Physics - Rotation Question with Solution | TestHub

Equilibrium of the ring along the smooth incline gives the string tension $T=Mg\sin \theta =30$ N.

Torque about the centre: $TR=M{R}^2\alpha$, so $\alpha =\frac{30}{5}=6rad/{\mathrm{s}}^2$ and the block's acceleration is $a=\alpha R=6m/{\mathrm{s}}^2$.

For the block: $mg-T=ma$ gives $10m-30=6m$, hence $m=7.5$ kg - option (B).