Physics · Rotational Dynamics

JEE Main 2024 — 6 April, Shift 2 — Question 54

Three balls of masses $2 \mathrm{~kg}, 4 \mathrm{~kg}$ and 6 kg respectively are arranged at centre of the edges of an equilateral triangle of side 2 m . The moment of inertia of the system about an axis through the centroid and perpendicular to the plane of triangle, will be $\qquad$ $\mathrm{kg} \mathrm{m}^{2}$ .

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

Moment of inertia about C and perpendicular to the plane is :

$\begin{array}{*{35}{r}}\text{I} & ~={{\text{r}}^{2}}\left[ 2+4+6 \right] \\{} & ~=\frac{1}{3}\times 12 \\\text{I} & ~=4\text{ }\!\!~\!\!\text{ kg}-{{\text{m}}^{2}} \\\end{array}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Physics
Chapter: Rotational Dynamics
Topic: Moment of Inertia