Two particles of equal mass $m$ have charges $+q$ and $+4q$. If they are accelerated through the same potential difference $V$,the ratio of their velocities $\frac{v_A}{v_B}$ is:

The two plates of a parallel plate capacitor are given charges $Q_1$ and $Q_2$. The capacity of the capacitor is $C$. When the switch $(S)$ is closed,mark the correct statement. (Assume both $Q_1$ and $Q_2$ to be positive)

In the figure below,what is the potential difference between the points $A$ and $B$ and between $B$ and $C$ respectively in steady state?

When two identical capacitors are charged individually to different potentials and then connected in parallel,after disconnecting from the source . . . . . .

When a potential difference of $10^{3} \, V$ is applied between $A$ and $B$, a charge of $0.75 \, mC$ is stored in the circuit. The value of $C$ is (in $\mu F$):

$A$ capacitor $C_1 = 1.0 \, \mu F$ is charged up to a voltage $V = 60 \, V$ by connecting it to a battery $B$ through switch $(1)$. Now,$C_1$ is disconnected from the battery and connected to a circuit consisting of two uncharged capacitors $C_2 = 3.0 \, \mu F$ and $C_3 = 6.0 \, \mu F$ in series through a switch $(2)$ as shown in the figure. The sum of the final charges on $C_2$ and $C_3$ is ...... $\mu C$.