When an air bubble rises from the bottom of lake to the surface, its radius is doubled. The atmospheric pressure is equal to that of a column of water of height ' H '. The depth of the lake

Question

When an air bubble rises from the bottom of lake to the surface, its radius is doubled. The atmospheric pressure is equal to that of a column of water of height ' H '. The depth of the lake is

TestHub · Accepted Answer

Let the radius of the bubble at the bottom be $r$ and at the surface be $2 r$ .

Volume of the bubble at the bottom, $V_{1} = \frac{4}{3} π r^{3}$

Volume of the bubble at the surface, $V_{2} = \frac{4}{3} π (2 r)^{3} = 8 V_{1}$

Let the depth of the lake be $h$ . The atmospheric pressure is given as $P_{2} = ρ g H$ , where $ρ$ is the density of water.

Pressure at the bottom of the lake, $P_{1} = P_{a t m} + ρ g h = ρ g H + ρ g h = ρ g (H + h)$

Assuming the temperature of the lake is constant, we apply Boyle's law:

$P_{1} V_{1} = P_{2} V_{2}$

$ρ g (H + h) \times V_{1} = ρ g H \times 8 V_{1}$

$H + h = 8 H$

$h = 7 H$

Answer: 7 H

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