• # question_answer Two blocks of same masses are attached with two rough pulleys of moment of inertia${{I}_{1}}$and${{I}_{2}}$. If both the blocks left at the same instant and let after time t the angular speeds of two pulleys are${{\omega }_{1}}$and${{\omega }_{2}}$then A)  ${{\omega }_{1}}={{\omega }_{2}}$                     B)  ${{\omega }_{1}}>{{\omega }_{2}}$ C)  ${{\omega }_{1}}<{{\omega }_{2}}$                     D)  Information insufficient

Idea Here, pulleys will rotate due to the tension of rope as pulleys are rough. The angular speeds will depend on the angular acceleration. So, just find angular acceleration to find relation between${{\omega }_{1}}$and${{\omega }_{2}}$. As, $\omega ={{\omega }_{0}}+\alpha t$ Since, ${{\omega }_{0}}=0$so, $\omega =\alpha t$ So, the pulley which has more angular acceleration would have more angular velocity. Now, consider a pulley mass system $ma=mg-T$ $\Rightarrow$$a=g-T/m$...(i) Since, $a=\alpha .r$ and as$l\alpha =T\times r$ i.e., $\alpha =\frac{Tr}{l}$ $\therefore$$a=\frac{T{{r}^{2}}}{l}$ $\Rightarrow$$\frac{T.{{r}^{2}}}{l}=g-\frac{T}{m}$ $\Rightarrow$$T\left[ \frac{1}{m}+\frac{{{r}^{2}}}{l} \right]=g$ $\Rightarrow$$T=\frac{g}{\frac{1}{m}+\frac{{{r}^{2}}}{l}}$ $\therefore$$\alpha =\frac{gr}{\left( \frac{1}{m}+\frac{{{r}^{2}}}{l} \right)l}=\frac{mgr}{l+m{{r}^{2}}}$ Since, relation between r2/l is not given. So, data is insufficient here. TEST Edge Rough pulley-block system could also be solved by applying conservation of mechanical energy. $\Delta KE+\Delta PE=0$