Air is pushed into a soap bubble to increase its radius from $R$ to $2R$. In this case, the pressure inside the bubble

Two small drops of mercury,each of radius $R$,coalesce to form a single large drop. The ratio of the total surface energies before and after the change is:

If the radius of a soap bubble is four times that of another,then the ratio of their excess pressures will be

The energy needed for breaking a liquid drop of radius $R$ into $n$ droplets each of radius $r$ is (where $T$ is the surface tension of the liquid).

The excess pressure inside a soap bubble of radius $0.5 \ cm$ is balanced by the pressure due to an oil column of height $4 \ mm$. If the density of the oil is $900 \ kg \ m^{-3}$, then the surface tension of the soap solution is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)

If the internal pressures of two soap bubbles are $1.01 \text{ atm}$ and $1.02 \text{ atm}$ respectively, find the ratio of their volumes.