Energy needed in breaking a drop of radius $R$ into $n$ drops of radii $r$ is given by

The excess pressure inside a spherical soap bubble of radius $1 \,cm$ is balanced by a column of oil (specific gravity $= 0.8$), $2 \,mm$ high. The surface tension of the bubble is: (in $\,N/m$)

What should be the radius of a water drop so that the excess pressure inside it is $72 \ Nm^{-2}$ (in $mm$)? (The surface tension of water is $7.2 \times 10^{-2} \ Nm^{-1}$)

$A$ spherical soap bubble inside an air chamber at pressure $P_0 = 10^5 \ Pa$ has a certain radius $R$ such that the excess pressure inside the bubble is $\Delta P = 144 \ Pa$. Now,the chamber pressure is reduced to $8P_0 / 27$ so that the bubble radius and its excess pressure change. In this process,all temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure $\Delta P$ in both cases to be much smaller than the chamber pressure. The new excess pressure $\Delta P'$ in $Pa$ is:

In a cylinder provided with a piston, air is under pressure $P_1$ at a constant temperature $t$. $A$ soap bubble with radius $r$ and surface tension $T$ is lying inside the cylinder. To reduce the radius of the soap bubble to half, the required air pressure inside the cylinder is

$A$ soap bubble in vacuum has a radius $3 \, cm$ and another soap bubble in vacuum has radius $4 \, cm$. If two bubbles coalesce under isothermal conditions,then the radius of the new bubble will be .............. $cm$.