1. Identify the forces on the bob

The pendulum bob has:

• वजन due to gravity: $mg$ downward
• electric force: $qE$ horizontally

So the bob experiences a constant resultant force which is the vector sum of these two.

2. Find the effective acceleration

The gravitational acceleration is $g$ downward and the electric acceleration is $a_e = \frac{qE}{m}$ horizontally.

Hence the magnitude of the resultant effective acceleration is $g_{\text{eff}} = \sqrt{g^2 + \left(\frac{qE}{m}\right)^2}$

3. Interpret the motion of the pendulum

A pendulum in a constant uniform field oscillates about the direction of the resultant acceleration. For small oscillations, the time period is the same as that of a simple pendulum with $g$ replaced by $g_{\text{eff}}$.

Thus, $T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}}$

Substitute $g_{\text{eff}}$: $T = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + \left(\frac{qE}{m}\right)^2}}}$

or equivalently, $T = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + \frac{q^2E^2}{m^2}}}}$

4. Match with the options

This matches Option C: $2\pi \sqrt {{L \over {\sqrt {{g^2} + {{{q^2}{E^2}} \over {{m^2}}}} }}}$

5. Comparison with stored answer

Stored correct answer is C, which matches our derived answer.