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JEE Advanced · Physics · 4. Motion in 2D

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

  1. A \(\theta=\frac{\pi}{4}\)
  2. B \(\frac{\pi}{4} < \theta < \frac{\pi}{2}\)
  3. C \(\frac{\pi}{2} < \theta < \frac{3 \pi}{4}\)
  4. D \(\frac{3 \pi}{4} < \theta < \pi\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{3 \pi}{4} < \theta < \pi\)

Step-by-step Solution

Detailed explanation

\(v=\sqrt{5 g L}\)
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).
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