question_answer

A bob of mass M is suspended by a massless string of length L. The horizontal velocity v at position A is just sufficient to make it reach the point B. The angle \[\theta \]at which the speed of the bob is half of that at A, satisfies

A) \[\theta =\frac{\pi }{4}\]

B) \[\frac{\pi }{4}<\theta <\frac{\pi }{2}\]

C) \[\frac{\pi }{2}<\theta <\frac{3\pi }{4}\]

D) \[\frac{3\pi }{4}<\theta <\pi \]

Correct Answer: D

Solution :

[d]: According to conservation of energy,       \[\frac{1}{2}M{{v}^{2}}=\frac{1}{2}M{{\left( \frac{v}{2} \right)}^{2}}+Mgh\] \[\frac{1}{2}M5gL=\frac{1}{2}M\frac{5gL}{4}+MgL(1-cos\theta )\] or\[\cos \theta =-\frac{7}{8}\]or\[\theta ={{\cos }^{-1}}\left( \frac{-7}{8} \right)={{151}^{o}}\] \[\therefore \]\[\frac{3\pi }{4}<\theta <\pi \]