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 :
JEE MAIN 2024, Diffcult
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Loss in kinetic energy $=$ Gain in potential energy
$ \Rightarrow \frac{1}{2} \mathrm{mv}^2=\mathrm{mg} \ell(1-\cos \theta)$
$ \Rightarrow \frac{\mathrm{v}^2}{\ell}=2 \mathrm{~g}(1-\cos \theta)$
Acceleration at lowest point $=\frac{\mathrm{v}^2}{\ell}$
Acceleration at extreme point $=g \sin \theta$
Hence, $\frac{\mathrm{v}^2}{\ell}=\mathrm{g} \sin \theta$
$ \therefore \sin \theta=2(1-\cos \theta) $
$ \Rightarrow \tan \frac{\theta}{2}=\frac{1}{2} \Rightarrow \theta=2 \tan ^{-1}\left(\frac{1}{2}\right)$
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