JEE MAIN - Physics (2024 - 27th January Evening Shift - No. 2)
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 :
$$\tan ^{-1}\left(\frac{1}{2}\right)$$
$$2 \tan ^{-1}\left(\frac{1}{2}\right)$$
$$2 \tan ^{-1}\left(\frac{1}{\sqrt{5}}\right)$$
$$\tan ^{-1}(\sqrt{2})$$
Explanation
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Loss in kinetic energy $=$ Gain in potential energy
$$\begin{aligned} & \Rightarrow \frac{1}{2} \mathrm{mv}^2=\mathrm{mg} \ell(1-\cos \theta) \\ & \Rightarrow \frac{\mathrm{v}^2}{\ell}=2 \mathrm{~g}(1-\cos \theta) \end{aligned}$$
Acceleration at lowest point $$=\frac{\mathrm{v}^2}{\ell}$$
Acceleration at extreme point $$=g \sin \theta$$
$$\begin{aligned} & \text { Hence, } \frac{\mathrm{v}^2}{\ell}=\mathrm{g} \sin \theta \\ & \therefore \sin \theta=2(1-\cos \theta) \\ & \Rightarrow \tan \frac{\theta}{2}=\frac{1}{2} \Rightarrow \theta=2 \tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}$$
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