Q.

A uniform ring of mass m and radius R is released from top of an inclined plane.  The plane makes an angle θ with horizontal.  The coefficient of friction between the ring and the plane is µ.  Initially, the point of contact of ring and plane is P.  Angular momentum of the ring about point P as a funtion of time t is

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a

mgR(sinθ)t - µmgR(cosθ)t

b

mgR(sinθ)t

c

mgR(sinθ)t + µmgR(cosθ)t

d

mgR(1 - µ2)(sinθ)t

answer is B.

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Detailed Solution

Force of friction passes through point P, Hence, its torque about  P will be zero.  Only (mgsinθ) will have torque about P. Thus τ.t = ΔL or (mgRsinθ)t = L
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