Consider an air bubble of radius 2 ,mm in a l | vedclass

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(B) The pressure inside an air bubble at a depth $h$ is given by $P_{in} = P_{atm} + \rho gh + \frac{2S}{R}$.The pressure at the free surface is $P_{a

Vedclass · Accepted Answer

$600$

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(B) The pressure inside an air bubble at a depth $h$ is given by $P_{in} = P_{atm} + ho gh + \frac{2S}{R}$.The pressure at the free surface is $P_{atm}$.Therefore,the pressure inside the bubble relative to the pressure at the free surface is $\Delta P = P_{in} - P_{atm} = ho gh + \frac{2S}{R}$.Given:Density $ho = 1000 \,kg/m^3$Depth $h = 5 \,cm = 0.05 \,m$Acceleration due to gravity $g = 10 \,m/s^2$Surface tension $S = 0.1 \,N/m$Radius $R = 2 \,mm = 0.002 \,m$Step $1$: Calculate the hydrostatic pressure due to the liquid column:$P_{hydro} = ho gh = 1000 	imes 10 	imes 0.05 = 500 \,Pa$.Step $2$: Calculate the excess pressure due to surface tension:$P_{excess} = \frac{2S}{R} = \frac{2 	imes 0.1}{0.002} = \frac{0.2}{0.002} = 100 \,Pa$.Step $3$: Calculate the total pressure difference:$\Delta P = 500 \,Pa + 100 \,Pa = 600 \,Pa$.

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