question_answer

A uniform disc of mass \[M\] and radius \[R\] is mounted on an axle supported in frictionless bearings. A light cord is wrapped around the rim of the disc and a steady downward pull \[T\] is exerted on the cord. The angular acceleration of the disc is

A) \[\frac{T}{MR}\]                      

B) \[\frac{MR}{T}\]

C) \[\frac{2T}{MR}\]        

D) \[\frac{MR}{2T}\]

Correct Answer: C

Solution :

[c]  Torque exerted on the disc \[\tau =TR\] Now \[\tau =I\alpha \] \[\alpha =\frac{\tau }{I}=\frac{TR}{\frac{1}{2}M{{R}^{2}}}\] \[=\frac{2TR}{M{{R}^{2}}}=\frac{2T}{MR}\]