Q 11-06-182JEE MainJEE Main 2020 (7 Jan, Shift 1)Easy
The radius of gyration of a uniform rod of length $l$, about an axis passing through a point $\dfrac l4$ away from the centre of the rod, and perpendicular to it, is:
Answer: (C) $\sqrt{\dfrac{7}{48}}\,l$
Parallel axis theorem:
$$I = \frac{ml^2}{12} + m\left(\frac l4\right)^2 = ml^2\left(\frac{1}{12} + \frac{1}{16}\right) = \frac{7}{48}ml^2$$
$$k = \sqrt{\frac Im} = \sqrt{\frac{7}{48}}\,l$$
Solution by Sreeraj P, M.Sc Physics