Let $n$ be the number of liquid drops,each with surface energy $E$. These drops join to form a single drop. 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

When two soap bubbles of radius $r_1$ and $r_2$ $(r_2 > r_1)$ coalesce,the radius of curvature of the common surface is:

In a water tank,an air bubble rises from the bottom to the top surface of the water. If the depth of the water in the tank is $7.28 \ m$ and atmospheric pressure is $10 \ m$ of water,then the ratio of the radii of the bubble at the bottom of the tank and at the top surface of the water is (Temperature of the water in the tank is constant).

$A$ big drop of radius $R$ is formed by $1000$ small droplets of water. The radius of each small droplet is:

Two soap bubbles combine to form a single bubble. In this process,the change in volume and surface area are respectively $V$ and $A$. If $P$ is the atmospheric pressure,and $T$ is the surface tension of the soap solution,the following relation is true :