If $n$ drops, each of capacitance $C$, coalesce to form a single big drop, then the ratio of the energy stored in the big drop to that in each small drop will be

In the circuit shown,$C_1 = 6\,\mu F, C_2 = 3 \,\mu F$ and battery $B = 20\,V$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge (in $\mu C$) finally on $C_2$?

$A$ capacitor of capacitance $20 \mu F$ is charged by a battery of potential $24.3 \ V$. The capacitor is then disconnected from the battery and is connected to another uncharged capacitor of capacitance $10 \mu F$. After some time,the second capacitor is disconnected,discharged fully,and is again connected to the first capacitor. If the process is repeated several times,the charge on the first capacitor at the end of the fifth process is . . . . . . $\mu C$.

$A$ sphere of capacitance $100 \text{ pF}$ is charged to a potential of $100 \text{ V}$. Another identical uncharged metal sphere is brought in contact with the charged sphere,then the change in the total energy stored on these spheres,when they touch is $\alpha \times 10^{-7} \text{ J}$. The value of $\alpha$ is . . . . . . . (combined capacitance of spheres is $200 \text{ pF}$)

$A$ capacitor of capacitance $4 \mu F$ is charged to a potential difference of $6 \ V$ with a battery. The battery is then removed and in its place another capacitor of capacitance $8 \mu F$ is introduced and the circuit is closed. The potential difference attained by each of the capacitors in $V$ is

Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals $a$ and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using a battery of $emf$ $V$ volt. Now,disconnecting $a$ and $b$,the terminals $b$ and $c$ are connected. Due to this,what will be the percentage loss of energy?