Question

An air bubbles starts rising from the bottom of a lake. Its diameter is $3.6mm$ at the bottom and $4mm$ at the surface. The depth of the lake is $250cm$ and temperature at the surface is ${40}^{\circ }C$. What is the temperature at the bottom of the lake? Given atmospheric pressure $=76cm$ of $Hg$ and $g=980cmse{c}^{-2}$. Neglect surface tension effect.

• A. ${9.37}^{o}C$
• B. ${10.37}^{o}C$
• C. ${11.37}^{o}C$
• D. ${7.37}^{o}C$

Answer 1

The correct option is B ${10.37}^{o}C$

At the bottom of the lake, the volume of the bubble

$V_1=\frac{4}{3}\pi r_1^3=\frac{4}{3}\pi (0.18{)}^{3}c{m}^3$

Pressure on the bubble ($P_1$ = Atmospheric pressure + pressure due to a column of 250cm of water

$=76\times 13.6\times 980+250\times 1\times 980$

$=(76\times 13.6+250)980dyne/c{m}^2$

At the surface of the lake, Volume of the bubble

$V_2=\frac{4}{3}\pi r_2^3=\frac{4}{3}\pi (0.2{)}^{3}c{m}^3$

pressure on the bubble, $P_2$=atmosphericpressure

$=(76\times 13.6\times 980)dyne/c{m}^2$

$T_2=273+{40}^{o}C$

$=313K$

Now $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

or $\frac{(76\times 13.6+250)980\times \left(\frac{4}{3}\right)\pi (0.18{)}^3}{T_1}=\frac{(76\times 13.6)\times 980\left(\frac{4}{3}\right)\pi (0.2{)}^3}{313}$

or $T_1=288.37K$

$\therefore T_1=283.37-273$

$={10.37}^{0}C$

Hence, the correct option is $\text{B}$