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 1.4 cm (hw) above the liquid–air interface. If the height of the original water mark is 23 cm (hl + hw), determine the density of the liquid.

Answer 1

Answer:

1064.8 kg/m³

Explanation:

Weight of the hydrometer = ρghA where ρ is the density, g is acceleration due to gravity, h is the submerged height and A is the cross sectional area.

W in water = ρwghwA

W in liquid = (ρliq)g hliq A where the cross sectional area is constant

W in water = W in liquid

(ρw)ghwA = (ρliq)g hliq A  where ρw is density of water, ρliq is the density of liquid and hw and hliq are the heights of the liquid and that water. g acceleration due to gravity cancel on both sides as well as the constant A

pliq = $\frac{hw}{hliq}$ × 1000 kg /m³ ( density of water) =( $\frac{23}{23-1.4}$) × 1000 = 1064.8 kg/m³