There are two conducting spheres of radii $10 \ cm$ and $20 \ cm$. One sphere is given a charge of $150 \ \mu C$ and is connected to the other sphere by a wire. What will be their common potential?

In the circuit shown in the figure,initially $K_1$ is closed and $K_2$ is open. What are the charges on each capacitor? Then $K_1$ was opened and $K_2$ was closed (order is important),what will be the charge on each capacitor now? [Given: $C = 1 \,\mu F$,$C_1 = 6C$,$C_2 = 3C$,$C_3 = 3C$,$E = 9 \, V$]

$A$ 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 $200 \text{ pF}$)

$A$ $5\, \mu F$ capacitor is charged fully by a $220\,V$ supply. It is then disconnected from the supply and is connected in series to another uncharged $2.5\, \mu F$ capacitor. If the energy change during the charge redistribution is $\frac{ X }{100} \, J$,then the value of $X$ to the nearest integer is $.....$

The circuit shows two capacitors $A$ and $B$ of capacitances $C$ and $2C$ respectively. When they are fully charged,the cell is removed and the capacitors are connected with their plates of opposite polarities touching each other. Then:
$(a)$ Charge on $A$ is $\frac{4CE}{9}$
$(b)$ Charge on $B$ is $\frac{8CE}{9}$
$(c)$ Loss of energy in this process is $\frac{CE^2}{3}$
The correct statement$(s)$ is/are:

Two hollow metallic spheres of radii $20 \ cm$ and $10 \ cm$ each carry a charge of $150 \ \mu C$. If they are connected by a conducting wire,the common potential is ...... $V$.