JEE Physics 2023 Question with Solution

Given: A solid sphere of radius $R = 5 \, \text{cm}$ rotates about axis PQ. The solution uses the radius of gyration relation $I = Mk^2$.

Find: The value of $x$ if $k = \sqrt{x} \, \text{cm}$.

For radius of gyration,

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

The moment of inertia about axis PQ is taken using the parallel axis theorem:

$$I = I_{\text{center}} + Md^2$$

where

$$I_{\text{center}} = \frac{2}{5}MR^2$$

and $d = 10 \, \text{cm}$.

Substituting $R = 5 \, \text{cm}$,

$$I = \frac{2}{5}M(5)^2 + M(10)^2$$ $$I = \frac{2}{5}M \cdot 25 + 100M$$ $$I = 10M + 100M = 110M$$

Therefore,

$$k = \sqrt{\frac{I}{M}} = \sqrt{\frac{110M}{M}} = \sqrt{110} \, \text{cm}$$

Comparing with $k = \sqrt{x} \, \text{cm}$, we get $x = 110$.

Therefore, the required numerical value is $110$.

Given: Radius of sphere $= 5 \, \text{cm}$.

Find: The number $x$ in $k = \sqrt{x} \, \text{cm}$.

The radius of gyration is defined by

$$I = Mk^2$$

so

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

From the extracted solution, the axis PQ is at a distance $d = 10 \, \text{cm}$ from the center. Using parallel axis theorem,

$$I = I_{\text{center}} + Md^2$$

For a solid sphere,

$$I_{\text{center}} = \frac{2}{5}MR^2$$

Hence,

$$I = \frac{2}{5}M(5)^2 + M(10)^2$$ $$I = \frac{2}{5}M \cdot 25 + M \cdot 100$$ $$I = 10M + 100M = 110M$$

Now,

$$k = \sqrt{\frac{110M}{M}} = \sqrt{110} \, \text{cm}$$

Thus in the form $\sqrt{x} \, \text{cm}$, the value is $x = 110$.