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?

$A$ 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

When a large bubble rises from the bottom of a lake to the surface,its radius doubles. If atmospheric pressure is equal to that of a column of water of height $H$,then the depth of the lake is:

$A$ water drop of radius '$r$' and volume '$V$' is kept between two identical glass plates such that it forms a thin layer of area '$A$' between the plates. $A$ force '$F$' is applied such that the two plates separate from each other. The surface tension '$T$' of the liquid is:

Two soap bubbles of radii $3r$ and $4r$ are in contact with each other. The radius of curvature of the interface between the bubbles is:

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: