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Question

Physics Question on System of Particles & Rotational Motion

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

A

tanθ=23\tan \theta = \frac{2}{\sqrt{3}}

B

tanθ=13\tan \theta = \frac{1}{3}

C

tanθ=12\tan \theta = \frac{1}{2}

D

tanθ=123\tan \theta = \frac{1}{2 \sqrt{3}}

Answer

tanθ=13\tan \theta = \frac{1}{3}

Explanation

Solution

Let mass of one rod is mm.
Balancing torque about hinge point.

mg(C1P)=mg(C2N)mg \left(C_{1}P\right) = mg \left(C_{2}N\right)
mg(L2sinθ)=mg(L2cosθLsinθ)mg \left(\frac{L}{2} \sin\theta\right) = mg \left(\frac{L}{2} \cos\theta - L \sin\theta\right)
32mgLsinθ=mgL2cosθ\Rightarrow \frac{3}{2} mg L \sin\theta = \frac{mgL}{2} \cos\theta
tanθ=13\Rightarrow \tan \theta = \frac{1}{3}