question_answer

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) \[\text{tan}\theta =\,\frac{2}{\sqrt{3}}\] 

B) \[\text{tan}\theta \,=\,\frac{1}{3}\]

C) \[\text{tan}\theta \,=\,\frac{1}{2}\]

D) \[\text{tan}\theta \,=\,\frac{1}{2\sqrt{3}}\]

Correct Answer: B

Solution :

\[{{\text{C}}_{1}}\text{P}\,=\,\frac{\text{L}}{2}\text{sin}\theta \]

and       \[{{\text{C}}_{2}}\text{N}\,=\,\frac{\text{L}}{2}\text{cos}\theta \,-\,L\text{sin}\theta \]

Let mass of one rod is m. Balancing torque about hinge point, \[mg\,({{\text{C}}_{1}}\text{P})\,=\,mg\,({{\text{C}}_{2}}\text{N})\]

\[mg\left( \frac{L}{2}\text{sin}\theta  \right)\,=\,mg\,\left( \frac{\text{L}}{2}\cos \theta -\text{Lsin}\theta  \right)\]

\[\Rightarrow \]   \[\frac{3}{2}mgL\text{sin}\theta \,=\,\frac{mgL}{2}\text{cos}\theta \]

\[\Rightarrow \]   \[\text{tan}\,\theta \,=\,\frac{1}{3}.\]