The surface tension and vapour pressure of a liquid at $25^{\circ} C$ are $8 \times 10^{-2} \text{ Nm}^{-1}$ and $2.5 \times 10^3 \text{ Pa}$ respectively. The radius of the smallest spherical water droplet which can form without evaporating at $25^{\circ} C$ is (in $\mu m$)

The excess pressure in a soap bubble is thrice that in another one. Then the ratio of their volume is

$A$ glass capillary tube of inner diameter $0.28 \ mm$ is lowered vertically into water in a vessel. The pressure to be applied on the water in the capillary tube so that the water level in the tube is the same as that in the vessel (in $N/m^2$) is:
Surface tension of water $= 0.07 \ N/m$
Atmospheric pressure $= 10^5 \ N/m^2$

The pressure of air in a soap bubble of $0.7 \ cm$ diameter is $8 \ mm$ of water above the pressure outside. The surface tension of the soap solution is ........ $dyne/cm$.

$A$ soap bubble is blown with the help of a mechanical pump at the mouth of a tube. The pump produces a constant increase per minute in the volume of the bubble,irrespective of its internal pressure. The graph between the pressure inside the soap bubble and time $t$ will be-

$A$ spherical drop of water has a radius of $1\, mm$. If the surface tension of water is $70 \times 10^{-3} \, N/m$,then the difference of pressure between the inside and outside of the spherical drop is: