The excess pressure inside a spherical water drop $A$ is four times that of another water drop $B$. Then, the ratio of the mass of water drop $A$ to that of drop $B$ is

$A$ big drop is formed by coalescing $1000$ small droplets of water. The surface energy will become :

$A$ 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:

$A$ large number of liquid drops each of radius $r$ coalesce to form a single drop of radius $R$. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given,surface tension of liquid $T$,density $\rho$):

$A$ soap bubble is in the form of a circular tube having a radius of curvature $R$ and a radius of curvature perpendicular to it of $5R$. Find the excess pressure in the bubble.

$A$ spherical liquid drop of radius $R$ is divided into eight equal droplets. If surface tension is $T$,then the work done in this process will be