1 : 13) 1 : 2)4) 4 : 1

Vikasana - CET 2012

∴Ans − 1 (1 : 4)

Vikasana - CET 2012

3. When a 1 kg mass hung from a spring 50 cmlong it stretches by 2 cm The mass is thenlong, it stretches by 2 cm. The mass is thenpulled down until the length of the springbecomes 60 cm The amount of elasticbecomes 60 cm. The amount of elasticenergy stored in the spring in this conditionwhen g is 10 ms−1 can be given aswhen g is 10 ms can be given as1) 3J2) 2J2) 2J3) 2.5J

Vikasana - CET 20124) 1.5J

∴Ans − 3 (2.5 J)Soln :- F = Kx

2= 2.5 J

Vikasana - CET 2012

4. Feeling of weightlessness in a satellite or hi i d tspace ship is due to

1) absence of inertia2) absence of gravity3) absence of acceleration due to gravity3) absence of acceleration due to gravity4) free fall of space ship

Vikasana - CET 2012

∴Ans − 2 (absence of gravity)Soln :- In the artificial satellite ‘g’ is zero so

weightlessness in a satellite is due tob f l i d iabsence of acceleration due to gravity.

Vikasana - CET 2012

5. A wheel rotating at 900 rpm about its axis.It comes to rest in 60 s The angularIt comes to rest in 60 s. The angularretardation in rad s−1 is

1) 2)

3) 4)

Vikasana - CET 2012

∴Ans − 4

Vikasana - CET 2012

6. A ring is rolling on a surface withoutslipping What is the ratio of itsslipping. What is the ratio of itstranslational to rotational kinetic energies?1) 5 71) 5 : 72) 2 : 53) 2 : 74) 1 : 1

Vikasana - CET 2012

∴Ans − 4 (1 : 1)Rotational K E But for a ringRotational K. E. But for a ring

Rotational K. E.

So translational K. E. = Rotational K. E.i.e. 1 : 1

Vikasana - CET 2012

7. A ring, a solid sphere and disc have thed di hi h f th hsame mass and radius, which of them have

the largest moment of inertia?1) Ri1) Ring2) Solid sphere3) Disc4) All have the same4) All have the same

moment of inertia

Vikasana - CET 2012

∴Ans − 1 (Ring)Soln :-

Vikasana - CET 2012

8. A fly wheel of mass 50 kg and radius ofti b t it i f t ti f 0 5 igyration about its axis of rotation of 0.5 m is

acted upon by a constant torque of 12.5N m Its angular velocity at t 5 sec is:N.m, Its angular velocity at t = 5 sec is:

1) 2.5 rad s−1

2) 5 rad s−1

3) 7 5 rad s−13) 7.5 rad s4) 10 rad s−1

Vikasana - CET 2012

∴Ans − 2 (5rads-2)2 2Soln :- I = MR2 = 50×(0.5)2

τ = 12.5 N.m

Vikasana - CET 2012

9. An electric fan has blades of length 30 cm asd f h i f i If hmeasured from the axis of rotation. If the

fan is rotating at 1200 rpm, the accelerationf i t th ti f th bl d i b tof point on the tip of the blade is about

1) 4740 ms−21) 4740 ms2) 5055 ms−2

3) 1600 23) 1600 ms−2

4) 2370 ms−2

Vikasana - CET 2012

∴Ans −1 (4740 ms-2)Soln :- r = 30 cm = 0.3 m

n = 1200 rpm = 20 rps

ω=2πn = 2π20 = 40 π rads−1

a = ω2r = (40π)20.3 = 4740 ms−2.

Vikasana - CET 2012

10. A homogeneous disc with a radius 0.2 m andmass 5 kg rotates around an axis passingmass 5 kg rotates around an axis passingthrough its centre. The angular velocity ofthe rotation of the disc as a function of timethe rotation of the disc as a function of timeis given by the formula ω = 2 + 6t. Thetangential force applied to the rim of the disctangential force applied to the rim of the discis

1) 1 N 2) 2 N1) 1 N 2) 2 N3) 3 N 4) 4 N

Vikasana - CET 2012

∴Ans − 3 (3N)S ln Gi th t 2 + 6tSoln :- Given that ω = 2 + 6t

But τ = Iα

Now τ = F . R = F × 0.2

F = 3 NVikasana - CET 2012

11. When a body starts rolling on an inclinedplane the potential energy had by it isplane, the potential energy had by it isconverted into

1) Translational K E1) Translational K. E.2) Translational and rotational K. E.3) Rotational energy4) None of these)

Vikasana - CET 2012

∴Ans −2 (Translational and rotational K. E.)

Vikasana - CET 2012

12. Height at which the value of g becomes 1/4th

to that on earth isto that on earth is1) R2) 4R3) 2R)

4)4)

Vikasana - CET 2012

∴Ans − 1 (R)Soln :-

∴R + h = 2R ∴h = 2R – R = R

Vikasana - CET 2012

13. A satellite in circular orbit of radius R has aperiod of 4 hr A satellite with orbital radiusperiod of 4 hr, A satellite with orbital radiusof 4R around the same planet will have aperiodperiod1) 8 hr2) 16 hr3) 32 hr)4) 64 hr

Vikasana - CET 2012

∴Ans − 3 (32 hr.)Soln :-

T2 = 16×64T = 32 hr.

Vikasana - CET 2012

14. Escape velocity from earth is about 11kms−1 The escape velocity from a planetkms 1. The escape velocity from a planethaving twice the radius and the same meandensity isdensity is1) 22 kms−1

2) 11 k 12) 11 kms−1

3) 0.55 kms−1

4) 15.5 kms−1

Vikasana - CET 2012

∴Ans − 4 (15.5 kms-1)Soln :-Escape velocity for planetEscape velocity for planet

Vikasana - CET 2012

15. The gravitational potential energy of a bodyin the gravitational field of earth isin the gravitational field of earth isminimum1) On the surface of the earth1) On the surface of the earth2) Below the surface of the earth3) At infinity4) between infinity ) y

and surface of the earth

Vikasana - CET 2012

∴Ans −1 (On the surface of earth)Soln :- Gravitational energy decreases

continuously from infinity to surface ofearth and again increases inside the earth.So it is minimum at the surface of theearth.

Vikasana - CET 2012

16. A force of 10+6 Nm−2 is required forbreaking a material If the density of thebreaking a material. If the density of thematerial is 3×103 kgm−3, then the length ofwire made of this material that breaks by itswire made of this material that breaks by itsown weight is1) 43 m1) 43 m2) 34 m3) 127 m4) data incomplete

Vikasana - CET 2012) p

∴Ans −2 (34 m)Soln :- Weight of the wire mg = FF = volume × density × g = ALρgy g ρgStressBreaking stress = 106 i e Lρg = 106Breaking stress = 106 i.e. Lρg = 106

34= 34 m

Vikasana - CET 2012

17. How does the Young’s modulus vary withthe increase of temperature?the increase of temperature?

1) Decreases1) Decreases2) Increases) R i   t t3) Remains constant4) First increases 

and then decreases

Vikasana - CET 2012

∴Ans −1 (decreases)Soln :- The interatomic binding energy decreases

with temperature, so more strain isd d hi hproduced at higher temperature.

Hence y decreases

Vikasana - CET 2012

18. If both the length and radius of the wire ared bl d h d th d l f l ti itdoubled, how does the modulus of elasticitychange?1) B f h1) Becomes one fourth2) Halved3) Doubled4) Remains unchanged4) Remains unchanged

Vikasana - CET 2012

∴Ans − 4 (Remains unchanged)Soln :‐Modulus elasticity is a property of

material. It does not depend on lengthor area of cross section.

Vikasana - CET 2012

19. According to the Hooke’s law, the forcei d t h th l th f i b ‘l’required to change the length of a wire by ‘l’

is proportional to1) l 21) l−2

2) l−1

3) l4) l24) l

Vikasana - CET 2012

∴Ans − 3 (l)Soln :- From Hookes law Since A and L are fixed, therefore F ∝ l.

Vikasana - CET 2012

20. Young’s modulus of the material of a wire is‘y’ If it is under a stress S, the energy storedper unit volume is given by

1) 2)

3) 4)3) 4)

Vikasana - CET 2012

∴Ans − 2

Soln :- Energy stored (U)

Strain

Vikasana - CET 2012

21. A force of one newton doubles the length ofa cord having cross sectional area 1mm2a cord having cross-sectional area 1mm2.The young’s modulus of the material of cordisis1) Nm−2

2) 0.5×106 Nm−2

3) 106 Nm−2

4) 2×106 Nm−2

Vikasana - CET 2012

∴Ans − 3 (106 Nm−2)Soln :-

Vikasana - CET 2012

22. In which case potential energy decrease on 1) compressing the spring1) compressing the spring2) stretching the spring3) moving a body against

gravitational pullg p4) raising of an

air bubble in waterair bubble in water

Vikasana - CET 2012

∴ Ans − 4 (raising of an air bubble in water)Soln :- P. E. increases in 1, 2 and 3

The upthrust makes the air bubble rise in the water so there is decrease in energy

Vikasana - CET 2012

23. A 0.5 kg ball is thrown vertically up with 141 I i h i h f 8 Thms−1. It attains a height of 8m. The energy

dissipated by air drag acting on the balld i iduring ascent is1) 9.8 J2) 4.9 J3) 10 J)4) 19.6 J

Vikasana - CET 2012

∴Ans − 1 (9.8 J)Soln :-

E`= mgh = 0.5 ×9.8×8 = 39.2 JEnergy lost = E − E`Energy lost E E = 49 − 39.2 = 9.8 J

Vikasana - CET 2012

24. If momentum of a body increases by 50%the kinetic energy will increase bythe kinetic energy will increase by

1) 50%)2) 150%3) 125%3) 125%4) 100%

Vikasana - CET 2012

∴Ans − 3 (125%) S lnSoln :-

Increase in K.EIncrease in K.E

Vikasana - CET 2012

25. A bullet is fired from a rifle. If rifle recoilsf l th K E f b ll t ifreely, then K.E. of bullet is1) less than that of rifle2) more than that of rifle3) same as that of rifle3) same as that of rifle4) equal or less

h h f iflthan that of rifle

Vikasana - CET 2012

∴Ans − 2 (more than that of rifle)Soln :-

Velocity of bullet is more because it is very light as compared to the rifle, so K.E. of bullet is more

Vikasana - CET 2012

26. An object of mass 40 kg and having velocity 4 1 llid ith th bj t f4 ms−1 collides with another object of mass 60 kg having velocity 2 ms−1. The loss of

h th lli i i f tlenergy when the collision is perfectly inelastic 1) 392 J1) 392 J2) 440 J3) 110 J4) 48 J

Vikasana - CET 20124) 48 J

∴Ans − 4 (48 J)Soln :Soln :-

Loss of energy gy

= 48 JVikasana - CET 2012

27. Graph between stretching force andextension of spring is shown in the figureextension of spring is shown in the figure.Change in the line PQ when same force isapplied to a spring of half the length is givenapplied to a spring of half the length is givenby

1) Same graph1) Same graph2) line of double the length3) hifti t d i3) shifting towards x – axis4) shifting towards y – axis

Vikasana - CET 2012

∴ Ans − 4 (shifting towards y – axis)Soln :- Extension ∝ original length

So for halved spring the extension will be lessfor the same force, so the PQ line will shifttowards y – axis.

Vikasana - CET 2012

28. Which of the following statement is true?1) g is less at the earth’s surface than at1) g is less at the earth s surface than at

a height above it or a depth below it2) g is the same at all places on the2) g is the same at all places on the

surface of the earth3) g has its maximum value at the3) g has its maximum value at the

equator4) g is greater at the places than at the4) g is greater at the places than at the

equator.

Vikasana - CET 2012

∴Ans − 4 (g is greater at the places than at the )equator)

Vikasana - CET 2012

29. Two particles of equal mass go round ai l f di R d th ti f th icircle of radius R under the action of their

mutual gravitational attraction. The speedf h ti l iof each particle is

1) 2)1) 2)

3) 4)

Vikasana - CET 2012

∴Ans − 3∴Ans 3

S lSoln :-Centripetal force = Gravitational force

Vikasana - CET 2012

30. The force of gravitation isg

1) repulsive2) l t t ti2) electrostatic3) conservative4) non-conservative

Vikasana - CET 2012

∴Ans − 3 (conservative)( )

Vikasana - CET 2012

31. If the spinning speed of the earth isincreased then weight of the body at theincreased then weight of the body at theequator1) i1) increases2) decreases3) doubles4) does not change4) does not change

Vikasana - CET 2012

∴Ans − 2 (decreases) S ln W k th t ` (R 2) t thSoln :-We know that g` = g − (Rω2) at the

equator. If spinning speed (angularl it ) i th l f `velocity - ω) increases the value of g

decreases i.e. weight of the body willddecrease

Vikasana - CET 2012

32. If radius of earth is ‘R’ then the height ‘h’ athi h th l f b f th iwhich the value of g becomes one fourth is

) )1) 2)

3) R 4)

Vikasana - CET 2012

∴Ans − 3 (R)Soln :-

Taking Rootor h = R

Vikasana - CET 2012

33. Distance of a geostationary satellite from the surface of earth issurface of earth is1) 6R2) 7R3) 5R)4) 3R

Vikasana - CET 2012

∴Ans − 1 (6R)S ln R di f bit f t ti t llitSoln :-Radius of orbit of geostationary satellite

is = 7RIts height from the surface of the earth = 7R − R = 6R

Vikasana - CET 2012

34. The angular momentum of a moving body remains constant ifremains constant if1) net external force is applied2) net pressure is applied2) net pressure is applied3) net external torque is applied4) t t l t i t li d4) net external torque is not applied

Vikasana - CET 2012

∴Ans − 4 (net external torque is not applied)

Vikasana - CET 2012

35. A hard boiled egg will spin faster than a raw b fegg because of

1) high moment of inertia2) low moment of inertia3) high angular velocity) g g y4) Both high moment of

inertia and high angularinertia and high angular velocity

Vikasana - CET 2012

∴Ans − 2 (low moment of inertia)

Vikasana - CET 2012

36. Rotational analogue of mass in linearti imotion is

1) weight) g2) moment of inertia3) torque3) torque4) Angular momentum

Vikasana - CET 2012

∴Ans − 2 (moment of inertia)

Vikasana - CET 2012

37. Which of the following affects the elasticity of a substance?of a substance?

1) hammering and annealing2) change in temperature3) impurity in substance4) all of these

Vikasana - CET 2012

∴Ans − 4 (all of these)Hint: Hammering – increases elasticityAnnealing – decreases elasticityg yIncreasing temperature – increase elasticityDecrease in temperature decrease elasticityDecrease in temperature – decrease elasticityImpurity – increases elasticity, So all are responsible for change in elasticity

Vikasana - CET 2012

38. The following four wires of length ‘L’ anddi ‘ ’ d f h i lradius ‘r’ are made of the same material

which of these will have the largestt i h th t i i li d?extension when the same tension is applied?

1) L = 50 cm, r = 0.25 mm2) L = 100 cm, r = 0.5 mm3) L = 200 cm r = 1 mm3) L 200 cm, r 1 mm4) L = 300 cm, r = 1.5 mm

Vikasana - CET 2012

∴Ans − 1 (L = 50 cm, r = 0.25 mm)Soln :- Check is maximum1) 8 × 106)2) 4 × 106

3) 2 × 1063) 2 × 106

4) 1.33 × 106

Vikasana - CET 2012

39. A rope made of steel has a diameter of 5 cm.The breaking strength of the rope is2 × 108 N. The breaking strength of asimilar rope of double diameter is1) 2 × 108 N)2) 4 × 108 N3) 1 × 108 N3) 1 × 108 N4) 8 × 108 N

Vikasana - CET 2012

∴Ans − 4 (8 × 108 N)Soln :- Breaking strength ∝Area of cross section

S ∝ πr2

S` = π(2r)2

= 4πr2 = 4×2×108= 4πr2 = 4×2×108

= 8 × 108

Vikasana - CET 2012

40. It is easier to draw up a wooden block along i li d l th h it ti llan inclined plane then have it up vertically,

because1) the mass becomes smaller2) g becomes smaller) g3) the friction is reduced4) only a part of weight4) only a part of weight

has to be overcome

Vikasana - CET 2012

∴Ans − 4 (only a part of weight has to be )overcome)

Soln :- Only a component of weight acts along the inclined plane i.e. only a part of weight has to be taken care of by the applied force.

Vikasana - CET 2012

41. A fixed volume of iron is drawn into a wiref l th ‘l’ Th t i d d i thiof length ‘l’. The extension produced in this

wire by a constant force F is proportional to

1) 2)

3) l2 4) l3) l 4) l.

Vikasana - CET 2012

∴Ans − 3 (l2)Hint:- If force is constant and volume is fixed,

then extension ∝ (length)2

Vikasana - CET 2012

42. The K. E. of a body becomes four times itsi iti l l Th t ill binitial value. The new momentum will be1) same as initial value2) t i i iti l l2) twice as initial value3) thrice as initial value4) f ti th4) four times the

initial value

Vikasana - CET 2012

∴Ans − 2 (twice as initial value)Hint:-K.E1. = K. E2. =K. E2. = 4 K. E1.∴v2 = 2v1.

Vikasana - CET 2012

43. A gymnast is sitting on a rotating stool with her arms outstretched Suddenly folds herher arms outstretched. Suddenly folds her arms near her body which of the following is correct?correct?

1) angular speed decreases2) t f i ti d2) moment of inertia decreases3) angular momentum decreases4) angular speed remains constant

Vikasana - CET 2012

∴Ans − 2 (moment of inertia decreases)

Vikasana - CET 2012

44. What is the velocity of the bob of a simplependulum at its mean position, if it is able topendulum at its mean position, if it is able torise to vertical height of 10cm(take g = 9.8ms−2)(take g 9.8ms )1) 0.6 ms−1

2) 1 4 ms−12) 1.4 ms 1

3) 1.9 ms−1

4) 2.2 ms−1

Vikasana - CET 2012

∴Ans − 2 (1.4 ms−1)

Hint:-

v = 1.4 ms−1

Vikasana - CET 2012

45. If central force is inversely proportional todi R h i i d ill bdistance R, then time period will beproportional to

1) R 2)1) R 2)

3) 4) R3) 4) R2

Vikasana - CET 2012

∴Ans − 1 (R)Hint:-

Vikasana - CET 2012

Vikasana - CET 2012