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 $0.1mm$, calculate its depth where its radius is $0.126\,mm$ given that the surface tension of mercury is$567\,dyne\,c{{m}^{-1}}$. Assume that the atmospheric pressure is $76\,cm$of mercury.
(A). ${{h}_{2}}=9.48\,cm$
(B). ${{h}_{2}}=18.48\,cm$
(C). ${{h}_{2}}=\,9.48\,m$
(D). ${{h}_{2}}=19.48\,cm$

Answer 1

Substitute corresponding values in Bernoulli’s equation for the two given heights and find the pressures at respective heights. Using Boyle’s law, equate the products of pressure and volume to find height. Convert the units as required to get the correct answer.
Formulas Used:
$P'=(P+h\rho g)+\,\dfrac{2T}{r}$
${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$

Complete answer:
The total pressure inside the bubble at height,$h$can be given by-
$P'=(P+h\rho g)+\,\dfrac{2T}{r}$ - (1)
Here,
$P'$is the pressure inside the bubble
$P$is the atmospheric pressure
$h$is the height at which the bubble is in the fluid
$\rho$is the density of the fluid
$g$is acceleration due to gravity
$T$is surface tension of the fluid
$r$is the radius of the bubble
According to Boyle’s Law, the Pressure is inversely proportional to the volume, when temperature is constant. It is denoted as-

& P\propto \dfrac{1}{V} \\\ & \\\ \end{aligned}$$ $$\therefore {{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$$ - (2) Substituting the values of $$\begin{aligned} & {{h}_{1}}=100cm=1m,\,{{r}_{1}}=0.1mm=\,{{10}^{-4}}\,m \\\ & {{h}_{2}}=h,\,{{r}_{2}}=0.126mm=1.26\times {{10}^{-4}}m \\\ \end{aligned}$$in eq (1), we get, $${{P}_{1}}=(76+1\times 13.6\times 10)+\dfrac{2\times 567}{{{10}^{-4}}}$$ - (3) $${{P}_{2}}=(76+h\times 13.6\times 10)+\dfrac{2\times 567}{1.26\times {{10}^{-4}}}$$ - (4) Substituting pressures from eq (3) and eq (4) in eq (2), we get, $$\begin{aligned} & \left( (76+1\times 13.6\times 10)+\dfrac{2\times 567}{{{10}^{-4}}} \right)\dfrac{4\pi {{r}_{1}}^{3}}{3}=\left( (76+h\times 13.6\times 10)+\dfrac{2\times 567}{1.26\times {{10}^{-4}}} \right)\dfrac{4\pi {{r}_{2}}^{3}}{3} \\\ & \Rightarrow \left( (76+1\times 13.6\times 10)+\dfrac{2\times 567}{{{10}^{-4}}} \right){{r}^{3}}_{1}=\left( (76+h\times 13.6\times 10)+\dfrac{2\times 567}{1.26\times {{10}^{-4}}} \right){{r}_{2}}^{3} \\\ \end{aligned}$$ Solving the above equations we get, $$h=0.0948\,m=9.48cm$$ The air bubble acquires a radius of $$0.126\,mm$$at$$h=9.48\,cm$$. **Therefore, the correct option is (A).** **Note:** The tendency of a fluid to take the smallest surface area possible is called surface tension. Most fluids have a concave or convex meniscus which is due to surface tension. Mercury tends to take up a concave meniscus. As we move deeper into a fluid, the pressure increases due to which the radius of the bubble decreases at a greater depth.