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It is possible to measure liquids density with an hydrometer:

hydrometer

I want to figure out how this work. Hence I wrote:

$$\sum F = 0\\[1em] F_b = F_w \\[1em] \large \delta^{l}\,g\,V_h^l = m_o\,g \hspace{1cm} (1)$$

where $V_h^l=A\,L$ is the volume of the hygroscope ($h$) on the liquid ($l$). Guided by the scale on the hygrometer, one could postulate:

$$\large\frac{d\delta}{dL}=c$$

where $L$ is the length measured on the stick. But if the equation (1) is differentiated we get $$\large \frac{d\delta}{dL}=-\frac{k}{L^2}$$ where $k$ is a constant.

What's the mistake? Any solution?

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  • $\begingroup$ I think that you made a mistake in assuming that the measurement scale is linear with density. In going through the equations, that doesn't appear to be strictly true. In fact, you should be able to find out for yourself what the functional relationship between $L$ and $\delta$ is if you go through your equations carefully. Suggest backing up and more carefully going through your derivation step by step. In particular, not sure how you got your last equation by differentiating your first equation. $\endgroup$ – Samuel Weir Oct 3 '18 at 0:40
  • $\begingroup$ The measurement scale may look linear over the limited range that the scale is printed on the float but, no, if you work forward from your equation #1, you should be able to see for yourself that the relationship between $L$ and $\delta$ is not a linear one. $\endgroup$ – Samuel Weir Oct 3 '18 at 0:46
  • $\begingroup$ @SamuelWeir do you think it is complete now? $\endgroup$ – santimirandarp Oct 3 '18 at 0:47
  • $\begingroup$ I still don't understand where the last equation comes from. BTW, $m_o$, the total mass of the float, is a constant. Go back to eqn.1. You have $\delta V = m_o$, where $\delta$ is the density of the liquid, V is the submerged volume of the float, and $m_o$ is the total mass of the float. So then $\delta A L = m_o$, where $A L = V$ and A is the cross-sectional area of the float and L is the submerged distance. From $\delta A L = m_o$, it's apparent that $L$ is proportional to the reciprocal of the density $\delta$. $\endgroup$ – Samuel Weir Oct 3 '18 at 0:57
  • $\begingroup$ @SamuelWeir it seems something is wrong yet, the conclusions go on the opposite direction respecto to how the instrument is supposed to work $\endgroup$ – santimirandarp Oct 3 '18 at 1:08
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Suppose the scale balances exactly at some mark when the density is some $\rho_0$ calibration density. That means that the total mass of the device $m_d$ is balanced by the displaced liquid at the calibration point. So let the volume of the device below that be $V_{sub}$. At density $\rho_0$ we have the following.

$$ m_d g = \rho_0 V_{sub} g $$

From now on I'll drop the $g$'s.

Now let's suppose the density is $\rho = r + \rho_0 > \rho_0$. Here $r$ is the change in density from the calibration point density. Then the device has to ride higher in the water such that a smaller volume of the denser liquid displaces the same weight of device. The decreased volume is proportional to the height as long as the device rides no higher than the straight stem is still in the liquid.

$$ m_d = \rho \left(V_{sub} - A h \right) = (r + \rho_0) \left(V_{sub} - A h \right) $$

where $h$ is how much it rides above the calibration point, and $A$ is the cross sectional area of the stem. So, using the form with $r$, and recalling $ m_d = \rho_0 V_{sub} $ , and solving for $r$ in terms of $h$ we get the following.

$$ r = \frac{\rho_0 A h}{V_{sub}- Ah} $$

When the scale has a large bulb on the bottom then $V_{sub}$ will be very much larger than $Ah$ for $h$ not too large. And so this is very close to linear for a useful range. For example, if the bulb has a radius 10 times that of the scale, then as long as you stay within one bulb-length of the calibration point, the non-linear nature is less than 10%.

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