When an air bubble of radius $r$ rises from the bottom to the surface of a lake. Its radius becomes $\frac{5…
When an air bubble of radius $r$ rises from the bottom to the surface of a lake. Its radius becomes $\frac{5 r}{4}$ (the pressure of the atmosphere is equal to the $10 \mathrm{~m}$ height to water column). If the temperature is constant and the surface tension is neglected the depth of the lake is
$5.53 \mathrm{~m}$
$6.53 \mathrm{~m}$
$9.53 \mathrm{~m}$
$12.53 \mathrm{~m}$
Solution
Initial pressure $p_1=$ atmospheric pressure + pressure of liquid column
$=h d g+h_1 d g$
$h=10 \mathrm{~m}$ of water,
$h_1=$ depth of lake
$p_1=d g\left(h+h_1\right)$
From,
$p_1 V_1=p_2 V_2$
$\begin{aligned} d g(10+h) \times \frac{4}{3} \pi r_1^3 & =h d g \times \frac{4}{3} \pi r_2^3 \\ (10+h) \times r^3 & =10 \times\left(\frac{5}{4} r\right)^3 \\ (10+h) & =10 \times \frac{125}{64} \\ h & =\frac{1250}{64}-10 \\ \Rightarrow \quad h & =\frac{610}{64}=9.53 \mathrm{~m}\end{aligned}$