• # question_answer Two Blocks are placed over a rough pulley of moment of inertia I when the block of mass m comes down by a distance h, the angular of the pulley is A)  $\sqrt{\frac{2gh}{(M+m){{r}^{2}}+l}}$                B)  $\sqrt{\frac{2(M+m)gh}{(M-m){{r}^{2}}+l}}$ C)  $\sqrt{\frac{2Mgh}{(M-m){{r}^{2}}+l}}$             D)  $\sqrt{\frac{2(M-m)gh}{(M+m){{r}^{2}}+l}}$

Idea In this question as mass moves and pulley rotates then by energy conservation. Loss of potential energy = gain in linear kinetic energy + gain in rotational kinetic energy Loss of PE = $(Mgh-mgh)$ $=(M-m)gh$ Gain in linear KE$=\frac{1}{2}(m+M){{v}^{2}}$ Gain in rotational KE$=\frac{1}{2}l{{\omega }^{2}}$ $(M-m)gh=\frac{1}{2}(m+M){{v}^{2}}+\frac{1}{2}l{{\omega }^{2}}$ We know$v=r\omega$ $(M-m)gh=\frac{1}{2}(m+M){{r}^{2}}{{\omega }^{2}}+\frac{1}{2}l{{\omega }^{2}}$ $\Rightarrow$${{\omega }^{2}}=\frac{2(M-m)gh}{[(M+m){{r}^{2}}+l]}$ $\omega =\sqrt{\frac{2(M-m)gh}{(M+m){{r}^{2}}+l}}$ TEST Edge Here, I tangential acceleration of pulley is equal to the I linear acceleration I of two blocks.