question_answer

A plumb bob is hung from the ceiling of a train compartment. The train moves on an inclined track of inclination \[30{}^\circ \] with horizontal. The acceleration of train up the plane is \[a\,=\,g/2\]. The angle which the string supporting the bob makes with normal to the ceiling in equilibrium is

A) \[30{}^\circ \]              

B) \[{{\tan }^{-1}}(2\sqrt[{}]{3})\]

C) \[{{\tan }^{-1}}(\sqrt[{}]{3}/2)\]          

D) \[{{\tan }^{-1}}(2)\]

Correct Answer: B

Solution :

[b] \[T\sin \theta -mg\sin 30{}^\circ =ma\] \[\Rightarrow T\sin \theta =mg\sin 30{}^\circ +mg/2(i)\] \[T\cos \theta =mg\cos 30{}^\circ (ii)\] Dividing Eqs. (i) by (ii), we get \[\tan \theta =\frac{2}{\sqrt{3}}\]