JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Rotational Mechanics

Take torques about the line of contact with the ground, where friction contributes nothing. The string is pulled horizontally from the top of the cylinder, a height $2R$ above the contact line, so its torque is $F\cdot 2R$. The moment of inertia of the hollow cylinder about the contact line is $M{R}^2+M{R}^2=2M{R}^2$ (its central value plus the parallel-axis shift). For rolling, $F\cdot 2R=2M{R}^2\alpha$, so $\alpha =\frac{F}{MR}=\frac{40}{5\times 0.5}=16$ rad/s${}^2$.