Physics · Rotational Dynamics

JEE Main 2024 — 1 February, Shift 2 — Question 35

$A$ disc of radius $R$ and mass $M$ is rolling horizontally without slipping with speed $v$ . It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is :

Option A:
$\frac{v^{2}}{g}$
Option B:
$\frac{3}{4} \frac{v^{2}}{g}$
Correct
Option C:
$\frac{1}{2} \frac{v^{2}}{g}$
Option D:
$\frac{2}{3} \frac{v^{2}}{g}$

Answer: B

Step-by-step solution

Only the translational kinetic energy of disc changes into gravitational potential energy.

And rotational KE remains unchanged as there is no friction.

$\frac{1}{2} \operatorname{mv}^{2}=\mathrm{mgh}$ $h=\frac{v^{2}}{2 g}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 2
Subject: Physics
Chapter: Rotational Dynamics
Topic: Rolling Motion

A uniform rod AB of mass 2 kg and Length 30 cm at rest on a smooth horizontal surface. An impulse of force 0.2 Ns is applied to end B. The…

Answer: B

Step-by-step solution

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