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

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