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Physics · Rotational Dynamics

JEE Main 2026 — 23 January, Morning Shift — Question 33

The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length L(R<L)\mathrm{L}(\mathrm{R}<\mathrm{L}) about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as M ) :

  1. Option A:

    38MR2+712ML2\frac{3}{8} \mathrm{MR}^{2}+\frac{7}{12} \mathrm{ML}^{2}

  2. Option B:

    34MR2+16ML2\frac{3}{4} M R^{2}+\frac{1}{6} M L^{2}

  3. Option C:

    34MR2+712ML2\frac{3}{4} \mathrm{MR}^{2}+\frac{7}{12} \mathrm{ML}^{2}

  4. Option D:

    38MR2+16ML2\frac{3}{8} \mathrm{MR}^{2}+\frac{1}{6} \mathrm{ML}^{2}

    Correct

Answer: D

Step-by-step solution

Inet =2(I1+I2)\mathrm{I}_{\text {net }}=2\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) =2(M′R24+M′ℓ212)+2(M′R22+M′(ℓ2)2)=2\left(\frac{\mathrm{M}^{\prime} \mathrm{R}^{2}}{4}+\frac{\mathrm{M}^{\prime} \ell^{2}}{12}\right)+2\left(\frac{\mathrm{M}^{\prime} \mathrm{R}^{2}}{2}+\mathrm{M}^{\prime}\left(\frac{\ell}{2}\right)^{2}\right) =M′R22+M′R26+M′R2+M′ℓ22=\frac{M^{\prime} R^{2}}{2}+\frac{M^{\prime} R^{2}}{6}+M^{\prime} R^{2}+\frac{M^{\prime} \ell^{2}}{2} =3M′R22+2M′ℓ23=\frac{3 \mathrm{M}^{\prime} \mathrm{R}^{2}}{2}+\frac{2 \mathrm{M}^{\prime} \ell^{2}}{3} Given masses M′=M4\mathrm{M}^{\prime}=\frac{\mathrm{M}}{4} So, I=3(M/4)R22+2(M/4)ℓ23I=\frac{3(M / 4) R^{2}}{2}+2 \frac{(M / 4) \ell^{2}}{3} I=38MR2+Mℓ26\mathrm{I}=\frac{3}{8} \mathrm{MR}^{2}+\frac{\mathrm{M} \ell^{2}}{6}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
23 January, Morning Shift
Subject
Physics
Chapter
Rotational Dynamics
Topic
Moment of Inertia

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