Q63 / 75 · JEE Main 2026 (23 January Shift 1) · Physics
The moment of inertia of a square loop made of four uniform solid cylinders, each having radius $R$ and length $L(\mathrm{R}<\mathrm{L})$ about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as $M$) :
- A$\frac{3}{4} M R^{2}+\frac{7}{12} M L^{2}$
- Correct answerB$\frac{3}{8} M R^{2}+\frac{1}{6} M L^{2}$
- C$\frac{3}{4} M R^{2}+\frac{1}{6} M L^{2}$
- D$\frac{3}{8} M R^{2}+\frac{7}{12} M L^{2}$
Solution
Each cylinder has mass $M/4$, radius $R$, length $L$.
Axis through midpoints of top and bottom sides (along y-direction).
Top/bottom cylinders (perpendicular to axis, axis through midpoint):
$I = \frac{M_c R^2}{4} + \frac{M_c L^2}{12}$ each.
Left/right cylinders (parallel to axis, distance $L/2$ from it):
$I = \frac{M_c R^2}{2} + M_c(L/2)^2$ each.
Total $= 2\left(\frac{M_c R^2}{4}+\frac{M_c L^2}{12}\right) + 2\left(\frac{M_c R^2}{2}+\frac{M_c L^2}{4}\right)$
$= \frac{M}{2}\left(\frac{3R^2}{4}+\frac{L^2}{3}\right) = \frac{3MR^2}{8}+\frac{ML^2}{6}$.
Axis through midpoints of top and bottom sides (along y-direction).
Top/bottom cylinders (perpendicular to axis, axis through midpoint):
$I = \frac{M_c R^2}{4} + \frac{M_c L^2}{12}$ each.
Left/right cylinders (parallel to axis, distance $L/2$ from it):
$I = \frac{M_c R^2}{2} + M_c(L/2)^2$ each.
Total $= 2\left(\frac{M_c R^2}{4}+\frac{M_c L^2}{12}\right) + 2\left(\frac{M_c R^2}{2}+\frac{M_c L^2}{4}\right)$
$= \frac{M}{2}\left(\frac{3R^2}{4}+\frac{L^2}{3}\right) = \frac{3MR^2}{8}+\frac{ML^2}{6}$.
Hence, the correct answer is option: 'B'.