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

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Solution:

Let mass of one rod is m.
Balancing torque about hinge point.
$mg \left(C_{1}P\right) = mg \left(C_{2}N\right) $
$ mg \left(\frac{L}{2} \sin\theta\right) = mg \left(\frac{L}{2} \cos\theta - L \sin\theta\right) $
$ \Rightarrow \frac{3}{2} mg L \sin\theta = \frac{mgL}{2} \cos\theta $
$ \Rightarrow \tan \theta = \frac{1}{3} $

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