The energy needed for breaking a liquid drop of radius $R$ into $216$ droplets, each of radius $r$, is $x$ times $TR^2$. The value of $x$ is [$T =$ surface tension of the liquid]. (in $\pi$)

The excess pressure inside the first soap bubble is three times that inside the second bubble. Then,the ratio of the volume of the first to the second bubble will be:

$A$ 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

The excess pressure due to surface tension in a spherical liquid drop of radius $r$ is directly proportional to

$A$ small soap bubble of radius $4 \ cm$ is trapped inside another bubble of radius $6 \ cm$ without any contact. Let $P_2$ be the pressure inside the inner bubble and $P_0$ be the pressure outside the outer bubble. The radius of another bubble with a pressure difference $P_2 - P_0$ between its inside and outside would be....... $cm$.

There is an air bubble of radius $1.0\,mm$ in a liquid of surface tension $0.075\,N\,m^{-1}$ and density $1000\,kg\,m^{-3}$ at a depth of $10\,cm$ below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is $....Pa \left( g = 10\,m\,s^{-2} \right)$