Radius of an air bubble at the bottom of the lake is $r$ and it becomes $ 2r $ when the air bubbles rises to the top surface of the lake. If $P $ $cm$ of water be the atmospheric pressure, then the depth of the lake is
Diffcult
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Initially the radius of the bubble is r. After reaching surface it becomes $2 r$. The atmospheric pressure is given as,
$P _{ atm }= P cm$ of water
What we can conclude from this process is that the volume is changing in the air bubble but the temperature remains unchanged.
For isothermal process,
$P _{1} V _{1}= P _{2} V _{2}$
Let the height of water surface be $x$.
$(P d g+x d g)\left(\frac{4}{3} \pi r^{3}\right)=P d g\left[\frac{4}{3} \pi(2 r)^{3}\right]$
$(P+x) r^{3}=P\left(8 r^{3}\right)$
$x=8 P-P$
$\Rightarrow x=7 P$
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