$A$ bubble of $8$ mm diameter is formed in the air. The surface tension of soap solution is $30$ dynes/cm. The excess pressure inside the bubble is ........ $dynes/cm^2$.

$Assertion :$ $A$ bubble comes from the bottom of a lake to the top.
$Reason :$ Its radius increases.

What is the ratio of the number of moles of air in two soap bubbles of radii $R_1$ and $R_2$?

An air bubble of radius $1 \ mm$ is at a depth of $8 \ cm$ below the free surface of a liquid column. If the surface tension and density of the liquid are $0.1 \ N \ m^{-1}$ and $2000 \ kg \ m^{-3}$ respectively, by what amount is the pressure inside the bubble greater than the atmospheric pressure (in $N \ m^{-2}$)? (Take $g = 10 \ m \ s^{-2}$)

Derive the formula for the excess of pressure (pressure difference) inside a liquid drop and a soap bubble.

$A$ big drop of radius $R$ is formed by $729$ small drops of water of radius $r$. Then the radius of each small drop will be: