Question

In the bottom of a vessel with mercury there is a round hole of diameter d = 70 µm. At what maximum thickness of the mercury layer will the liquid still not flow out through this hole?

• A. 0.199 m
• B. 0.070 m
• C. 0.035 m
• D. 0.35 m

Answer 1

Answer: (A)

0.199 m

Answer 2

The condition for the liquid not to flow out is when the hydrostatic pressure is balanced by the pressure due to surface tension. The hydrostatic pressure is $P_{hydrostatic} = h\rho g$. For a convex meniscus, the pressure difference across the surface is $P_{surface\_tension} = \frac{2T}{r}$, where $T$ is the surface tension and $r$ is the radius of the hole.

Thus, $h\rho g = \frac{2T}{r}$, which gives the maximum height $h = \frac{2T}{\rho g r}$.

Given: Diameter, $d = 70 \text{ µm}$, so radius $r = 35 \text{ µm} = 35 \times 10^{-6} \text{ m}$. Surface tension of mercury, $T \approx 0.465 \text{ N/m}$. Density of mercury, $\rho \approx 13600 \text{ kg/m}^3$. Acceleration due to gravity, $g \approx 9.8 \text{ m/s}^2$.

$h = \frac{2 \times 0.465 \text{ N/m}}{(13600 \text{ kg/m}^3) \times (9.8 \text{ m/s}^2) \times (35 \times 10^{-6} \text{ m})}$ $h = \frac{0.930 \text{ N/m}}{4.6648 \text{ N/m}^2}$ $h \approx 0.199365 \text{ m}$

Therefore, the maximum thickness of the mercury layer is approximately $0.199 \text{ m}$.