A hydrometer of mass y kg and volume 2y x 10−5 − 5 m3 3 floats in a fluid with 20% of its volume above the fluid, what is the density of the fluid?

A hydrometer of mass y kg and volume 2y x 10−5 ${}^{-5}$m3 ${}^3$ floats in a fluid with 20% of its volume above the fluid, what is the density of the fluid?

Answer Details

To find the density of the fluid, we need to apply the principle of floatation, which states that the weight of the fluid displaced by the submerged part of the object is equal to the weight of the object. Let's walk through the steps:

Step 1: Understand the volume submerged

The hydrometer has a total volume of 2y x 10^-5 m³. It floats with 20% of its volume above the fluid. Hence, 80% of its volume is submerged in the fluid.

Submerged Volume, V_sub = (0.80) x (2y x 10^-5 m³) = 1.6y x 10^-5 m³

Step 2: Apply the principle of floatation

The weight of the fluid displaced equals the weight of the hydrometer.

Weight of hydrometer = Mass x Gravity = y kg x g (where g is the acceleration due to gravity). For the purpose of calculations, g can be considered as 9.81 m/s².

Weight of displaced fluid = Density of fluid (ρ_fluid) x Submerged Volume x g

According to the principle of floatation:

y x g = ρ_fluid x 1.6y x 10^-5 m³ x g

g is common on both sides and can be canceled out:

y = ρ_fluid x 1.6y x 10^-5

Step 3: Solving for the density of the fluid

ρ_fluid = y / (1.6y x 10^-5)

The y on both numerator and denominator cancels out:

ρ_fluid = 1 / (1.6 x 10^-5)

ρ_fluid = 6.25 x 10⁴ kg/m³

Thus, the density of the fluid is 6.25 x 10⁴ kg/m³.