$(a)\;\theta=tan^{-1} (\large\frac{qE}{mg})\qquad(b)\;\theta=\pi\qquad(c)\;\theta=0^{0}\qquad(d)\;\theta=\pi+tan^{-1} (\large\frac{qE}{mg})$

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Answer : (d) $\;\theta=\pi+tan^{-1} (\large\frac{qE}{mg})$

Explanation :

Let the velocity at bottom most point is $\;V_{0}$

Then at angle $\;\theta\;$ let the velocity of particle is V

$T-mg cos \theta-qE sin \theta=\large\frac{mV^2}{l}--(1)$

Using energy conservation

$\large\frac{1}{2}\;mV^2-\large\frac{1}{2}\;mV_{0}^2=-mgl (1-cos \theta) + qEl sin \theta$

$\large\frac{mV^2}{l}=\large\frac{mV_{0}^2}{l}-2mg(1-cos \theta) + 2qE sin \theta$

Putting in equation (1) we get

$T=mg cos \theta +qE sin \theta+\large\frac{mV_{0}^2}{l}-2mg(1-cos \theta) +2qE sin \theta$

$T=3mg cos \theta +3qE sin \theta+\large\frac{mV_{0}^2}{l}-2mg $

For T to be minimum

$\large\frac{d T}{d \theta}=0$ & $\;\large\frac{d^2T}{d \theta ^2} > 0$

Thus $\;\large\frac{dT}{d \theta}=-3 mg sin \theta + 3 qE cos \theta =0$

$tan \theta= \large\frac{qE}{mg}$

$\theta=tan^{-1}\;(\large\frac{qE}{mg})\quad \; or \quad \; \pi+tan^{-1}\;(\large\frac{qE}{mg})$

$\large\frac{d^2 T}{d \theta^2}=-3\;(mg cos \theta+qE sin\theta) > 0$

At $\;\theta=tan^{-1}(\large\frac{qE}{mg})\quad\;\large\frac{d^2T}{d \theta^2} < 0$

At $\;\theta=\pi+tan^{-1}(\large\frac{qE}{mg})\quad\;\large\frac{d^2T}{d \theta^2} > 0$

Thus the desired $\;\theta \;$ at which tension is minimum is $\;\pi+tan^{-1} (\large\frac{qE}{mg})\;.$

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