Question

A air bubble rises from bottom of a lake to surface. If its radius increases by $200\%$ and atmospheric pressure is equal to water coloumn of height $H_1$ then depth of lake is

• A. 21 H
• B. 8 H
• C. 9 H
• D. 26 H

Answer 1

Answer: (D)

Let initial radius be $=r$
Final radius $=r+200\%$ of $r$
$=3r$
Atmospheric pressure $=\rho gH$
Let depth of the lake be $h$
So, pressure at the bottom of lake $=\rho gH+\rho gh$
Using $P_1{V}_1=P_2{V}_2$
$\rho gH\times \frac{4}{3}\pi (3r{)}^3=(\rho gH+\rho gh)\times \frac{4}{3}\pi r^3$
$\rho gH\times \frac{4}{3}\pi \times 27r^3=(\rho gH)\frac{4}{3}\pi r^3+\rho gh\times \frac{4}{3}\pi r^3$
Solving this equation we get
$26H=h$