Two soap bubbles of radii $r_1$ and $r_2$ coalesce in vacuum under isothermal conditions to form a bigger bubble of radius $R$. What is the radius of the bigger bubble?

$A$ container,whose bottom has round holes with diameter $0.1 \ mm$ is filled with water. The maximum height in $cm$ up to which water can be filled without leakage will be ........ $cm$. (Surface tension $= 75 \times 10^{-3} \ N/m$ and $g = 10 \ m/s^2$)

$A$ spherical liquid drop of radius $R$ is divided into eight equal droplets. If surface tension is $T$,then the work done in this process will be

Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is . . . . . . . ($S$ is the surface tension of mercury). (in $\pi r^2 S$)

If a soap bubble expands, the pressure inside the bubble:

The excess pressure inside a bubble in water is $P_1$. The excess pressure inside a drop of the same radius is $P_2$. Then: