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 -

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

$A$ spherical drop of radius $r$ is divided into $8$ equal droplets. If the surface tension is $S$, then the work done in the process will be

$A$ soap bubble $(S.T = 30\, dyne/cm)$ has a radius of $1\, cm$. The work done in doubling its radius would be ........ $ergs$.

$A$ shell having a hole of radius $r$ is dipped in water. It holds the water up to a depth of $h$. Then the value of $r$ is: