A uniform ring of mass m and radius R is released from top of an inclined plane. The plane makes an angle `theta` with horizontal. The cofficent of friction between the ring and plane is `mu`. Initially, the point of contact of ring and plane is P. Angular momentum of ring about an axis passing from point P and perpendicular to plane of motion as a function of time t isA. `mgR(sintheta)t-mumgR(costheta)t`B. `mgR(sintheta)t`C. `mgR(sintheta)t+mumgR(costheta)t`D. `mgR(1-mu^(2))(sintheta)t`

A uniform ring of mass m and radius R is released from top of an incli

1.	A uniform ring of mass m and radius R is released from top of an inclined plane. The plane makes an angle `theta` with horizontal. The cofficent of friction between the ring and plane is `mu`. Initially, the point of contact of ring and plane is P. Angular momentum of ring about an axis passing from point P and perpendicular to plane of motion as a function of time t isA. `mgR(sintheta)t-mumgR(costheta)t`B. `mgR(sintheta)t`C. `mgR(sintheta)t+mumgR(costheta)t`D. `mgR(1-mu^(2))(sintheta)t`
Answer» Correct Answer - B

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