Physics · Rotational Dynamics

JEE Main 2024 — 27 January, Shift 1 — Question 53

Four particles each of mass 1 kg are placed at four corners of a square of side 2 m . Moment of inertia of system about an axis perpendicular to its plane and passing through one of its vertex is $\_\_\_\_\_\_\_$ $\mathrm{kgm}^{2}$

Answer: 16

Numerical answer — enter this value.

Step-by-step solution

$I=m a^{2}+m a^{2}+m(\sqrt{2} a)^{2}$ $=4 \mathrm{ma}^{2}$ $=4 \times 1 \times(2)^{2}=16$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 1
Subject: Physics
Chapter: Rotational Dynamics
Topic: Moment of Inertia