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(C) Let the radius of the air bubble at the bottom be $r_1 = r$ and at the surface be $r_2 = 2r$.Let the height of the water reservoir be $h$.The pres

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$70$

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(C) Let the radius of the air bubble at the bottom be $r_1 = r$ and at the surface be $r_2 = 2r$.Let the height of the water reservoir be $h$.The pressure at the bottom of the reservoir is $P_1 = P_{atm} + h ho g$,where $P_{atm}$ is the atmospheric pressure.Given $P_{atm} = 10 \, m$ of water,we can write $P_1 = (10 + h) ho g$.The pressure at the surface is $P_2 = P_{atm} = 10 ho g$.According to Boyle's law,for a constant temperature,$P_1 V_1 = P_2 V_2$.The volume of the bubble is $V = \frac{4}{3} \pi r^3$.Substituting the values: $(10 + h) ho g 	imes \frac{4}{3} \pi r^3 = 10 ho g 	imes \frac{4}{3} \pi (2r)^3$.$(10 + h) = 10 	imes 8$.$10 + h = 80$.$h = 70 \, m$.

An air bubble doubles its radius on rising from the bottom of a water reservoir to the surface. If the atmospheric pressure is equal to $10 \, m$ of water,the height of the water in the reservoir is ..... $m$.

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