$A$ 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:

The excess pressure inside an air bubble of radius $r$ just below the surface of water is $p_1$. The excess pressure inside a drop of the same radius just outside the surface is $p_2$. If $T$ is surface tension,then

$A$ liquid column of height $0.04 \,cm$ balances excess pressure of a soap bubble of a certain radius. If the density of the liquid is $8 \times 10^3 \,kg \,m^{-3}$ and the surface tension of the soap solution is $0.28 \,N \,m^{-1}$, then the diameter of the soap bubble is . . . . . . $cm$.
$(g = 10 \,m \,s^{-2})$

Two soap bubbles of radius $2 \ cm$ and $4 \ cm$,respectively,are in contact with each other. The radius of curvature of the common surface,in $cm$,is . . . . . . .

$Assertion :$ 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.

Which of the following statements are true when two water drops coalesce to form a bigger drop?