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, \[mg\,({{\text{C}}_{1}}\text{P})\,=\,mg\,({{\text{C}}_{2}}\text{N})\] |
\[mg\left( \frac{L}{2}\text{sin}\theta \right)\,=\,mg\,\left( \frac{\text{L}}{2}\cos \theta -\text{Lsin}\theta \right)\] |
\[\Rightarrow \] \[\frac{3}{2}mgL\text{sin}\theta \,=\,\frac{mgL}{2}\text{cos}\theta \] |
\[\Rightarrow \] \[\text{tan}\,\theta \,=\,\frac{1}{3}.\] |
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