Q 11-13-058JEE MainJEE Main 2024 (27 Jan, Shift 2)Hard
A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle ($\theta$) of thread deflection in the extreme position will be:
Answer: (B) $2\tan^{-1}\left(\dfrac12\right)$
At the extreme position the speed is zero, so only tangential acceleration acts: $a_1 = g\sin\theta$.
At the lowest position the tangential acceleration is zero and $a_2 = \dfrac{v^2}{L}$, with $v^2 = 2gL(1-\cos\theta)$, so $a_2 = 2g(1-\cos\theta)$.
Setting $a_1 = a_2$:
$$2\sin\frac\theta2\cos\frac\theta2 = 4\sin^2\frac\theta2 \;\Rightarrow\; \tan\frac\theta2 = \frac12 \;\Rightarrow\; \theta = 2\tan^{-1}\left(\frac12\right)$$
Solution by Sreeraj P, M.Sc Physics