A bob of mass $M$ is suspended by a massless string of length $L$. The horizontal velocity $v$ at position…
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
$\theta=\frac{\pi}{4}$
$\frac{\pi}{4} < \theta < \frac{\pi}{2}$
$\frac{\pi}{2} < \theta < \frac{3 \pi}{4}$
$\frac{3 \pi}{4} < \theta < \pi$
Solution
$v=\sqrt{5 g L}$
Solving Eqs. (i), (ii) and (iii) we get,
$
\cos \theta=-\frac{7}{8} \quad \text { or } \theta=\cos ^{-1}\left(-\frac{7}{8}\right)=151^{\circ}
$
$\therefore$ correct option is (d)
