Question

When an air bubble of radius $r$ rises from the bottom to the surface of a lake, its radius becomes $\frac{5r}{4}$. Taking the atmospheric pressure to be equal to $10m$ height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature):

• A. $10.5m$
• B. $8.7m$
• C. $11.2m$
• D. $9.5m$

Answer 1

The correct option is D $9.5m$

Pressure at bottom

$\left(P_1\right)=P_{atm}+\rho gh+\frac{4T}{R_1}$__________(1)

Pressure at top

$\left(P_2\right)=P_{atm}+\frac{4T}{R_2}$____________(2)

Given $R_1=r$

$R_2=\frac{5r}{4}$

So $P_1{V}_1=P_2{V}_2$

$\left(P_1\right)\frac{4}{3}\pi r^3=\left(P_2\right)\frac{4}{3}\frac{125r^3}{64}$

Dividing (1) and (2)

$\frac{P_1}{P_2}=\frac{P_{atm}+\rho gh+\frac{4T}{r}}{P_{atm}+\frac{4T\times 4}{5r}}=\frac{125}{64}$

$\frac{\rho g\left(10\right)+\rho gh}{\rho g\left(10\right)}=\frac{125}{64}$

$640+64h=1250$

On solving we get

$h=9.5m$