First slide

Rotational motion

Question

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 $θ$ , then :

Difficult

A

$\tan θ = \frac{2}{\sqrt{3}}$
B

$\tan θ = \frac{1}{3}$
C

$\tan θ = \frac{1}{2}$
D

$\tan θ = \frac{1}{2 \sqrt{3}}$

Solution

Let mass of one rod is ‘m’.
Balancing torque about hinge point.
$\begin{array}{l} m g (C_{1} P) = m g (C_{2} N) \dots . . (1) \\ C_{1} P = \frac{L}{2} \sin θ and C_{2} N = C_{2} L - L N \\ C_{2} L = \frac{L}{2} \cos θ \\ L N = A A^{'} = L \sin θ \\ \Rightarrow C_{2} N = \frac{L}{2} \cos θ - L \sin θ \\ From (1) \Rightarrow m g (\frac{L}{2} \sin θ) = m g (\frac{L}{2} \cos θ - L \sin θ) \\ \Rightarrow \frac{3}{2} m g L \sin θ = \frac{1}{2} m g L \cos θ \\ \Rightarrow \tan θ = \frac{1}{3} \end{array}$
Therefore, the correct answer is (B).

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