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 Multiple Choice QuestionsMultiple Choice Questions

1.

A smooth block is released at rest on a 45° incline and then slides a distance d. The time taken to slide is n times as much to slide on rough incline than on a smooth incline. The coefficient of friction is

  • straight mu subscript straight k space equals space 1 minus 1 over straight n squared
  • straight mu subscript straight k space equals space square root of 1 minus 1 over straight n squared end root
  • straight mu subscript straight s space equals space 1 minus 1 over straight n squared
  • straight mu subscript straight s space equals space 1 minus 1 over straight n squared


A.

straight mu subscript straight k space equals space 1 minus 1 over straight n squared
straight d space equals space 1 half fraction numerator straight g over denominator square root of 2 end fraction straight t subscript 1 superscript 2
straight d equals space 1 half fraction numerator straight g over denominator square root of 2 end fraction left parenthesis 1 minus straight mu subscript straight k right parenthesis straight t subscript 2 superscript 2
fraction numerator straight t subscript 2 superscript 2 over denominator straight t subscript 1 superscript 2 end fraction space equals space straight n squared space equals fraction numerator 1 over denominator 1 minus straight mu subscript straight k end fraction
288 Views

2.

A body A of mass M while falling vertically downwards under gravity breaks into two parts; a body B of mass 1/3 M and a body C of mass 2/3 M. The centre of mass of bodies B and C taken together shifts compared to that of body A towards

  • depends on height of breaking

  • does not shift

  • body C

  • body C


B.

does not shift

No horizontal external force is acting
∴ = acm =0
since vcm= 0
∴∆xcm = 0

208 Views

3.

An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 cm. If the car is going twice as fast, i.e 120 km/h, the stopping distance will be

  • 20 m

  • 40 m

  • 60 m

  • 60 m


D.

60 m

Third equation of motion gives
v2 = u2 + 2as ⇒ 2
s ∝ u (since v = 0)
where a = retardation of body in both the cases

therefore, straight s subscript 1 over straight s subscript 2 space equals space fraction numerator straight u subscript 1 superscript 2 over denominator straight u subscript 2 superscript 2 end fraction .... (i)

Here, s1 = 20 m, u1 = 60 km/h, u2 = 120 km/h. Putting the given values in eq. (i), we get

20 over straight s subscript 2 space equals space open parentheses 60 over 120 close parentheses squared
space straight s subscript 2 space equals space 20 space straight x space open parentheses 120 over 60 close parentheses squared
space equals space 20 space straight x space 4 space
space equals space 80 space straight m

168 Views

4.

The upper half of an inclined plane with inclination φ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by

  • 2sinφ

  • 2cosφ

  • 2tanφ
  • 2tanφ

C.

2tanφ
181 Views

5.

A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?

  • 3.0 cm

  • 2.0 cm

  • 1.5 cm

  • 1.5 cm


D.

1.5 cm

straight F.3 space equals 1 half mv squared minus 1 half straight m straight v squared over 4
straight F left parenthesis 3 plus straight x right parenthesis space equals 1 half mv squared
space straight x space equals space 1 space cm
217 Views

6.
A block rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (take g = 10 m/s2 )
  • 2.0

  • 4.0

  • 1.6 

  • 1.6 


A.

2.0

Let the mass of block be m.
Frictional force in rest position
F = mg sin 30



10 space equals space straight m space straight x space 10 space straight x space 1 half
space straight m space equals space fraction numerator 2 space straight x space 10 over denominator 10 end fraction space equals space 2 space kg

429 Views

7.

Out of the following pair, which one does NOT have identical dimensions is

  • angular momentum and Planck’s constant

  • impulse and momentum

  • moment of inertia and moment of a force

  • moment of inertia and moment of a force


C.

moment of inertia and moment of a force

221 Views

8.

A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds?

  • h/9 metres from the ground

  • 7h/9 metres from the ground

  • 8h/9 metres from the ground

  • 8h/9 metres from the ground


C.

8h/9 metres from the ground



second law of motion
straight s space equals ut space plus space 1 half space plus gT squared
or space straight h equals 0 plus 1 half gT squared space left parenthesis because space straight u equals 0 right parenthesis
therefore space straight T equals square root of open parentheses fraction numerator 2 straight h over denominator straight g end fraction close parentheses end root
At space straight t space equals space straight T over 3 straight s comma
straight s space equals space 0 plus space 1 half straight g space open parentheses straight T over 3 close parentheses squared
rightwards double arrow space straight s space equals space 1 half space straight g. straight T squared over 9
rightwards double arrow space straight s space equals space straight g over 18 space straight x space fraction numerator 2 straight h over denominator straight g end fraction space space space space open parentheses therefore space equals space square root of fraction numerator 2 straight h over denominator straight g end fraction end root close parentheses
therefore s = h/9 m
Hence, the position of ball from the ground= h- h/9 = 8h/9 m

267 Views

9.

A machine gun fires a bullet of mass 40 g with a velocity 1200 ms−1. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most?

  • one 

  • Four

  • Two

  • Two


D.

Two

The force exerted by machine gun on man's hand in firing a bullet = change in momentum per second on a bullet or rate of change of momentum

space equals open parentheses 40 over 1000 close parentheses space straight x space 1200 space equals space 48 space straight N

The force exerted by man on machine gun = 144 N Hence, number of bullets fired =144/48 = 3

265 Views

10.

An annular ring with inner and outer radii R1 and R2 is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, F1/F2 is

  • R1/R2

  • (R1/R2)2

  • 1

  • 1


A.

R1/R2

straight F subscript 1 over straight F subscript 2 space equals space fraction numerator straight R subscript 1 straight omega squared over denominator straight R subscript 2 straight omega squared end fraction space equals straight R subscript 1 over straight R subscript 2
298 Views