Question

The diameter of a bubble at the surface of a lake is $4\text{mm}$ and at the bottom of the lake is $1\text{mm}$. If the atmospheric pressure is $1\text{atm}$ and the temperature of the lake and the atmosphere are equal, what is the depth of the lake?
(The density of the lake and mercury are $1\text{g/ml}$ and $13.6\text{g/ml}$ respectively. Ignore the surface tension)

• A. $453\text{m}$
• B. $651\text{m}$
• C. $784\text{m}$
• D. $102\text{m}$

Answer 1

The correct option is B $651\text{m}$

We know that pressure $=\rho gh$
Also the volume of spherical bubble $=\frac{4}{3}\pi r^3$
As temperature of lake is constant, using Boyle's law:
$P_1{V}_1=P_2{V}_2$
$\therefore (760\text{mm}\times 13.6\times g)\frac{4}{3}\pi {\left(\frac{4}{2}\text{mm}\right)}^3=(760\text{mm}\times 13.6\times g+h\times 1\times g)\frac{4}{3}\pi {\left(\frac{1}{2}\text{mm}\right)}^3$
$760\times 13.6\times 64=(760\times 13.6+h)$
$h=64\times 760\times 13.6-760\times 13.6$
$h=63\times 760\times 13.6\text{mm}$
$h=\frac{63\times 760\times 13.6}{1000\times 1000}\text{km}=0.6511\text{km}=651.1\text{m}$