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

One end of a uniform glass capillary tube of radius $r = 0.025 \ cm$ is immersed vertically in water to a depth $h = 1 \ cm$. The excess pressure in $N/m^2$ required to blow an air bubble out of the tube is: (Surface tension of water $T = 7 \times 10^{-2} \ N/m$,Density of water $\rho = 10^3 \ kg/m^3$,Acceleration due to gravity $g = 10 \ m/s^2$)

The excess pressure inside a bubble in water is $P_1$. The excess pressure inside a drop of the same radius is $P_2$. Then:

Pressure inside two soap bubbles is $1.01 \, atm$ and $1.03 \, atm$. The ratio between their volumes is (Pressure outside the soap bubble is $1 \, atm$). (in $ : 1$)

$A$ big drop is formed by coalescing $1000$ small identical drops of water. If $E_1$ be the total surface energy of $1000$ small drops of water and $E_2$ be the surface energy of the single big drop of water,then the ratio $E_1 : E_2$ is $x : 1$,where $x = . . . . . . $.

Air is pushed into a soap bubble to increase its radius from $R$ to $2R$. In this case, the pressure inside the bubble