Q 11-06-108JEE MainJEE Main 2023 (24 Jan, Shift 1)Easy
Solid sphere $A$ is rotating about an axis $PQ$. If the radius of the sphere is $5\ \text{cm}$, then its radius of gyration about $PQ$ will be $\sqrt x$ cm. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 110
By the parallel axis theorem:
$$Mk^2=\frac25MR^2+Md^2\ \Rightarrow\ k^2=\frac25(25)+100=110\ \text{cm}^2$$
So $k=\sqrt{110}\ \text{cm}$ and $x=110$.
Solution by Sreeraj P, M.Sc Physics