The excess pressure inside a soap bubble of radius $2.5 \text{ cm}$ is $48 \text{ dyne/cm}^2$. The surface tension of the soap solution in $\text{dyne/cm}$ is:

Two water drops each of radius $r$ coalesce to form a bigger drop. If $T$ is the surface tension,the surface energy released in this process is

$A$ spherical soap bubble has internal pressure $P_0$ and radius $r_0$ and is in equilibrium in an enclosure with pressure $P_1 = \frac{8P_0}{9}$. The enclosure is gradually evacuated. Assuming temperature and surface tension of the soap bubble to be fixed, find the value of $\frac{\text{final radius}}{\text{initial radius}}$ of the soap bubble.

$A$ container,whose bottom has round holes with diameter $0.1 \ mm$ is filled with water. The maximum height in $cm$ up to which water can be filled without leakage will be ........ $cm$. (Surface tension $= 75 \times 10^{-3} \ N/m$ and $g = 10 \ m/s^2$)

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:

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