A solid sphere rolls without slipping on an inclied plane at an angle θ. The ratio of total kinetic energy to its rotational kinetic energy is (a) 5/2 (b) 7/2 (c) 5/4 (d) 5/7

Answer: (b). Solid sphere: Krot = 1/2·2/5mR²ω² = 1/5mv², Ktrans = 1/2mv². Ktotal = 7/10mv² KtotalKrot = (7/10)/(1/5) = 7/2 So the answer is (b) 7/2.

A solid sphere rolls without slipping on an inclied plane at an angle… | MHT CET 2026 (18 Apr, Shift 2)

Q1MHT CET 2026 · 18 APR · SHIFT 2Easy

Rolling Motion

A solid sphere rolls without slipping on an inclied plane at an angle $\theta$ . The ratio of total kinetic energy to its rotational kinetic energy is

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Answer: (b)

Solid sphere: $K_{rot} = \frac{1}{2}\cdot\frac{2}{5}mR^2\omega^2 = \frac{1}{5}mv^2$ , $K_{trans} = \frac{1}{2}mv^2$ .
$K_{total} = \frac{7}{10}mv^2$
$\frac{K_{total}}{K_{rot}} = \frac{7/10}{1/5} = \frac{7}{2}$
So the answer is (b) $\frac{7}{2}$ .

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