Work Power and Energy 4 Question 2
3. 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 $A$, statisfies
$(2008,3 M)$
(a) $\theta=\frac{\pi}{4}$
(b) $\frac{\pi}{4}<\theta<\frac{\pi}{2}$
(c) $\frac{\pi}{2}<\theta<\frac{3 \pi}{4}$
(d) $\frac{3 \pi}{4}<\theta<\pi$
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Answer:
Correct Answer: 3. (d)
Solution:
- As,
$$ \begin{gathered} v=\sqrt{5 g L} \\ \frac{v^{2}}{2}=v^{2}-2 g h \\ h=L(1-\cos \theta) \end{gathered} $$
Solving Eqs. (i), (ii) and (iii), we get
$$ \begin{aligned} \cos \theta & =-\frac{7}{8} \\ \text { or } \quad \theta & =\cos ^{-1}-\frac{7}{8}=151^{\circ} \end{aligned} $$
