The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is $n$ times the volume of the second,where $n$ is:

$A$ glass tube of uniform internal radius has a valve separating two identical ends. Initially,the valve is in a tightly closed position. End $1$ has a hemispherical soap bubble of radius $r$. End $2$ has a soap bubble with a radius of curvature $R$ $(R > r)$ as shown in the figure. Just after opening the valve,

One end of a uniform glass capillary tube of radius $r = 0.025 \ cm$ is immersed vertically in water to a depth $h = 1 \ cm$. The excess pressure in $N/m^2$ required to blow an air bubble out of the tube is: (Surface tension of water $T = 7 \times 10^{-2} \ N/m$,Density of water $\rho = 10^3 \ kg/m^3$,Acceleration due to gravity $g = 10 \ m/s^2$)

$64$ small drops,each of radius $r$ and charge $q$,coalesce to form a single large drop. If the charge is conserved,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 $\pi TR^2$)

$A$ capillary tube of radius $r$ is dipped in a liquid of density $\rho$ and surface tension $S$. If the angle of contact is $\theta$,what is the pressure difference between the two surfaces in the beaker and the capillary?