An air bubble doubles in radius when it rises from the bottom of the sea to the surface. If the atmospheric pressure is equal to the pressure exerted by a $10 \, m$ column of water,then the depth of the sea is $... \, m$. (Assume surface tension is negligible.)

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:

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	$(ii)$ $\frac{2T}{R}$
	$(iii)$ $\frac{2R}{T}$

What is the excess pressure inside a bubble of soap solution of radius $5.00 \; mm$,given that the surface tension of soap solution at the temperature $(20 \; ^{\circ}C)$ is $2.50 \times 10^{-2} \; N m^{-1}$? If an air bubble of the same dimension were formed at a depth of $40.0 \; cm$ inside a container containing the soap solution (of relative density $1.20$),what would be the pressure inside the bubble? ($1$ atmospheric pressure is $1.01 \times 10^{5} \; Pa$).

If the shape of the liquid surface is curved, then

$A$ glass capillary tube of inner diameter $0.28 \ mm$ is lowered vertically into water in a vessel. The pressure to be applied on the water in the capillary tube so that the water level in the tube is the same as that in the vessel (in $N/m^2$) is:
Surface tension of water $= 0.07 \ N/m$
Atmospheric pressure $= 10^5 \ N/m^2$