The excess pressure inside a soap bubble of radius $2 \ cm$ is $50 \ dyne/cm^2$. The surface tension is

$64$ small mercury droplets,each of radius $r$ and charge $q$,coalesce to form a single large drop. What is the ratio of the surface charge density of a small droplet to that of the large drop?

Two water drops each of radius $r$ coalesce to form a bigger drop. If $T$ is the surface tension,the surface energy released in this process is

The excess pressure inside an air bubble of radius $r$ just below the surface of water is $p_1$. The excess pressure inside a drop of the same radius just outside the surface is $p_2$. If $T$ is surface tension,then

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

$A$ 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: