Question

Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kg when weighed in air. The density of air is 1.29 kg/m3, and the density of aluminum is 2.7×103kg/m3.'

Answer 1

Answer:

m = 4.5021 kg

Explanation:

given,

Apparent mass of aluminium = 4.5 kg

density of air = 1.29 kg/m³

density of aluminium = 2.7 x 10⁷ kg/m³

true mass of the aluminium = ?

Weight in Vacuum

W = m g

W = ρV g

Air buoyancy acting on aluminium

B = ρ₀V g

Volume is the same in both cases since the volume of the aluminum

displaces an equal amount of volume air.

Apparent weight:

ρV g − ρ₀V g = 4.5 g

ρV − ρ₀V = 4.5

$V = \dfrac{4.5}{\rho - \rho_0}$

m = ρV

$m = \dfrac{4.5\times \rho}{\rho - \rho_0}$

$m = \dfrac{4.5\times 2700}{2700 - 1.29}$

m = 4.5021 kg