A simple pendulum with a bob of mass $m=1\text{kg}$, charge $q=5\mu \text{C}$ and string length $l=1\text{m}$ is given a horizontal velocity $u$ in a uniform electric field $E=2\times {10}^6\text{V/m}$ at its bottom most point $\text{A}$ as shown in figure. It is given that the speed $u$ is such that the string slacks at point $\text{C}$. Find the speed $u$.
[Take $g=10{\text{m/s}}^2$]

The correct option is C $\sqrt{31}\text{m/s}$
Given:
$m=1\text{kg};q=5\mu \text{C}=5\times {10}^{-6}\text{C},$
$l=1\text{m};E=2\times {10}^6\text{V/m}$


Let the speed of the bob at point $\text{C}$ is $v.$

Now apply work-energy theorem between points $\text{A}$ and $\text{C}$.

From work energy theorem,

$\Rightarrow \frac{1}{2}m(v^2-u^2)=-mgl(1+\sin {60}^{\circ })+qEl\cos {60}^{\circ }$

$\Rightarrow \frac{1}{2}\times 1\times (v^2-u^2)=[-1\times 10\times 1(1+\sin {60}^{\circ }\left)\right]+[5\times {10}^{-6}\times 2\times {10}^6\times 1\times \cos {60}^{\circ }]$

$\Rightarrow u^2-v^2=27.3.........\left(1\right)$

Further, at point $\text{C}$, tension in the string will be zero.

Hence,

$\frac{m{v}^2}{l}=mg\sin {60}^{\circ }-qE\cos {60}^{\circ }$

$\Rightarrow \frac{1\times v^2}{1}=1\times 10\times \sin {60}^{\circ }-5\times {10}^{-6}\times 2\times {10}^6\times \cos {60}^{\circ }$

$\Rightarrow v^2=3.7...........\left(2\right)$

From eqs. (1) and (2)

$u=\sqrt{31}\text{m/s}$

Hence, option $\left(c\right)$ is the correct answer.