Two droplets merge with each other and form a large droplet. In this process,

Let $R_{1}$ and $R_{2}$ be the radii of two mercury drops. $A$ big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is

Pressure inside two soap bubbles is $1.01 \, atm$ and $1.03 \, atm$. The ratio between their volumes is (Pressure outside the soap bubble is $1 \, atm$). (in $ : 1$)

The excess pressure inside the first soap bubble is three times that of a second soap bubble. What is the ratio of the volumes of the first bubble to the second bubble?

$A$ 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......$cm$.

$64$ 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?