Question

A metal object is suspended from a spring scale. The scale reads 920 N when the object is suspended in air, and 750N when the object is completely submerged in water. Find the volume of the object

Answer 1

17.3 L.

Explanation

The object appears $920 - 750 = 170\;\text{N}$ lighter in water than in the air. Water has supplied that 170 N of buoyant force.

The size of the buoyant force on an object in water is the same as the weight of water that the object has displaced. The buoyant force on the metal object here is 170 N. The object must have displaced water of the same weight.

$g = 9.81 \;\text{N}\cdot\text{kg}^{-1}$.

Mass of water displaced:

$\text{Mass} = \dfrac{\text{Weight}}{g} = \dfrac{170}{9.81} = 17.3 \;\text{kg}$.

Volume of water displaced:

The density of water at room temperature is $1.000\;\text{kg}\cdot\text{dm}^{-3}$. Each kilogram of water will occupy a volume of 1 dm³ (one cubic decimeter), which is the same as 1 L (one liter).

$V = \dfrac{\text{Mass}}{\text{Density}} =\dfrac{m}{\rho} = \dfrac{17.3\;\text{N}}{1.000\;\text{kg}\cdot\text{dm}^{-3}} = 17.3\;\text{dm}^{-3}=17.3\;\text{L}$.

Volume of the object:

The object is completely under water. As a result, the volume of the object will be the same as the volume of water displaced. The volume of the object is also 17.3 L.