Physics · Rotational Dynamics

JEE Main 2024 — 9 April, Shift 2 — Question 50

A circular disc reaches from top to bottom of an inclined plane of length ll. When it slips down the plane, if takes t s . When it rolls down the plane then it takes (α2)1/2ts\left(\frac{\alpha}{2}\right)^{1 / 2} \mathrm{t} s, where α\alpha is

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

For slipping

a=gsin⁡θ\mathrm{a}=\mathrm{g} \sin \theta

ℓ=12at2⇒t=2ℓgsin⁡θ\ell=\frac{1}{2} a t^{2} \Rightarrow t=\sqrt{\frac{2 \ell}{g \sin \theta}}

For rolling

a′=gsin⁡θ1+k2R2[k=R2]\mathrm{a}^{\prime}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}}\left[\mathrm{k}=\frac{\mathrm{R}}{\sqrt{2}}\right]

⇒a′=2 gsin⁡θ3\Rightarrow \mathrm{a}^{\prime}=\frac{2 \mathrm{~g} \sin \theta}{3}

ℓ=12a′(t′)2\ell=\frac{1}{2} \mathrm{a}^{\prime}\left(\mathrm{t}^{\prime}\right)^{2}

⇒t′=6ℓ2gsin⁡θ=α22ℓgsin⁡θ\Rightarrow \mathrm{t}^{\prime}=\sqrt{\frac{6 \ell}{2 g \sin \theta}}=\sqrt{\frac{\alpha}{2}} \sqrt{\frac{2 \ell}{g \sin \theta}}

⇒α=3\Rightarrow \alpha=3

Answer key and solution verified before publishing.

Practise Rotational Dynamics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2024
Subject
Physics
Chapter
Rotational Dynamics
Topic
Rolling Motion