If a section of a soap bubble of radius $r$ by a plane through its centre is considered,the force on the half due to surface tension is

$A$ drop of water with a volume of $0.05 \ cm^3$ is pressed between two glass plates,causing it to spread between the plates. The area of contact with each plate is $40 \ cm^2$. If the surface tension of water is $70 \ dyne/cm$,the minimum normal force required to separate the two glass plates in newtons is approximately: (assuming the angle of contact is zero)

The work done in breaking a drop of liquid of radius $R$ (Surface tension $T$) into $64$ equal drops is

$A$ 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?

An air bubble rises from the bottom to the top of a water tank in which the temperature of the water is uniform. The surface area of the bubble at the top of the tank is $125 \%$ more than its surface area at the bottom of the tank. If the atmospheric pressure is equal to the pressure of $10 \ m$ water column, then the depth of water in the tank is (in $m$)

$64$ 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?