A solid sphere of mass 1 kg rolls without slipping on a plane surface. Its kinetic energy is 7 × 10⁻³ J. The speed of the center of mass of the sphere in cm s ⁻¹ is (a) 1 (b) 10 (c) 100 (d) 1000

Answer: (b). Rolling solid sphere: KE = 1/2mv²(1 + 2/5) = 7/10mv². 7/10× 1 × v² = 7× 10⁻³ ⇒ v² = 10⁻² v = 0.1 m/s = 10 cm/s. So the answer is (b).

A solid sphere of mass 1 kg rolls without slipping on a plane… | MHT CET 2026 (13 Apr, Shift 2)

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Rolling Motion

A solid sphere of mass $1$ kg rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3}$ J. The speed of the center of mass of the sphere in cm s $^{-1}$ is

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Answer: (b)

Rolling solid sphere: $KE = \frac{1}{2}mv^2\left(1 + \frac{2}{5}\right) = \frac{7}{10}mv^2$ .
$\frac{7}{10}\times 1 \times v^2 = 7\times 10^{-3} \Rightarrow v^2 = 10^{-2}$
$v = 0.1$ m/s $= 10$ cm/s.
So the answer is (b).

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