A) \[\frac{2g}{3}\]and\[\frac{mg}{3}\]
B) \[g\]and \[\frac{mg}{2}\]
C) \[\frac{g}{3}\]and \[\frac{mg}{2}\]
D) \[\frac{g}{2}\]and\[\frac{mg}{3}\]
Correct Answer: A
Solution :
Let a be the acceleration of the cylinder and \[\alpha \]be its angular acceleration. Invoking \[\tau =I\alpha \]about the center of the cylinder, we get \[TR=Ia=\left( \frac{m{{R}^{2}}}{2} \right)\alpha \] As the string unwinds without slipping, so \[a=R\alpha \] Also, \[{{F}_{ext}}=m{{a}_{cm}}\] \[\therefore \]\[mg-T=ma\] Solving these equations, we have \[a=\frac{2}{3}g\]and \[T=\frac{mg}{3}\] Hence, the correction option is [a].You need to login to perform this action.
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