There is an air bubble of radius 1.0,mm in a | vedclass

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(A) The pressure inside an air bubble at a depth $h$ in a liquid is given by the formula:$P = P_0 + h \rho g + \frac{2T}{r}$where $P_0$ is the atmosph

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(A) The pressure inside an air bubble at a depth $h$ in a liquid is given by the formula:$P = P_0 + h ho g + \frac{2T}{r}$where $P_0$ is the atmospheric pressure,$h$ is the depth,$ho$ is the density of the liquid,$g$ is the acceleration due to gravity,$T$ is the surface tension,and $r$ is the radius of the bubble.We need to find the excess pressure inside the bubble over the atmospheric pressure,which is $P - P_0 = h ho g + \frac{2T}{r}$.Given values:$h = 10\,cm = 0.1\,m$$ho = 1000\,kg\,m^{-3}$$g = 10\,m\,s^{-2}$$T = 0.075\,N\,m^{-1}$$r = 1.0\,mm = 10^{-3}\,m$Substituting these values into the equation:$P - P_0 = (0.1 	imes 1000 	imes 10) + \frac{2 	imes 0.075}{10^{-3}}$$P - P_0 = 1000 + \frac{0.15}{10^{-3}}$$P - P_0 = 1000 + 150 = 1150\,Pa$.

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