Physics · Rotational Dynamics

JEE Main 2025 — 3 April, Morning Shift — Question 54

A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg , kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is

Option A:
$2.5 \mathrm{~m} / \mathrm{s}^{2}$
Option B:
$0.25 \mathrm{~m} / \mathrm{s}^{2}$
Option C:
$3.5 \mathrm{~m} / \mathrm{s}^{2}$
Correct
Option D:
$0.35 \mathrm{~m} / \mathrm{s}^{2}$

Answer: C

Step-by-step solution

$\tau=l \alpha \Rightarrow 49 \times 2 r=\frac{7}{5} m r^{2} \alpha$

$49 \times 2 r=\frac{7}{5} m r a$

$\frac{49 \times 2 \times 5}{7 \times 20}=a$

$a=3.5 \mathrm{~m} / \mathrm{s}^{2}$

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.

Solve a similar one free

Exam: JEE Main 2025
Paper: 3 April, Morning Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: Torque, Equation of Motion and Toppling