A body of mass 50g is weight in water and then in a liquid of unknown density. If apparent weight in water and liquid are 46g and 45.5g respectively find the density of a body and density of the liquid
The apparent weight of an object in a fluid is equal to the weight of the displaced fluid.
Let's assume the density of the body is ρ1 and the density of the liquid is ρ2.
The weight of the body in air is 50g.
The weight of the body in water is 46g. This means that the weight of the displaced water is 50g - 46g = 4g.
The weight of the body in the unknown liquid is 45.5g. This means that the weight of the displaced liquid is 50g - 45.5g = 4.5g.
Since the weight of the displaced fluid is equal to the weight of the body, we can set up the following equation for the density of the body:
ρ1 * V = 4g ...(1)
where V is the volume of the body.
Similarly, for the density of the liquid, we can set up the following equation:
ρ2 * V = 4.5g ...(2)
Dividing equation (2) by equation (1), we get:
ρ2/ρ1 = 4.5g / 4g
Simplifying, we find:
ρ2/ρ1 = 9/8
So the ratio of the densities is 9:8.
Since the density of water is 1g/cm^3, we can use this ratio to determine the density of the body and the density of the liquid:
ρ1 = (9/8) * 1g/cm^3 = 1.125g/cm^3
ρ2 = 1g/cm^3 / (9/8) = 8/9g/cm^3 ≈ 0.89g/cm^3
Therefore, the density of the body is approximately 1.125g/cm^3 and the density of the liquid is approximately 0.89g/cm^3.