JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2026
- Hard
- MCQ
The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length \(L ( R < L )\) about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as M ) :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \( \frac{3}{8}MR^{2}+\frac{1}{6}ML^{2} \)
Step-by-step Solution
Detailed explanation

\(I _{ net }=2\left( I _1+ I _2\right)\)
\(=2\left(\frac{ M ^{\prime} R ^2}{4}+\frac{ M ^{\prime} \ell^2}{12}\right)+2\left(\frac{ M ^{\prime} R ^2}{2}+ M ^{\prime}\left(\frac{\ell}{2}\right)^2\right)\)
\(=\frac{M^{\prime} R^2}{2}+\frac{M^{\prime} R^2}{6}+M^{\prime} R^2+\frac{M^{\prime} \ell^2}{2}\)
\(=\frac{3 M ^{\prime} R ^2}{2}+\frac{2 M ^{\prime} \ell^2}{3}\)
Given masses \(M ^{\prime}=\frac{ M }{4}\)
So, \(I=\frac{3(M / 4) R^2}{2}+2 \frac{(M / 4) \ell^2}{3}\)
\(I =\frac{3}{8} MR ^2+\frac{ M \ell^2}{6}\)
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