If two soap bubbles of equal radii $r$ coalesce,then the radius of curvature of the interface between the two bubbles will be:

When a big drop of water is formed from $n$ small drops of water, the energy loss is $3E$, where $E$ is the energy of the bigger drop. If $R$ is the radius of the bigger drop and $r$ is the radius of the smaller drop, then the number of smaller drops $(n)$ is:

$64$ small drops of mercury,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 drop to that of the large drop?

Under isothermal conditions, two soap bubbles of radii $a$ and $b$ coalesce to form a single bubble of radius $c$. If the external pressure is $P$, then the surface tension of the bubbles is:

$A$ shell having a hole of radius $r$ is dipped in water. It holds the water up to a depth of $h$. Then the value of $r$ is:

$64$ small drops,each of radius $r$ and charge $q$,coalesce to form a single large drop. If the charge is conserved,what is the ratio of the surface charge density of a small drop to that of the large drop?