Physics · Rotational Dynamics

JEE Main 2025 — 24 January, Evening Shift — Question 55

A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be $t_1$ and $t_2$ , respectively, then

Option A:
$\mathrm{t}_{1}<\mathrm{t}_{2}$
Correct
Option B:
$\mathrm{t}_{1}=\mathrm{t}_{2}$
Option C:
$t_{1}=2 t_{2}$
Option D:
$t_{1}>t_{2}$

Answer: A

Step-by-step solution

$\mathrm{t}=\sqrt{\frac{2 \ell}{\mathrm{a}_{\mathrm{cm}}}}$

$\mathrm{a}_{\mathrm{cm}}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}_{\mathrm{cm}}}{\mathrm{MR}^{2}}}$

$\mathrm{a}_{1}=\mathrm{a}_{\mathrm{cm}_{1}}=\frac{5 \mathrm{~g} \sin \theta}{7} \ldots$ . Solid

$\mathrm{a}_{2}=\mathrm{a}_{\mathrm{cm}_{2}}=\frac{3 \mathrm{~g} \sin \theta}{5} \ldots$ . Hollow

$a_{1}>a_{2}$

$\implies \mathrm{t}_{1}<\mathrm{t}_{2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: 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…

Answer: A

Step-by-step solution

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