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**Q 1. Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is: [2019]A: 1/9B: 8/9C: 4/9D: 5/9Ans:** B

A: 3 J

B: 30 kJ

C: 2 J

D: 1 J

Ans:

Initial KE

|Î”KE| = 3 J

A: 30 J

B: 5 J

C: 25 J

D: 20 J

Ans

A: mgR

B: 2mgR

C:

D:

Ans:

Initial potential energy at earths surface is

Final potential energy at height h = R

As work done = Change in PEâˆ´ W = U

(âˆµ GM = gR

A: 0.5

B: 0.25

C: 0.8

D: 0.4

Ans

A: 3/2 D

B: D

C: 7/5 D

D: 5/4 D

Ans:

A: KA < KB < KC

B: KA > KB > KC

C: KB < KA < KC

D: KB > KA > KC

Ans:

A: 1 : 6

B: 1 : 9

C: 1 : 11

D: 1 : 14

Ans:

Force constant of spring

after cutting the spring in the ratio 1:2:3, the force constants will be in the ratio 6:3:2

Force constant in series K' is given by

Force constant in parallel K" : 6k+3k+2k = 11k;

hence K'/K" = 1/11

(i) gravitational force and the

(ii) resistive force of air is :-

A: (i) 1.25 J (ii) â€“ 8.25 J

B: (i) 100 J (ii) 8.75 J

C: (i) 10 J (ii) â€“ 8.75 J

D: (i) â€“ 10 J (ii) â€“ 8.25 J

Ans:

A: (2t

B: (2t

C: (2t

D: (2t

Ans:

A:

B:

C:

D:

Ans:

A: W

B: W

C: W

D: W

Ans:

**Q 13. A mass m moves in a circle on a smooth horizontal plane with velocity v _{0} at a radius R_{0}. The mass is attached to a string which passes through a smooth hole in the plane as shown.**

**The tension in the string is increased gradually and finally m moves in a circle of radius R _{0}/2. The final value of the kinetic energy is : [2015]A: B: C: D:Ans: **D

When a mass moves in a circle of radius R

The tension in the string is gradually increased and the radius of the circle decreased to R_{0}/2. When the radius of the circle is R the tension in the string is the same as the centripetal force.

where L = mRv is the angular momentum which is conserved.

Work done in reducing the radius of the circle from R_{0} to R_{0}/2 is

**Q 14. A block of mass 10 kg, moving in x direction with a constant speed of 10 ms-1, is subjected to a retarding force F = 0.1 x J/m during its travel from x = 20 m to 30 m. Its final KE will be : [2015]A: 250 JB: 475 JC: 450 JD: 275 JAns: **B

Solution:

It is subjected to a retarding force F = 0.1 x J/m during its travel from x = 20 m to 30 m.

Work done is given by

Final kinetic energy is, KE

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