JEE Physics 2023 Question with Solution

Given: A solid sphere and a solid cylinder have the same mass and radius.

Find: The value of $x$ if $k_{sphere} : k_{cylinder} = 2 : \sqrt{x}$.

Radius of gyration is defined as

$$k = \sqrt{\frac{I}{M}}$$

where $I$ is the moment of inertia and $M$ is the mass.

For a solid sphere about its centre,

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

Hence,

$$k_{sphere} = \sqrt{\frac{I_{sphere}}{M}} = \sqrt{\frac{2}{5}}\,R$$

For a solid cylinder about its central axis,

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

Hence,

$$k_{cylinder} = \sqrt{\frac{I_{cylinder}}{M}} = \sqrt{\frac{1}{2}}\,R$$

Now,

$$\frac{k_{sphere}}{k_{cylinder}} = \frac{\sqrt{\frac{2}{5}}}{\sqrt{\frac{1}{2}}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$$

Given,

$$\frac{2}{\sqrt{x}} = \frac{2}{\sqrt{5}}$$

So,

$$\sqrt{x} = \sqrt{5}$$

Therefore,

$$x = 5$$

The value of $x$ is $5$.

Given: Standard moments of inertia for a solid sphere and a solid cylinder.

Find: The value of $x$.

Using $k = \sqrt{I/M}$, radius of gyration is proportional to the square root of the coefficient of $MR^2$ in the moment of inertia.

So directly,

$$k_{sphere} : k_{cylinder} = \sqrt{\frac{2}{5}} : \sqrt{\frac{1}{2}} = 2 : \sqrt{5}$$

Comparing with $2 : \sqrt{x}$, we get

$$\sqrt{x} = \sqrt{5}$$

Hence, the value of $x$ is $5$.