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Q.

A uniform circular disc of radius R lying on a frictionless horizontal plane is rotating with an angular velocity ω about its own axis. Another identical circular disc is gently placed on the top of first disc co-axially. The loss in rotational kinetic energy due to friction between two discs as they acquire common angular velocity is (I is moment of inertia of the disc)

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a

12Iω2

b

Iω2

c

18Iω2

d

14Iω2

answer is C.

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Detailed Solution

Apply Law of conservation of angular momentum :

I1ω1+0=(I1+I2)ω2

1=22

KEi=12I1ω12

KEf=12(I1+I2)ω22=12(2I)ω22

ΔKE=1212-222

Therefore, loss in energy is 142

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