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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 $θ$ then

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$\tan θ = \frac{1}{2 \sqrt{3}}$

$\tan θ = \frac{1}{2}$

$\tan θ = \frac{1}{\sqrt{3}}$

$\tan θ = \frac{1}{3}$

answer is D.

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Detailed Solution

$C_{1} P = \frac{L}{2} \sin θ$

$and C_{2} N = C_{2} M - MN = \frac{L}{2} \cos θ - Lsin θ$

Balancing torque about hinge point.

$mg (C_{1} P) = mg (C_{2} N)$

$\Rightarrow mg (\frac{L}{2} sinθ) = mg (\frac{L}{2} cosθ - Lsinθ)$

$\Rightarrow tanθ = \frac{1}{3}$

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