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

 

Let mass of one rod is ‘m’.
Balancing  torque about hinge point.
$\begin{array}{l}mg\left(C_{1}P\right)=mg\left(C_{2}N\right)\dots ..\left(1\right)\\ C_{1}P=\frac{L}{2}\sin \theta \text{ and }{C}_{2}N=C_{2}L-LN\\ C_{2}L=\frac{L}{2}\cos \theta \\ LN=A{A}^{\mathrm{\prime }}=L\sin \theta \\ \Rightarrow C_{2}N=\frac{L}{2}\cos \theta -L\sin \theta \\ \text{ From }\left(1\right)\Rightarrow mg\left(\frac{L}{2}\sin \theta \right)=mg\left(\frac{L}{2}\cos \theta -L\sin \theta \right)\\ \Rightarrow \frac{3}{2}mgL\sin \theta =\frac{1}{2}mgL\cos \theta \\ \Rightarrow \tan \theta =\frac{1}{3}\end{array}$