When a large bubble rises from bottom of a water lake to its surface, then its radius doubles. If the atmospheric pressure is equal to the pressure of height H of a certain water column, then the dept

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When a large bubble rises from bottom of a water lake to its surface, then its radius doubles. If the atmospheric pressure is equal to the pressure of height H of a certain water column, then the depth of lake will be

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When a large bubble rises from bottom of water lake to its surface, then its radius becomes double.
i.e., $r_{2} = 2 r_{1}$
Since, volume $V \propto r^{3}$
$∴ \frac{V_{2}}{V_{1}} = {(\frac{r_{2}}{r_{1}})}^{3} = (\frac{2 r_{1}}{r_{1}}) = 8$
$\Rightarrow V_{2} = 8 V_{1}$ … (i)
From Boyle’s law,
$p_{1} V_{1} = p_{2} V_{2}$
Where, $p_{2}$ = atmospheric pressure.
$p_{1} V_{1} = p_{2} 8 V_{1}$ [from Eq. (ii)]
$p_{2} = \frac{p_{1}}{8}$
where, $p_{2} =$ atmospheric pressure
and $p_{1}$ = pressure at depth d.
$∴ p_{1} = p_{a t o m s p h e r i c} + p g d$
$p_{1} = p_{2} + p g d$ … (iii)
From Eqs. (ii) and (iii), we have
$p_{2} = \frac{p_{2} + p g d}{8}$
$7 p_{2} = p g d$
$7 . p g H = p g d$
$7 H = d \Rightarrow d = 7 H$

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