Q82 / 90 · JEE Main 2022 (26 Jul Shift 2) · Physics

The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the center will be _____ $m .$ Given, the length of the rod is $10 \sqrt{3} m$ .

Correct answer: 5

Solution

Moment of inertia of the rod about an axis perpendicular to its length and passing through the centre is $I = \frac{m l^{2}}{12}$ .
Using the relation, $I = m k^{2}$ , we have
$\Rightarrow k^{2} = \frac{l^{2}}{12} \Rightarrow k = \frac{l}{\sqrt{12}} = \frac{l}{2 \sqrt{3}} = \frac{10 \sqrt{3}}{2 \sqrt{3}} = 5 m$

Hence, the correct answer is 5.

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Solution

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