A hydrometer floats in water of density 1.0g/cm³ with 6.0cm of its graduated stem above the water level, and in oil of density 0.8g/cm³ with 4.0cm above the oil level. Calculate the length of stem above the liquid level when the hydrometer placed in a liquid of density 0.9g/cm³.

Answer 1

Let x be the length of the stem above the liquid level when placed in a liquid of density 0.9g/cm³.

Using the principle of flotation:

Weight of hydrometer in water = Weight of water displaced
Weight of hydrometer in oil = Weight of oil displaced
Weight of hydrometer in liquid of density 0.9g/cm³ = Weight of liquid displaced

The weight of the hydrometer is the same in all three cases. Let's assume it is w grams.

In water:
Weight of hydrometer = w
Weight of water displaced = volume of water displaced x density of water x gravity
= (total volume of hydrometer - submerged volume of hydrometer) x 1.0g/cm³ x 9.8 m/s²
= (6 + x) cm x 1.0g/cm³ x 9.8 m/s²

Therefore:
w = (6 + x) cm x 1.0g/cm³ x 9.8 m/s²

In oil:
Weight of hydrometer = w
Weight of oil displaced = volume of oil displaced x density of oil x gravity
= (total volume of hydrometer - submerged volume of hydrometer) x 0.8g/cm³ x 9.8 m/s²
= (4 cm) x 0.8g/cm³ x 9.8 m/s²

Therefore:
w = 4 cm x 0.8g/cm³ x 9.8 m/s²

Equating the weight of the hydrometer in water and in oil:

(6 + x) cm x 1.0g/cm³ x 9.8 m/s² = 4 cm x 0.8g/cm³ x 9.8 m/s²

Simplifying:
(6 + x) x 1 = 3.2
6 + x = 3.2
x = -2.8

This result doesn't make sense, since x represents a length and cannot be negative. Therefore, the hydrometer would sink in the liquid of density 0.9g/cm³.