Pressure inside two soap bubbles are $1.01$ and $1.02$ atmosphere, respectively. The ratio of their volumes is (in $ : 1$)

The surface tension of a soap solution is $25 \times 10^{-3} \, N/m$. The excess pressure inside a soap bubble of diameter $1 \, cm$ is ....... $Pa$.

The excess pressure inside a soap bubble $A$ in air is half the excess pressure inside another soap bubble $B$ in air. If the volume of the bubble $A$ is $n$ times the volume of the bubble $B$,then the value of $n$ is . . . . . . .

An air bubble of radius $r$ in water is at depth $h$ below the water surface. If $P$ is the atmospheric pressure, and $d$ and $T$ are the density and surface tension of water respectively, then the pressure inside the bubble will be:

$A$ big drop of radius $R$ is formed by $1000$ small droplets of water. The radius of each small droplet is:

$A$ spherical drop of water has a radius of $1\, mm$. If the surface tension of water is $70 \times 10^{-3} \, N/m$,then the difference of pressure between the inside and outside of the spherical drop is: