The bob of a simple pendulum possesses a mass of 40 g and a...

The initial time period of the simple pendulum is given by $T_1 = 2\pi \sqrt{\dfrac{L}{g}}$.

When the vertical upward electric field is switched on, the electric force acts upwards. The net downward acceleration of the bob becomes $g_{eff} = g - \dfrac{qE}{m}$.

Given $m = 40 \times 10^{-3} \text{ kg}$, $q = 4.0 \times 10^{-6} \text{ C}$, $E = 2.5 \times 10^4 \text{ N C}^{-1}$, and taking $g = 10 \text{ m s}^{-2}$:

$\dfrac{qE}{m} = \dfrac{4.0 \times 10^{-6} \times 2.5 \times 10^4}{40 \times 10^{-3}} = \dfrac{0.1}{0.04} = 2.5 \text{ m s}^{-2}$

$g_{eff} = 10 - 2.5 = 7.5 \text{ m s}^{-2}$

The new time period is $T_2 = 2\pi \sqrt{\dfrac{L}{g_{eff}}}$.

The ratio of the time taken for $20$ oscillations is equal to the ratio of their time periods:

$\dfrac{t_2}{t_1} = \dfrac{T_2}{T_1} = \sqrt{\dfrac{g}{g_{eff}}} = \sqrt{\dfrac{10}{7.5}} = \sqrt{\dfrac{4}{3}} = \dfrac{2}{\sqrt{3}}$

Substituting $t_1 = 45 \text{ s}$:

$t_2 = 45 \times \dfrac{2}{\sqrt{3}} = 30\sqrt{3} \approx 30 \times 1.732 = 51.96 \text{ s}$

Rounding to the nearest integer, we get $t_2 \approx 52 \text{ s}$.