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JEE Main 2026 (28 Jan, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2026 (28 Jan, Shift 1)

Q24JEE Main 2026 · 28 JAN · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

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 $\sqrt{n}\text{ cm}$ . The value of $n$ is

Numerical answer type. Work it out, then open the answer.

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Answer: 265

Parallel axis theorem: $k^2 = \frac{2}{5}R^2 + d^2$ .
$k^2 = \frac{2}{5}(10)^2 + (15)^2 = 40 + 225 = 265\ \text{cm}^2$
$k = \sqrt{265}$ cm, so $n = 265$ .

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