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

$$\begin{gathered} \text{ Let initial radius be }=\mathrm{r},\text{ Final radius }=\mathrm{r}+200\mathrm{\%}\text{ of }\mathrm{r}=3\mathrm{r} \\ \text{ Atmospheric pressure }=\mathrm{\rho gH} \\ \text{ Let depth of the lake be }\mathrm{h} \\ \text{ So, pressure at the bottom of lake= }\mathrm{\rho gH}+\mathrm{\rho gh} \\ \text{ Using }={\mathbf{P}}_1{\mathbf{V}}_1={\mathbf{P}}_2{\mathbf{V}}_2 \end{gathered}$$

$$\begin{gathered} \rho gH\times \frac{4}{3}\pi (3r{)}^3=(\rho gH+\rho gh)\times \frac{4}{3}\pi r^3 \\ \rho gH\times \frac{4}{3}\pi 27r^3=(\rho gH)\times \frac{4}{3}\pi r^3+\rho gh\times \frac{4}{3}\pi r^3 \\ \text{ Solving this equation we get }26\mathrm{H}=\mathrm{h} \end{gathered}$$