A neutral pendulum oscillates in a uniform electric field as shown in figure. If a positive charge is given to the pendulum, then its time period
Assume that $mg>qE$

The correct option is A will increase.
The time period of the simple pendulum is given by,
$T=2\pi \sqrt{\frac{l}{g_{\text{eff}}}}\dots \dots \left(1\right)$
Here, $g_{\text{eff}}$ is the net acceleration and $l$ is length of the string.

Upward force due to electric field and charge $=qE$
So, net force, $F_{\text{net}}=mg-qE$
$\Rightarrow a_{\text{net}}=g-\frac{qE}{m}=g_{\text{eff}}$
Here, $g_{\text{eff}}$ decreases from $g$ to $g-\frac{qE}{m}$. Hence, time period will increase. [from (1)]