Question

An air bubble rises from the bottom of a lake and its radius is doubled on reaching the surface. If the temperature is constant the depth of the lake is (Pressure is 1 atm above the lake)

Answer 1

Let,

$P_1=$ Pressure under water at asked depth.

$P_2=$ Pressure at surface of water $=$ atmosph $P$.

Let,

$V_1=$ Volume of bubble at original depth.

$V_2=$ Volume of bubble at surface at water.

Let original radius of bubble $=r$

$\Rightarrow$ new radius $=2r$

Now, $P_1{V}_1=P_2{V}_2$ (for constant temperature)

$\Rightarrow P_1\left(\frac{4}{3}\pi r^3\right)=P_2\left(\frac{4}{3}\pi \right(2r{)}^3)$

$\Rightarrow P_1\left(\frac{4}{3}\pi r^3\right)=P_2\left(\frac{4}{3}\pi .8r^3\right)$

$\Rightarrow P_1=8P_2$

But, $P_2=1atm$

$\Rightarrow P_1=7atm=P_1+h\rho g$

$\Rightarrow 1atm+h\left(\rho g\right)=8atm$

$\Rightarrow h\left(\rho g\right)=7atm=P$

$\Rightarrow h=70m$

[$10m$ column of water produces $1atm$ of $pr$]