The excess pressure in a soap bubble is double that in another one. The ratio of their volumes is .............

If the excess pressure inside a soap bubble of radius $3 \,mm$ is equal to the pressure of a water column of height $0.8 \,cm$, then the surface tension of the soap solution is ( $\rho_{\text{water}} = 1000 \,kg/m^3, g = 9.8 \,m/s^2$ ).

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

Two soap bubbles $A$ and $B$ are kept in a closed chamber where the air is maintained at pressure $8 \ N/m^2$. The radii of bubbles $A$ and $B$ are $2 \ cm$ and $4 \ cm$,respectively. The surface tension of the soap solution is $0.04 \ N/m$. Find the ratio $n_B / n_A$,where $n_A$ and $n_B$ are the number of moles of air in bubbles $A$ and $B$,respectively. [Neglect the effect of gravity.]

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 $g = 10 \ m/s^2$)

$1000$ droplets of water, each having a diameter of $2 \ mm$, coalesce to form a single drop. Given the surface tension of water is $0.072 \ N/m$. The energy loss in the process is