The excess pressure inside a soap bubble is three times that of another soap bubble. The ratio of their volumes is:

$A$ water drop of $0.01 \ cm^3$ is squeezed between two glass plates and spreads into an area of $10 \ cm^2$. If the surface tension of water is $70 \ dyne/cm$,then the normal force required to separate the glass plates from each other will be: (in $N$)

The excess pressure inside a soap bubble of radius $2 \ cm$ is $50 \ dyne/cm^2$. The surface tension is

Two mercury droplets of radii $0.1 \ cm$ and $0.2 \ cm$ coalesce into one single drop. What amount of energy is released? The surface tension of mercury is $T = 435.5 \times 10^{-3} \ N \ m^{-1}$.

What is the excess pressure inside a bubble of soap solution of radius $5.00 \; mm$,given that the surface tension of soap solution at the temperature $(20 \; ^{\circ}C)$ is $2.50 \times 10^{-2} \; N m^{-1}$? If an air bubble of the same dimension were formed at a depth of $40.0 \; cm$ inside a container containing the soap solution (of relative density $1.20$),what would be the pressure inside the bubble? ($1$ atmospheric pressure is $1.01 \times 10^{5} \; Pa$).

The energy needed for breaking a liquid drop of radius $R$ into $216$ droplets, each of radius $r$, is $x$ times $TR^2$. The value of $x$ is [$T =$ surface tension of the liquid]. (in $\pi$)