The excess pressure inside a soap bubble is three times that of another soap bubble. The ratio of their volumes 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

There is a small hole in a hollow sphere. Water enters the sphere when it is taken to a depth of $40 \, cm$ in water. The diameter of the hole is ....... $mm$ (Surface tension of water $= 0.07 \, N/m$):

Two soap bubbles of radii $r_1$ and $r_2$ in vacuum coalesce under isothermal conditions. The resulting bubble has a radius equal to

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:

The pressures inside two soap bubbles $A$ and $B$ are $1.02 \text{ atm}$ and $1.04 \text{ atm}$ respectively. The ratio of the volume of bubble $A$ to that of bubble $B$ is (outside pressure = $1 \text{ atm}$)