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(B) Let the mass of each rod be $m$. The center of mass of rod $AB$ is at a distance $a/2$ from $A$,and the center of mass of rod $BC$ is at a distanc

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$	an 	heta = \frac{1}{3}$

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(B) Let the mass of each rod be $m$. The center of mass of rod $AB$ is at a distance $a/2$ from $A$,and the center of mass of rod $BC$ is at a distance $a/2$ from $B$. For rotational equilibrium,the net torque about the suspension point $A$ must be zero. The torque due to the weight of rod $AB$ is $	au_1 = mg \cdot (a/2) \sin 	heta$. The horizontal distance of the center of mass of rod $BC$ from $A$ is $(a \sin 	heta + (a/2) \cos 	heta)$. The torque due to the weight of rod $BC$ is $	au_2 = mg \cdot (a \sin 	heta + (a/2) \cos 	heta)$. Equating the torques for equilibrium: $mg(a/2) \sin 	heta = mg(a \sin 	heta + (a/2) \cos 	heta)$ is incorrect based on the geometry. Correcting the geometry: Let $A$ be the origin $(0,0)$. Rod $AB$ makes angle $	heta$ with the vertical. Its $CM$ is at $(a/2 \sin 	heta, a/2 \cos 	heta)$. Rod $BC$ is perpendicular to $AB$. Its $CM$ is at $(a \sin 	heta + (a/2) \cos 	heta, a \cos 	heta - (a/2) \sin 	heta)$. For equilibrium,the $x$-coordinate of the center of mass of the system must be zero relative to the suspension point $A$. $x_{cm} = \frac{m(a/2 \sin 	heta) + m(a \sin 	heta + a/2 \cos 	heta)}{2m} = 0$ $(a/2) \sin 	heta + a \sin 	heta + (a/2) \cos 	heta = 0$ $(3a/2) \sin 	heta = -(a/2) \cos 	heta$ Taking magnitudes,$3 \sin 	heta = \cos 	heta \Rightarrow 	an 	heta = 1/3$.

An $L$-shaped object,made of two thin rods of uniform mass density and equal length $a$,is suspended from a string as shown in the figure. If $AB = BC = a$,and the angle made by $AB$ with the downward vertical is $\theta$,then:

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