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

The energy needed for breaking a liquid drop of radius $R$ into $216$ droplets, each of radius $r$, is $x$ times $TR^2$. The value of $x$ is [$T =$ surface tension of the liquid]. (in $\pi$)

$A$ soap bubble,having a radius of $1\; mm$,is blown from a detergent solution having a surface tension of $2.5 \times 10^{-2}\; N/m$. The pressure inside the bubble is equal to the pressure at a point $Z_{0}$ below the free surface of water in a container. Taking $g=10\; m/s^{2}$ and the density of water $\rho = 10^{3}\; kg/m^{3}$,the value of $Z_{0}$ is......$cm$.

$Assertion :$ Smaller drops of liquid resist deforming forces better than the larger drops.
$Reason :$ Excess pressure inside a drop is directly proportional to its surface area.

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

$64$ small mercury droplets,each of radius $r$ and charge $q$,coalesce to form a single large drop. What is the ratio of the surface charge density of a small droplet to that of the large drop?