JEE Main 2023 — Rotational Motion Question with Solution

The angular momentum about the diameter is 

$L_{\text{diameter }}=\frac{2}{5}M{R}^2\omega ;$

The moment of inertia of a sphere is $I_{CM}=\frac{2}{5}M{R}^2$. The moment of inertia about the tangent is 

$I_{\text{tangent }}=I_{CM}+M{R}^2=\frac{2}{5}M{R}^2+M{R}^2=\frac{7}{5}M{R}^2$.

It is given that $\omega =10 \mathrm{rad} {\mathrm{s}}^{-1}$

Using the given relation between the moment of inertia and the angular momentum,

$$\begin{gathered} \frac{7}{5}M{R}^2=x\times {10}^{-2}\times \frac{2}{5}M{R}^2\omega \\ \Rightarrow x=\frac{{\displaystyle 3.5\times {10}^2}}{\omega }=3.5\times 10=35 \end{gathered}$$