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

The correct option is A $\tan \theta =\frac{1}{3}$

Given that, the rod is of uniform mass density and $AB=BC$

Let the mass of one rod be $m$.

As the system is in equilibrium, the net torque about the hinged point is zero,

$\Rightarrow mg\left(C_{1}P\right)=mg\left(C_{2}N\right)$

$mg\left(\frac{L}{2}\sin \theta \right)=mg(\frac{L}{2}\cos \theta -L\sin \theta )$

$\Rightarrow \frac{3}{2}mgL\sin \theta =\frac{mgL}{2}\cos \theta$

$\Rightarrow \tan \theta =\frac{1}{3}$

Hence, option $\left(D\right)$ is correct.