$A$ ball of mass $1 \text{ g}$ having a charge of $20 \mu\text{C}$ is tied to one end of a string of length $0.9 \text{ m}$ and can rotate in a vertical plane in a uniform electric field of $100 \text{ NC}^{-1}$ directed upwards. The minimum horizontal velocity that must be given to the ball at the lowest position so that it completes the vertical circle is (Let $g = 10 \text{ ms}^{-2}$): (in $\text{ ms}^{-1}$)

An electron of mass $m$ and charge $e$ is accelerated from rest through a potential difference $V$ in vacuum. The final speed of the electron will be

$A$ liquid drop having $6$ excess electrons is kept stationary under a uniform electric field of $25.5 \times 10^3 \, Vm^{-1}$. The density of the liquid is $1.26 \times 10^3 \, kg \, m^{-3}$. The radius of the drop is (neglect buoyancy):

$A$ simple pendulum of length $L$ is placed between the plates of a parallel plate capacitor having an electric field $E,$ as shown in the figure. Its bob has mass $m$ and charge $q.$ The time period of the pendulum is given by:

An electron falls through a distance of $1.5 \, cm$ in a uniform electric field of magnitude $2.0 \times 10^4 \, N/C$ as shown in the figure. The time taken by the electron to fall through this distance is ($m_e = 9.1 \times 10^{-31} \, kg$,neglect gravity).

$A$ particle of charge $q$ and mass $m$ is projected from origin with an initial velocity $\vec{v} = (\frac{v_0}{\sqrt{2}}\hat{i} + \frac{v_0}{\sqrt{2}}\hat{j})$. There exists a uniform magnetic field $\vec{B} = B_0\hat{z}$ and a space varying electric field $\vec{E} = E_0 e^{-\lambda x}\hat{x}$ within the region $0 \leq x \leq L$. After travelling a distance such that $x$-coordinate has changed from $x = 0$ to $x = L$, the change in the kinetic energy is . . . . . . .