An air bubble of volume 50, cm^3 is formed at | vedclass

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(B) Let $P_1$ and $V_1$ be the pressure and volume at the bottom of the lake,and $P_2$ and $V_2$ be the pressure and volume at the surface.According t

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$300$

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(B) Let $P_1$ and $V_1$ be the pressure and volume at the bottom of the lake,and $P_2$ and $V_2$ be the pressure and volume at the surface.According to Boyle's Law,$P_1 V_1 = P_2 V_2$.The pressure at the bottom is $P_1 = P_0 + h ho_w g$,where $P_0$ is atmospheric pressure,$h = 47.6\, m = 4760\, cm$,$ho_w = 1\, g/cm^3$,and $g$ is acceleration due to gravity.The atmospheric pressure $P_0 = 70\, cm$ of $Hg = 70 	imes 13.6 	imes g$ (in $dyne/cm^2$).The pressure at the bottom $P_1 = (70 	imes 13.6 	imes g) + (4760 	imes 1 	imes g) = g(952 + 4760) = 5712g$.The pressure at the surface $P_2 = P_0 = 70 	imes 13.6 	imes g = 952g$.Using $P_1 V_1 = P_2 V_2$:$5712g 	imes 50 = 952g 	imes V_2$$V_2 = \frac{5712 	imes 50}{952} = 6 	imes 50 = 300\, cm^3$.

An air bubble of volume $50\, cm^3$ is formed at the bottom of a lake $47.6\, m$ deep. What will be its volume when it reaches the surface? (Atmospheric pressure $= 70\, cm$ of $Hg$ and density of $Hg = 13.6\, g/cm^3$)

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