JEE Main 2023 — Rotational Motion Question with Solution

Let the mass of both the solid sphere and the solid cylinder be $M$, and let their common radius be $R$. The moment of inertia $I$ of a solid sphere and a solid cylinder are given by:

$I_{sph}=\frac{2}{5}M{R}^2$

$I_{cyl}=\frac{1}{2}M{R}^2$

The radius of gyration $k$ is related to the moment of inertia $I$ by the formula $I=M{k}^2$. Therefore, we can find the radius of gyration for both the solid sphere and the solid cylinder using their respective moments of inertia:

$k_{sph}^2=\frac{I_{sph}}{M}=\frac{2}{5}{R}^2$

$k_{cyl}^2=\frac{I_{cyl}}{M}=\frac{1}{2}{R}^2$

Now, let's find the ratio of their radius of gyrations:

$\frac{k_{sph}}{k_{cyl}}=\frac{2}{\sqrt{x}}$

Squaring both sides:

$\frac{k_{sph}^2}{k_{cyl}^2}=\frac{4}{x}$

Substituting the expressions for $k_{sph}^2$ and $k_{cyl}^2$:

$\frac{\frac{2}{5}{R}^2}{\frac{1}{2}{R}^2}=\frac{4}{x}$

Simplifying and solving for $x$:

$\frac{2}{5}\cdot \frac{2}{1}=\frac{4}{x}$

$\frac{4}{5}=\frac{4}{x}$

Thus, $x=5$.