Question

The density of a liquid is to be determined by an old 1-cm-diameter cylindrical hydrometer whose division marks are completely wiped out. The hydrometer is first dropped in water, and the water level is marked. The hydrometer is then dropped into the other liquid, and it is observed that the mark for water has risen 0.3 cm above the liquid–air interface (Fig. P11–31). If the height of the original water mark is 12.3 cm, determine the density of the liquid.

Answer 1

Answer:

The density of the unknown liquid is 1.025 kg/m³ (considering the density of the water as 1.000 kg/m³)

Explanation:

The hydrometer works by the Archimedes principle. The cylinder floats in the liquid because the hydrostatic thrust is equal to the weight force. This means:

$Tr-W=0N\\Tr=W\\\delta_{fl} \cdot Vol \cdot g =W_{hydr}$

If we measure 2 fluids, the weight of the hydrometer is the same, so:

$\delta_{fl1} \cdot Vol_1 \cdot g =W_{hydr}=\delta_{fl2} \cdot Vol_2 \cdot g\\\delta_{fl1} \cdot Vol_1 =\delta_{fl2} \cdot Vol_2\\\delta_{fl1} H_1 \pi R^2 =\delta_{fl2} H_2 \pi R^2\\\delta_{fl1}\frac{H_1}{H_2}=\delta_{fl2}$

If the original watermark height is 12.3cm (H₁) and the mark for water has risen 0.3 cm above the unknown liquid–air interface, the height of the unknown liquid mark is 12cm (H₂). Therefore:

$delta_{fl1}\frac{H_1}{H_2}=\delta_{fl2}\\1.000\frac{kg}{m^3} \frac{12.3cm}{12cm}=\delta_{fl2}=1.025\frac{kg}{m^3}$