The excess pressure inside a first spherical drop of water is three times that of a second spherical drop of water. Then the ratio of the mass of the first spherical drop to that of the second spherical drop is

The excess pressure in a first soap bubble is three times that of another soap bubble. Then the ratio of the volume of the first bubble to the other is:

The excess pressure inside a soap bubble $A$ in air is half the excess pressure inside another soap bubble $B$ in air. If the volume of the bubble $A$ is $n$ times the volume of the bubble $B$,then the value of $n$ is . . . . . . .

An air bubble of radius $r$ in water is at depth $h$ below the water surface. If $P$ is the atmospheric pressure, and $d$ and $T$ are the density and surface tension of water respectively, then the pressure inside the bubble will be:

If the surface tension of a soap solution is $0.03 \, N/m$,then the excess pressure inside a soap bubble of diameter $6 \, mm$ over the atmospheric pressure will be:

The lower end of a capillary tube of diameter $2.00 \; mm$ is dipped $8.00 \; cm$ below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at the temperature of the experiment is $7.30 \times 10^{-2} \; N m^{-1}$. Atmospheric pressure $= 1.01 \times 10^{5} \; Pa$,density of water $= 1000 \; kg m^{-3}$,$g = 9.80 \; m s^{-2}$. Also,calculate the excess pressure.