An air bubble doubles in radius when it rises | vedclass

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(C) Assuming the temperature remains constant,we use Boyle's Law: $P_1 V_1 = P_2 V_2$.Let $h$ be the depth of the sea.The pressure at the surface is $

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$70$

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(C) Assuming the temperature remains constant,we use Boyle's Law: $P_1 V_1 = P_2 V_2$.Let $h$ be the depth of the sea.The pressure at the surface is $P_2 = P_{atm} = 10 \, dg$ (where $d$ is the density of water and $g$ is the acceleration due to gravity).The pressure at the bottom is $P_1 = P_{atm} + h \, dg = (10 + h) \, dg$.The volume of the bubble at the bottom is $V_1 = \frac{4}{3} \pi r^3$.The volume of the bubble at the surface is $V_2 = \frac{4}{3} \pi (2r)^3 = 8 	imes \frac{4}{3} \pi r^3 = 8 V_1$.Applying $P_1 V_1 = P_2 V_2$:$(10 + h) \, dg 	imes V_1 = 10 \, dg 	imes 8 V_1$.Dividing both sides by $dg 	imes V_1$:$10 + h = 80$.$h = 80 - 10 = 70 \, m$.

An air bubble doubles in radius when it rises from the bottom of the sea to the surface. If the atmospheric pressure is equal to the pressure exerted by a $10 \, m$ column of water,then the depth of the sea is $... \, m$. (Assume surface tension is negligible.)

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