Question 16

The moment of inertia of a rod of mass $m$ and length $l$ about an axis passing through one of its ends at an angle $\theta$ with the rod is given by $I = \frac{1}{3} ml^2 \sin^2\theta$.

For (P): There are two rods, each making an angle $\theta = 45^\circ$ with the axis. The axis passes through one end ($C$) of both rods.

$I_P = 2 \times \left( \frac{1}{3} ml^2 \sin^2 45^\circ \right) = 2 \times \frac{1}{3} ml^2 \times \frac{1}{2} = \frac{1}{3} ml^2$ This matches with (5).

For (Q): Rods $CA$ and $CB$ make an angle of $60^\circ$ with the axis and have the axis passing through end $C$. Rod $AB$ is parallel to the axis at a distance $h = l \sin 60^\circ = \frac{\sqrt{3}}{2}l$.

$I_Q = I_{CA} + I_{CB} + I_{AB} = 2 \times \left( \frac{1}{3} ml^2 \sin^2 60^\circ \right) + m h^2$

$I_Q = 2 \times \frac{1}{3} ml^2 \times \frac{3}{4} + m \left( \frac{3}{4} l^2 \right) = \frac{1}{2} ml^2 + \frac{3}{4} ml^2 = \frac{5}{4} ml^2$

This matches with (1).

For (R): There are four rods. Each rod makes an angle of $45^\circ$ with the diagonal axis and has the axis passing through one of its ends (either $A$ or $C$).

$I_R = 4 \times \left( \frac{1}{3} ml^2 \sin^2 45^\circ \right) = 4 \times \frac{1}{3} ml^2 \times \frac{1}{2} = \frac{2}{3} ml^2$ This matches with (4).

For (S): There are two rods, each making an angle of $30^\circ$ with the axis. The axis passes through one end ($C$) of both rods.

$I_S = 2 \times \left( \frac{1}{3} ml^2 \sin^2 30^\circ \right) = 2 \times \frac{1}{3} ml^2 \times \frac{1}{4} = \frac{1}{6} ml^2$ This matches with (2).

Comparing with options, the correct match is P-5, Q-1, R-4, S-2. Correct Option: A