$A$ proton and an $\alpha$-particle are placed in a uniform electric field. The accelerations produced in them are $a_p$ and $a_\alpha$ respectively. Then $a_p : a_\alpha$ will be . . . . . . .

An electron enters an electric field with its velocity in the direction of the electric lines of force. Then:

An inclined plane making an angle of $30^{\circ}$ with the horizontal is placed in a uniform horizontal electric field of $200 \, N/C$ as shown in the figure. $A$ body of mass $1 \, kg$ and charge $5 \, mC$ is allowed to slide down from rest at a height of $1 \, m$. If the coefficient of friction is $0.2$,find the time (in $s$) taken by the body to reach the bottom. $\left[ g = 9.8 \, m/s^2, \sin 30^{\circ} = 0.5, \cos 30^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866 \right]$

Kinetic energy of an electron accelerated in a potential difference of $100 \, V$ is

$A$ proton is $1840$ times heavier than an electron. When it is accelerated through a potential difference of $1 \ kV$,what will be its kinetic energy in $keV$?

$A$ sphere carrying charge $Q$ and having weight $w$ falls under gravity between a pair of vertical plates at a distance $d$ from each other. When a potential difference $V$ is applied between the plates,the acceleration of the sphere changes as shown in the figure,along the line $BC$. The value of $Q$ is: