Rotational Kinematics & Dynamics question from JEE Main 2019 (10 Jan, Shift 1)

Answer: (d)

1. Pressure on mop $= \dfrac{F}{\pi R^2}$; a ring of radius $r$, width $dr$ carries friction $dF = \mu \dfrac{F}{\pi R^2} 2\pi r\,dr$.
2. Torque of the ring $= r\,dF$, so $\tau = \dfrac{2\mu F}{R^2}\int_0^R r^2 dr = \dfrac{2\mu F}{R^2}\cdot\dfrac{R^3}{3}$
3. $\tau = \dfrac{2}{3}\mu F R$ (machine torque balances friction at constant ω).
4. So the answer is (d).