When a mercury drop of radius $R$ breaks into $n$ droplets of equal size,the radius $r$ of each droplet is

Two soap bubbles having radii $r_1$ and $r_2$ have inside pressures $P_1$ and $P_2$ respectively. If $P_0$ is the external pressure,then the ratio of their volumes is:

The excess pressure inside a soap bubble of radius $2.5 \text{ cm}$ is $48 \text{ dyne/cm}^2$. The surface tension of the soap solution in $\text{dyne/cm}$ is:

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

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.

The pressure of air in a soap bubble of $0.7 \ cm$ diameter is $8 \ mm$ of water above the pressure outside. The surface tension of the soap solution is ........ $dyne/cm$.