Laws Of Motion Question 134
Question: 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
Options:
A) $ 30{}^\circ $
B) $ {{\tan }^{-1}}(2\sqrt[{}]{3}) $
C) $ {{\tan }^{-1}}(\sqrt[{}]{3}/2) $
D) $ {{\tan }^{-1}}(2) $
Show Answer
Answer:
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}} $
BETA
