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:

Pure water rises through a height $h$ in a capillary tube of internal radius $r$. Surface tension of water is $T$. The pressure difference between the water level in the container and the lowest point of the concave meniscus is

$1000$ small water drops,each of radius $r$ and charge $q$,coalesce together to form one large spherical drop. The potential of the big drop is larger than that of the smaller drop by a factor of:

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 :

$A$ big water drop is divided into $8$ equal droplets. $\Delta P_{S}$ and $\Delta P_{B}$ are the excess pressure inside a smaller and bigger drop respectively. The relation between $\Delta P_{S}$ and $\Delta P_{B}$ is

When two soap bubbles of radii $a$ and $b$ $(b > a)$ coalesce,the radius of curvature of the common surface is: