Question

A minute spherical air bubble is rising slowly through a column of mercury contained in a deep jar. If the radius of the bubble at a depth of 100 cm is 0.1 mm, calculate its depth where its radius is 0.126 mm given that the surface tension of mercury is 567 dyne/cm. Assume that the atmospheric pressure is 76 cm of mercury

• A. $h_2=9.48cm$
• B. $h_2=18.48cm$
• C. $h_2=9.48m$
• D. $h_2=19.48cm$

Answer 1

The correct option is A $h_2=9.48cm$

The total pressure inside the bubble at depth $h_1$ is (P is atmospheric pressure)
= $(P+h_1\rho g)+\frac{2T}{r_1}=P_1$
and the total pressure inside the bubble at depth $h_2$ is = $(P+h_2\rho g)+\frac{2T}{r_2}=P_2$
Now according to Boyle's Law
$P_1{V}_1=P_2{V}_2$ where $V_1=\frac{4}{3}\pi r{{}_1}^3$ and $V_2=\frac{4}{3}\pi r{{}_2}^3$
Hence we get $\left[\right(P+h_1\rho g)+\frac{2T}{r_1}]\frac{4}{3}\pi r{{}_1}^3=\left[\right(P+h_2\rho g)+\frac{2T}{r_2}]\frac{4}{3}\pi r{{}_2}^3$
or $\left[\right(P+h_1\rho g)+\frac{2T}{r_1}]r{{}_1}^3\left[\right(P+h_2)+\frac{2T}{r_2}]r{{}_2}^3$
Given that : $h_1=100cm$ $r_1=0.1mm=0.01cm,r_2=0.126mm=0.0126cm,T=567$ dyne/cm, P = 76 cm
of mercury Substituting all the values we get $h_2=9.48cm$