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JEE Main 2025 (24 Jan, Shift 2) · Rolling Motion

Rolling Motion question from JEE Main 2025 (24 Jan, Shift 2)

Q10JEE Main 2025 · 24 JAN · SHIFT 2Easy

Rolling Motion

A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is:

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Answer: (c)

$\frac{K_{trans}}{K_{rot}} = \frac{R^2}{k^2}$ for rolling.
Solid sphere: $\frac{k^2}{R^2} = \frac{2}{5}$ , so $\frac{K_{trans}}{K_{rot}} = \frac{5}{2}$
So the answer is (c) $\frac{5}{2}$ .

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