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

The correct option is P + hdg + 2T/r.

Let us determine the pressure inside the air bubble step-by-step:
1. Pressure outside the bubble at depth h:
The air bubble is inside the water at a depth $h$ below the surface. The pressure at the water surface is the atmospheric pressure $P$. Therefore, the hydrostatic pressure of the water column of height $h$ above the bubble is $hdg$, where $d$ is the density of water and $g$ is the acceleration due to gravity.
So, the total pressure outside the bubble in the water at depth $h$ is:
$P_{\text{out}}=P+hdg$

2. Excess pressure inside the air bubble:
An air bubble in water has only one spherical liquid-gas interface (unlike a soap bubble in air, which has two surfaces). Due to surface tension $T$, the pressure inside a spherical bubble of radius $r$ is greater than the pressure outside it. The excess pressure $\Delta P$ is given by:
$\Delta P=P_{\text{in}}-P_{\text{out}}=\frac{2T}{r}$

3. Total pressure inside the bubble:
By substituting $P_{\text{out}}$ into the excess pressure equation, we get the absolute pressure inside the bubble $P_{\text{in}}$:
$P_{\text{in}}=P_{\text{out}}+\frac{2T}{r}$
$P_{\text{in}}=P+hdg+\frac{2T}{r}$