The volume of an air bubble is doubled as it | vedclass

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(A) Let $V$ be the initial volume of the air bubble at the bottom of the lake and $2V$ be its volume at the surface.Let $P_0$ be the atmospheric press

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(A) Let $V$ be the initial volume of the air bubble at the bottom of the lake and $2V$ be its volume at the surface.Let $P_0$ be the atmospheric pressure and $h$ be the depth of the lake.According to Boyle's Law,since the temperature is assumed to be constant,$P_1 V_1 = P_2 V_2$.At the surface,the pressure is $P_2 = P_0$.At the bottom,the pressure is $P_1 = P_0 + ho_w g h$,where $ho_w$ is the density of water.Given $P_1 V = P_2 (2V) \Rightarrow P_1 = 2 P_0$.Substituting $P_1$,we get $P_0 + ho_w g h = 2 P_0 \Rightarrow ho_w g h = P_0$.We know $P_0 = h_m ho_m g$,where $h_m = 0.75 \, m$ and $ho_m$ is the density of mercury.So,$ho_w g h = h_m ho_m g \Rightarrow h = h_m \left( \frac{ho_m}{ho_w} ight)$.Given $\frac{ho_m}{ho_w} = \frac{40}{3}$ and $h_m = 0.75 \, m = \frac{3}{4} \, m$.$h = \frac{3}{4} 	imes \frac{40}{3} = 10 \, m$.

The volume of an air bubble is doubled as it rises from the bottom of a lake to its surface. The atmospheric pressure is $75 \, cm$ of mercury. The ratio of the density of mercury to that of lake water is $\frac{40}{3}$. The depth of the lake in metres is:

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