Q 11-06-058JEE MainJEE Main 2026 (28 Jan, Shift 1)Easy
A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is $\sqrt n$ cm. The value of $n$ is
Numerical value type. Enter your answer.
Answer: 265
By the parallel axis theorem:
$$I = \frac25MR^2 + Md^2 = M\left(\frac25(10)^2 + (15)^2\right) = M(40 + 225) = 265M\ \text{(cm}^2)$$
$$k = \sqrt{\frac IM} = \sqrt{265}\ \text{cm}\Rightarrow n = 265$$
Solution by Sreeraj P, M.Sc Physics