Ametal object is suspended from a spring scale. the scale reads 930 n when the object is suspended in air, and 731 n when the object is completely submerged in water. find the volume of the object. the acceleration of gravity is 9.8 m/s 2 . answer in units of m3 .

The volume of the object = 0.0203 m³

Explanation:

From Archimedes principle,

Upthrust = lost in weight = amount of water displaced.

U = W₁ - W₂

Where U = upthrust, W₁ = Weight in air, W₂ = weight in water.

U = 930- 731

U = 199 N.

but mass of water displaced,

m = U/g where g = 9.8 m/s²

m = 199/9.8

m = 20.31 kg.

Recall,

Density = mass/volume.

Density of water = mass of water displaced/ volume of water displaced

D = m/V

V = m/D Equation 1

Where V = volume of water displaced, m = 20.31 kg, D = 1000 kg/m³

Substitute into equation 1

V = 20.31/1000

V = 0.0203 m³

the volume of water displaced = 0.0203 m³

Note: That when an object is immersed in water, the volume of water it displaced is equal to it's volume.

Hence, Volume of water displaced = volume of the object.

Hence the volume of the object = 0.0203 m³

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