JEE Main 2026 — Jan 23, Shift 1 — Physics solution

JEE Main2026Jan 23, Shift 1Physics

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) :

A
$\frac{3}{8} M R^{2} + \frac{7}{12} M L^{2}$
B
$\frac{3}{4} M R^{2} + \frac{1}{6} M L^{2}$
C
$\frac{3}{4} M R^{2} + \frac{7}{12} M L^{2}$
D
$\frac{3}{8} M R^{2} + \frac{1}{6} M L^{2}$

Solution

The square loop has four cylinders. The axis passes through midpoints of two opposite sides. Let each cylinder have mass $M^{'} = M /4$ .

Two cylinders lie along the axis (the two whose midpoints the axis passes through). For each, the axis is the cylinder's own perpendicular bisector, so the contribution is the standard $\frac{M ^{'} R ^{2}}{4} + \frac{M ^{'} L ^{2}}{12}$ .

Two cylinders lie perpendicular to the axis, with their centres at distance $L /2$ from the axis. Each contributes $\frac{M ^{'} R ^{2}}{2}$ (axis along its length) plus the parallel-axis term $M^{'} (L /2)^{2}$ .

Summing:

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

Substituting $M^{'} = M /4$ :

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

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