Physics · Rotational Dynamics

JEE Main 2025 — 28 January, Evening Shift — Question 60

A uniform rod of mass 250 g having length 100 cm is balanced on a sharp edge at 40 cm mark. A mass of 400 g is suspended at 10 cm mark. To maintain the balance of the rod, the mass to be suspended at 90 cm mark, is

Option A:
300 g
Option B:
190g
Correct
Option C:
200g
Option D:
290g

Answer: B

Step-by-step solution

$\tau_{\text {Net }}=0 \Rightarrow(400 \mathrm{~g} \times 30)=(250 \mathrm{~g} \times 10)(\mathrm{mg} \times 50)$

$\mathrm{m}=\frac{12000-2500}{50}=\frac{9500}{50}$

$\mathrm{M}=190 \mathrm{~g}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: Torque, Equation of Motion and Toppling