Two spherical soap bubbles of radii $r_{1}$ and $r_{2}$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to -

An air bubble of radius $1 \ mm$ is at a depth of $8 \ cm$ below the free surface of a liquid column. If the surface tension and density of the liquid are $0.1 \ N \ m^{-1}$ and $2000 \ kg \ m^{-3}$ respectively,by what amount is the pressure inside the bubble greater than the atmospheric pressure (in $N \ m^{-2}$)? (Take $g = 10 \ m \ s^{-2}$)

$A$ vessel having a small hole in the bottom must hold water without leakage when water is poured to a height of $7 \text{ cm}$. What is the radius of the hole (in $\text{ mm}$)? [Surface tension of water is $0.07 \text{ N/m}$, angle of contact is $0^{\circ}$, and $g = 10 \text{ m/s}^2$]

When two small bubbles join to form a bigger one,energy is

$A$ long cylindrical glass vessel has a pinhole of diameter $0.2 \,mm$ at its bottom. The depth to which the vessel can be lowered vertically in a deep water bath without the water entering into the vessel is (surface tension of water,$T=0.07 \,N/m$,acceleration due to gravity,$g=10 \,m/s^2$) (in $\,cm$)

The radii of two mercury drops are $R_1$ and $R_2$. Under isothermal conditions,a single drop of radius $R$ is formed from them. The relation between $R, R_1$ and $R_2$ is