Pressure inside a soap bubble is greater than the pressure outside by an amount: (given: $R =$ Radius of bubble,$S =$ Surface tension of 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:

Formation of bubbles is in Column-$I$ and the pressure difference between them is given in Column-$II$. Match them appropriately.

Column-$I$	Column-$II$
$(a)$ Liquid drop in air	$(i)$ $\frac{4T}{R}$
$(b)$ Bubble of liquid in air	$(ii)$ $\frac{2T}{R}$
	$(iii)$ $\frac{2R}{T}$

Assume that a drop of liquid evaporates by a decrease in its surface energy,such that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is $T$,the density of the liquid is $\rho$,and $L$ is its latent heat of vaporization.

The pressure inside a small air bubble of radius $0.1 \, mm$ situated just below the surface of water will be equal to. [Take surface tension of water $T = 70 \times 10^{-3} \, N/m$ and atmospheric pressure $P_0 = 1.013 \times 10^5 \, N/m^2$]

$A$ cylinder with a movable piston contains air under a pressure $p_1$ and a soap bubble of radius $r$. The pressure $p_2$ to which the air should be compressed by slowly pushing the piston into the cylinder for the soap bubble to reduce its size by half will be: (The surface tension is $\sigma$, and the temperature $T$ is maintained constant)