An air bubble of radius r in water is at dept | vedclass

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(B) The pressure at a depth $h$ in a liquid is given by $P_{liquid} = P + hdg$, where $P$ is the atmospheric pressure, $d$ is the density of the liqui

Vedclass · Accepted Answer

$P + hdg + (2T/r)$

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(B) The pressure at a depth $h$ in a liquid is given by $P_{liquid} = P + hdg$, where $P$ is the atmospheric pressure, $d$ is the density of the liquid, and $g$ is the acceleration due to gravity.An air bubble in water has only one surface in contact with the liquid. The excess pressure inside an air bubble of radius $r$ is given by $\Delta P = \frac{2T}{r}$, where $T$ is the surface tension.Therefore, the total pressure inside the bubble is the sum of the pressure of the surrounding liquid and the excess pressure due to surface tension:$P_{in} = P_{liquid} + \Delta P$$P_{in} = P + hdg + \frac{2T}{r}$

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:

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