$A$ 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:

The excess pressure in a first soap bubble is three times that of another soap bubble. Then the ratio of the volume of the first bubble to the other is:

$A$ water drop is divided into $8$ equal droplets. The pressure difference between the inner and outer side of the big drop will be

$A$ vessel having a small hole in the bottom must hold water without leakage when water is poured to a height of $7 \text{ cm}$. What is the radius of the hole (in $\text{ mm}$)? [Surface tension of water is $0.07 \text{ N/m}$, angle of contact is $0^{\circ}$, and $g = 10 \text{ m/s}^2$]

An air bubble of radius $0.1 \ cm$ lies at a depth of $20 \ cm$ below the free surface of a liquid of density $1000 \ kg/m^3$. If the pressure inside the bubble is $2100 \ N/m^2$ greater than the atmospheric pressure,then the surface tension of the liquid in $SI$ unit is (use $g=10 \ m/s^2$)

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: