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 Multiple Choice QuestionsMultiple Choice Questions

1.

Two particles are seen to collide and move jointly together after the collision. During such a collision, for the total system.

  • both the mechanical energy and the linear momentum are conserved

  • linear momentum is conserved but not the mechanical energy

  • neither the mechanical energy nor the linear momentum is conserved

  • mechanical energy is conserved but not the linear momentum


B.

linear momentum is conserved but not the mechanical energy

This is a case of a totally inelastic collision,in which linear momentum is conserved but the totalmechanical energy is not conserved.


2.

Maximum energy transfer for an elastic collisionwill occur if one body is at rest when

  • m1 = m2

  • m212m1

  • m>>  m2

  • m2 >> m1


A.

m1 = m2

During elastic collision between two equalmasses, the velocity of the two bodies get interchanged.So if one body is at rest, energy transfer will be maximum
for m1 = m2.


3.

A car and a container, moving with equal kinetic energies are stopped by applying a negative acceleration. Then

  • container will move less distance before being stopped

  • both will become stationary after moving some distance

  • car will move less distance before being stopped

  • none of the above


C.

car will move less distance before being stopped

Since, a car and a container have equal kinetic energies, the velocity of car will be greater and when same retarded force is applied to both, the car will move less distance, because retardation will be more on car


4.

The kinetic energy of a body becomes four timesits initial value. The new linear momentum will be

  • same as the initial value

  • four times the initial value

  • twice of the initial value

  • eight times of the initial value


C.

twice of the initial value

Initial kinetic energy ( E1) = E

and final kinetic energy (E2 ) = 4E

The kinetic energy of a body =12mv2

Since the value of m remains constant

Therefore for the kinetic energy to be 4 times, the new value velocity (v) should be 2 times the initial value

Initial linear momentum

(p1) = mv

Therefore new linear momentum

(p2) = m× 2v

= 2mv

(p2) = 2p1


5.

Which of the following is path dependent?

  • U

  • PdV

  • P

  • V


B.

PdV

Dependent on path taken to establish property or value.

PdV is the path dependent.


6.

The momentum of two masses m1 and m2 are same. The ratio of their kinetic energies E1 and E2 is

  • m1 : m2

  • m1 : m2

  • m2 : m1

  • m12 : m22 


C.

m2 : m1

Momentum of first body = momentum of second body = mv

E =12mv2

=12m(mv)2

Hence, E1E2=m2m1


7.

If forceF=5i^+3j^+4k^makesadisplacementofs=6i^-5k^workdonebytheforceis

  • 10units

  • 1225units

  • 5122units

  • 20units


A.

10units

Workdone=Force×displacement=F.s=5i^+3j^+4k^.6i^-5k^=5i^+3j^+4k^.6i^+0j^-5k^=30+0-20=10units


8.

n balls each of mass m and velocity v collide with a wall, assume that the collisions are perfectly elastic. The pressure exerted per second on the wall is

  • 3 mnv

  • mnv

  • 2mnv

  • none


C.

2mnv

Pressure = change in momentum

Change in momentum for a ball = mv − (− mv) = 2mv

For n balls change in momentum = 2mnv


9.

Assertion: Centripetal force does no work.

Reason: Force and displacement areperpendicular to each other.

  • If both assertion and  reason are true and reason is the correct explanation of assertion.

  • If both assertion and reason are true but reason is not the correct explanation of assertion.

  • If assertion is true but reason is false.

  • If assertion is false but reason is true.


A.

If both assertion and  reason are true and reason is the correct explanation of assertion.

Since centripetal force is perpendicular to the displacement of the body, work done is zero.

W =F·d= Fd cosθ


10.

A body weighed 250 N on the surface assumingthe earth to be a sphere of uniform mass density,how much would it weigh halfway down to thecentre of the earth?

  • 195 N

  • 240 N

  • 125 N

  • 210 N


C.

125 N

At a depth 'd' below the surface of the earth.

mg' =GM'mR-d2

M' =43πR-d3ρ

⇒ g' =GM'R-d2

g' =43R-dρ

On the surface of the earth

g =GmRV

g = 43πGRρ

∴ g' =g2

∴ The body weighed 250 N on the surface of the earth would weigh12× 250 = 125 N, halfway down towards the centre of the earth.