Probably due to surface tension the surface tends to curve near the hydrometer forming a meniscus. The question is whether it's correct to read the hydrometer at top of the meniscus or at the liquid plane? I've seen conflicting information so an explanation why would be appreciated (how do Archimedes principle work when surface tension is taken into account for example).


The hydrometer is "calibrated" with the reading taken at the bottom of the meniscus.
Taking the reading at the top is very difficult as it will depend on how well the liquid wets the hydrometer stem.
The surface tension effect is negligible compared with the Archimedian upthrust.

Update as the result of a comment.

The bulb of my hydrometer is approximately a cylinder has a radius of $9\,\rm mm$ and a length of $120\,\rm mm$ which makes its volume about $30,000\,\rm mm^3$.
Lets assume that the meniscus cross-section is a right angled triangle of side $4\,\rm mm$.
The stem of my hydrometer has a radius of about $3\,\rm mm$.
This gives an approximate volume for the meniscus of $\frac 12\, 4^2 \times 2\, \pi \,3 \approx 150\,\rm mm^3$which is very much less than the volume of the bulb.

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  • $\begingroup$ Could you be a bit more elaborate on the "why"? I think there ought to be some more physics than that is only "calibrated" that way. For example to me it seems that the hydrometer displacec the liquid inside the meniscus as well - doesn't that contribute to Archimedian boyancy for example? $\endgroup$ – skyking Sep 20 '17 at 9:19
  • $\begingroup$ @skyking I have updated my answer. $\endgroup$ – Farcher Sep 20 '17 at 10:06
  • $\begingroup$ But what about the volume of the stem inside the meniscus, with a volume of $4\times 3^2\pi/4 =28mm^3$? It looks like this liquid is displaced and should (or not) contribute to boyancy. This would exactly correspond to reading $4mm$ of on the stem. $\endgroup$ – skyking Sep 20 '17 at 10:16
  • $\begingroup$ @skyking All these small volumes are very much smaller than the volume of the bulb. The important thing is the calibration of the hydrometer is such that the reading should be taken at the bottom of the meniscus just like one would do when using a measuring cylinder. $\endgroup$ – Farcher Sep 20 '17 at 10:24
  • $\begingroup$ I understand the measuring cylinder (I think), but here the difference between reading corresponds to the liquid displaced by the stem. Ignoring the volume of the stem with the argument that it's small compared to the bulb is to basically ignore the reading alltogether. $\endgroup$ – skyking Sep 20 '17 at 10:58

It's the bottom of the meniscus or the main surface of the liquid that marks the reading point.

One can see this by looking at the Lagrangian of a hydrometer in an infinitely sized sample. The liquide displaced by the hydrometer is effectively raised to the surface that is not moved by this. Since it's a static situation only the potential has to be considered.

The potential energy of the hydrometer is:

$$V_{hydr}(h) = -mgh$$

And the displaced water is given by the same formula, but has to be integrated over the depth $h-x$, where $x$ is the distance from the bottom and $A(x)$ is the cross section.

$$V_{liq}(h) = \int_0^h (h-x) \rho A(x) dx$$

Finally the surface tension is proportional to the surface area of the liquid which is constant (because the meniscus doesn't change depending on how deep the hydrometer sinks). We have the same situation with the potential energy of the meniscus due to gravity. So we have

$$V(h) = \int_0^h (h-x) \rho A(x) dx - mgh$$

and the equilibrum would happen when $V'(h)=0$ which means

$$V'(h) = \int_0^h \rho A(x) dx + \left.(h-x) \rho A(x)\right\vert_{x=h} - mg = \rho \int_0^h A(x) dx - mg = 0$$

That is that the mass of the hydrometer and the mass of the liquid that would have occupied the space of the hydrometer below the original surface of the liquid. And the original surface is the main surface in this example.

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