QJEEM23Ph87f1: JEE Main 2023 Physics - Class 11 JEEM23Ph87f1 Solution: A solid sphere and a solid cylinder of same mass and radius are …

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:

$I_{sph} = \frac{2}{5}MR^2$

$I_{cyl} = \frac{1}{2}MR^2$

The radius of gyration $k$ is related to the moment of inertia $I$ by the formula $I = Mk^2$ . 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, $x = 5$ .