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

Consider a water droplet of diameter $0.2 \,mm$ where the outside pressure is $1.5 \,N / cm^2$ at $25^{\circ} C$. The pressure inside the droplet, when the surface tension at $25^{\circ} C$ is $0.08 \,N / m$ is

$A$ 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

An air bubble of radius $0.1\, cm$ is in a liquid having surface tension $0.06\, N/m$ and density $10^3\, kg/m^3$. The pressure inside the bubble is $1100\, N/m^2$ greater than the atmospheric pressure. At what depth $h$ (in $m$) is the bubble below the surface of the liquid? $(g = 9.8\, m/s^2)$

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 $\pi r^2 S$)

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: