Due to surface tension, the excess pressure inside a smaller drop is $9$ units. If $27$ smaller drops combine, then the excess pressure inside the bigger drop is:

The radii of two mercury drops are $R_1$ and $R_2$. Under isothermal conditions,a single drop of radius $R$ is formed from them. The relation between $R, R_1$ and $R_2$ is

When a big drop of water is formed from $n$ small drops of water,the energy loss is $3E$,where $E$ is the energy of the bigger drop. The radius of the bigger drop is $R$ and that of the smaller drop is $r$. Then the value of $n$ is:

The pressure inside a soap bubble $A$ is $1.01 \text{ atm}$ and that in a soap bubble $B$ is $1.02 \text{ atm}$. The ratio of the volume of bubble $A$ to that of $B$ is (Surrounding pressure $= 1 \text{ atm}$)

$A$ big liquid drop splits into $n$ similar small drops under isothermal conditions,then in this process

Two soap bubbles of radii $r_1 = 4 \ cm$ and $r_2 = 5 \ cm$ are touching each other over a common surface $S_1S_2$ (as shown in the figure). The radius of curvature of this common surface is...... $cm$.