Question 127 of 280

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:

A$\tan ⁡\theta =\dfrac{2}{\sqrt{3}}$B$\tan ⁡\theta =\dfrac{1}{3}$C$\tan ⁡\theta =\dfrac{1}{2}$D$\tan ⁡\theta =\dfrac{1}{2\sqrt{3}}$

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Answer: (B) $\tan ⁡\theta =\dfrac{1}{3}$

At equilibrium, the centre of mass will be below the point of suspension.

Taking torque about point O,

$mg\dfrac{l}{2}\sin ⁡\theta +mg\left(l\sin ⁡\theta -\dfrac{l}{2}\cos ⁡\theta \right)=0$

$\Rightarrow \dfrac{3l}{2}\sin ⁡\theta =\dfrac{l}{2}\cos ⁡\theta$

$\Rightarrow \tan ⁡\theta =\dfrac{1}{3}$

$\Rightarrow \theta ={tan}^{-1}\dfrac{1}{3}$