One end of a silk thread of length 10 cm is attached to the large...

In equilibrium, the forces acting on the ball are the tension in the thread $T$, the gravitational force $mg$, and the horizontal electric force $qE$.

From the equilibrium condition, the effective acceleration due to gravity $g_{\text{eff}}$ is the vector sum of the acceleration due to gravity $g$ and the acceleration due to the electric field $\dfrac{qE}{m}$.

The angle made by the thread with the vertical is \theta = 60^{}^{{}^{\circ}}. The effective gravity is given by:

g_{\text{eff}} = \dfrac{g}{\cos \theta} = \dfrac{g}{\cos 60^{}^{{}^{\circ}}} = 2g

The time period of small oscillations for the pendulum is:

$T = 2\pi \sqrt{\dfrac{l}{g_{\text{eff}}}}$

Substituting the given values $l = 10 \text{ cm} = 0.1 \text{ m}$ and $g = 9.8 \text{ m/s}^2$:

$T = 2\pi \sqrt{\dfrac{0.1}{2 \times 9.8}} = 2\pi \sqrt{\dfrac{0.1}{19.6}} = 2\pi \sqrt{\dfrac{1}{196}}$

$T = 2\pi \left(\dfrac{1}{14}\right) = \dfrac{\pi}{7} \text{ s}$

Using $\pi \approx 3.14$:

$T \approx \dfrac{3.14}{7} \approx 0.45 \text{ s}$