question_answer

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):                                            [JEE Online 15-04-2018 (II)]

A)  \[10.5m\]         

B)       \[8.7m\]

C)  \[11.2m\]         

D)       \[9.5m\]

Correct Answer: D Solution : [d] Pressure at bottom

\[({{P}_{1}})={{P}_{atm}}+\rho gh+\frac{4T}{{{R}_{1}}}..........(1)\]

Pressure at top

\[({{P}_{2}})={{P}_{atm}}+\frac{4T}{{{R}_{2}}}..........(2)\]

Given\[{{R}_{1}}=r\]

\[{{R}_{2}}=\frac{5r}{4}\]

So \[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\]

\[({{P}_{1}})\frac{4}{3}\pi {{r}^{3}}=({{P}_{2}})\frac{4}{3}\frac{125{{r}^{3}}}{64}\]

Dividing (1) and (2)

\[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{P}_{atm}}+\rho gh+\frac{4T}{r}}{{{P}_{atm}}+\frac{4T\times 4}{5r}}=\frac{125}{64}\]

\[\frac{\rho g(10)+\rho gh}{\rho g(10)}=\frac{125}{64}\]

\[640+64h=1250\]

On solving we get \[h=9.5m\]