In the circuit shown in the figure,$K_1$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now,the key $K_1$ is closed and at steady state,the charge on $C$ is $q_2$. The ratio of charges $q_1/q_2$ is

$125$ identical drops,each charged to a potential of $50\;V$,are combined to form a single drop. The potential of the new drop will be......$V$.

Calculate the charge on the capacitor in the steady state. (in $\mu C$)

Two capacitors of capacities $1 \mu F$ and $C \mu F$ are connected in series and the combination is charged to a potential difference of $120 \ V$. If the charge on the combination is $80 \mu C$,the energy stored in the capacitor of capacity $C$ in $\mu J$ is

Four identical thin, square metal sheets, $S_1, S_2, S_3$, and $S_4$, each of side $a$ are kept parallel to each other with equal distance $d( < < a)$ between them, as shown in the figure. Let $C_0 = \varepsilon_0 a^2 / d$, where $\varepsilon_0$ is the permittivity of free space.
Match the quantities mentioned in $List-I$ with their values in $List-II$ and choose the correct option.

$List-I$	$List-II$
$(P)$ The capacitance between $S_1$ and $S_4$, with $S_2$ and $S_3$ not connected, is	$(1)$ $3 C_0$
$(Q)$ The capacitance between $S_1$ and $S_4$, with $S_2$ shorted to $S_3$, is	$(2)$ $C_0 / 2$
$(R)$ The capacitance between $S_1$ and $S_3$, with $S_2$ shorted to $S_4$, is	$(3)$ $C_0 / 3$
$(S)$ The capacitance between $S_1$ and $S_2$, with $S_3$ shorted to $S_1$, and $S_2$ shorted to $S_4$, is	$(4)$ $2 C_0 / 3$
	$(5)$ $2 C_0$

Initially $n$ identical capacitors are joined in parallel and are charged to potential $V$. Now they are separated and joined in series. Then