$64$ small drops of mercury,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 drop to that of the large drop?

The work done in splitting a water drop of radius $R$ into $64$ droplets is ($T=$ Surface tension of water). (in $\pi TR^2$)

Consider an air bubble of radius $2 \,mm$ in a liquid at a depth of $5 \,cm$ below the free surface. The density of the liquid is $1000 \,kg/m^3$ and the surface tension is $0.1 \,N/m$. Find the pressure inside the air bubble relative to the pressure at the free surface of the liquid. (Take $g = 10 \,m/s^2$) (in $\,Pa$)

If two soap bubbles of different radii are connected by a tube,then

$A$ capillary type tube $AB$ is connected to a manometer $M$ as shown in the figure. Stopper $S$ controls the flow of air. $AB$ is dipped into a soap solution where surface tension is $T$. On opening the stopper for a while,a bubble is formed at $B$ end of the manometer and the level difference in manometer limbs is $h$. If $\rho$ is the density of the soap solution and $r$ is the radius of curvature of the bubble,then the surface tension $T$ of the liquid is given by:

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