$A$ circuit is connected as shown in the figure with the switch $S$ open. When the switch is closed,the total amount of charge that flows from $Y$ to $X$ is

Two neutral conducting spheres of diameters $8 \ cm$ and $2 \ cm$ separated with a distance of $15 \ cm$ between their centres are joined by a thin conducting wire. $A$ charge of $100 \ nC$ is given to one of the spheres and the system is allowed to reach electrostatic equilibrium. The electric potential at a point on the line joining the centres of the two spheres where the net electric field becomes zero is . . . . . . $V$. (Neglect the charge acquired by the wire and $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \ Nm^2 C^{-2}$)

$A$ radioactive substance in the form of a metallic sphere of diameter $10^{-3} \ m$ emits particles at a constant rate of $6.25 \times 10^{10}$ particles per second. If the sphere is electrically isolated,how long will it take for its potential to increase by $1.0 \ V$? Assume that $80\%$ of the emitted particles escape from the surface. The time taken is $... \mu s$.

In the figure,a system of four capacitors connected across a $10\, V$ battery is shown. The charge that will flow from switch $S$ when it is closed is:

In the circuit shown in the figure,for section $AB$,$C_1 = 1\ \mu F$,$C_2 = 2\ \mu F$,$E = 10\ V$,and the potential difference $V_A - V_B = -10\ V$. Find the charge on capacitor $C_1$.

$27$ identical drops of mercury are charged simultaneously with the same potential of $10 \ V$. Assuming the drops to be spherical,if all the charged drops are made to combine to form one large drop,then its potential will be . . . . . . volt.