JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Rotational Mechanics

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):

Answer: (D)

Solution

Each cylinder has mass

M^{'} = \frac{M}{4}

and length

L

. The two cylinders lying across the axis (the axis passes through their midpoints, perpendicular to their length) each contribute

\frac{M ^{'} R ^{2}}{4} + \frac{M ^{'} L ^{2}}{12}

. The two cylinders parallel to the axis, each a distance

\frac{L}{2}

from it, contribute

\frac{M ^{'} R ^{2}}{2} + M^{'} (\frac{L}{2})^{2}

by the parallel-axis theorem. Adding all four,

I = 2 (\frac{M ^{'} R ^{2}}{4} + \frac{M ^{'} L ^{2}}{12}) + 2 (\frac{M ^{'} R ^{2}}{2} + \frac{M ^{'} L ^{2}}{4}) = \frac{3}{2} M^{'} R^{2} + \frac{2}{3} M^{'} L^{2}

. Substituting

M^{'} = \frac{M}{4}

gives

I = \frac{3}{8} M R^{2} + \frac{1}{6} M L^{2}

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option