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

To find the volume of the object, subtract the weight of the object from the scale reading in water. Then, divide the weight of the water displaced by the density of water. The volume of the object is 170 mL.

Step-by-step explanation:

To find the volume of an object, we can use Archimedes' principle, which states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. In this case, the scale reading in air is 920 N, which is the weight of the object. When the object is completely submerged in water, the scale reading is 750 N, which is the weight of the object plus the buoyant force.

We can calculate the volume of water displaced by the object by subtracting the weight of the object from the scale reading in water: 750 N - 920 N = -170 N.

Since the buoyant force is equal to the weight of the fluid displaced, we can conclude that the object displaces 170 N of water. Using the given information that 1 N is equivalent to 1 g of weight, we can calculate the volume of water displaced using the formula:

Volume = weight of water displaced / density of water = 170 g / 1 g/mL = 170 mL

Therefore, the volume of the object is 170 mL.

Answer 2

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} = \frac{\text{Weight}}{g} = (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 = \frac{\text{Mass}}{\text{Density}} =(m)/(\rho) = \frac{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.