Question

A glass tube of $1$mm bore is dipped vertically into a container of mercury with its lower end $2$ cm below the mercury surface. What must be the guage pressure of air in the tube in order to blow a hemispherical bubble at its lower end? Given density of mercury = $13600kg{m}^{-3}$ and surface tension of mercury = $35\times {10}^{-3}N{m}^{-1}$.

Answer 1

R.E.F image

Consider the bubble

Now, the FBD can be

concisely repentant as

Applying Newtons Laws

Buoyant force $+$ Surface tension $=$ Pressure face

$\frac{2}{3}\pi r^3{p}_g+\sigma \times 2\pi r=p\times \pi r^2$

$p\times \frac{2}{3}t{r}^2\pi 10+\sigma \times 2t=p+\pi r$

$\frac{2}{3}p{r}^{2}p+\sigma =pr$

$\frac{2}{3}prg+\frac{\sigma }{r}=p$

$10\times \frac{2}{3}\times 13600\times 0.5\times {10}^{-3}\frac{35\times {10}^{-3}}{0.5\times {10}^{-3}}$

$10\times \frac{13.6}{3}+70=p$

$p=70+45.3$

$p=115.3\frac{N}{m^2}$