A cylinder is rolling down on an inclined plane of inclination 60 ° . It's acceleration during rolling down will be frac x √(3) m / s 2 , where x = \ \ \ \ \ \ \ . (use g =10 m / s 2 ).

Physics · Rotational Dynamics

JEE Main 2024 — 29 January, Shift 1 — Question 52

A cylinder is rolling down on an inclined plane of inclination $60^{\circ}$ . It's acceleration during rolling down will be $\frac{\mathrm{x}}{\sqrt{3}} \mathrm{~m} / \mathrm{s}^{2}$ , where $\mathrm{x}=$ $\_\_\_\_\_\_\_$ . (use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ).

Answer: 10

Numerical answer — enter this value.

Step-by-step solution

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

$\mathrm{a}=\frac{\mathrm{g} \sin \theta}{1+\frac{1}{2}}$

$=\frac{2 \times 10 \times \frac{\sqrt{3}}{2}}{3}$ $=\frac{10}{\sqrt{3}}$

Therefore $\mathrm{x}=10$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Physics
Chapter: Rotational Dynamics
Topic: Rolling Motion