If two soap bubbles each of radius $2 \ cm$ combine in vacuum under isothermal conditions,then the radius of the new bubble formed is

If the internal pressures of two soap bubbles are $1.01 \text{ atm}$ and $1.02 \text{ atm}$ respectively, find the ratio of their volumes.

Consider an air bubble of radius $2 \,mm$ in a liquid at a depth of $5 \,cm$ below the free surface. The density of the liquid is $1000 \,kg/m^3$ and the surface tension is $0.1 \,N/m$. Find the pressure inside the air bubble relative to the pressure at the free surface of the liquid. (Take $g = 10 \,m/s^2$) (in $\,Pa$)

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 $1.01 \times 10^{5} \; Pa$).

$A$ spherical soap bubble has internal pressure $P_0$ and radius $r_0$ and is in equilibrium in an enclosure with pressure $P_1 = \frac{8P_0}{9}$. The enclosure is gradually evacuated. Assuming temperature and surface tension of the soap bubble to be fixed, find the value of $\frac{\text{final radius}}{\text{initial radius}}$ of the soap bubble.

The work done in splitting a water drop of radius $R$ into $64$ droplets is ($T=$ Surface tension of water). (in $\pi TR^2$)