Question

A copper ba l with hollow center weighs 264 gf in air and 221 gf in water. determine the volume of the hollow portion of the ball. Given, the density of copper = 8.8 g cm³. (3m)​

Answer 1

13 cm³

Step-by-step explanation:

• Weight of the copper ball in air m = 264 g

• Density of copper in air μ = 8.8 g/cm³

• Using the relationship V = m/μ where V = volume,
  
• Volume of copper ball = 264/8.8 = 30 cm³

• Weight of copper ball in water = 221 g
  
• Weight of water displaced = 264 - 221 = 43 g
  
• Volume of water displaced = Weight of water displaced/density of water
  =
  `(43\;g)/(1 \;g/cm^3) = 43 \;cm^3`

• Volume of the copper ball including the hollow part
  
  `= 43 \;cm^3`

• Therefore volume of the hollow portion
  = volume of ball including hollow part - volume of copper part
  = 43 - 30
  = 13 cm³

Answer 2

the volume of the hollow portion of the ball is 0.425 cm³

Step-by-step explanation:

To determine the volume of the hollow portion of the ball, we need to use the principle of buoyancy. When the ball is placed in water, it displaces a volume of water equal to its own volume. The weight of the water displaced is equal to the weight of the ball in water, which is less than the weight of the ball in air by an amount equal to the weight of the hollow portion of the ball.

Let V be the total volume of the ball and Vh be the volume of the hollow portion. We can write two equations based on the weight of the ball in air and water:

Weight of ball in air = Weight of copper + Weight of air = Density of copper x Volume x g

Weight of ball in water = Weight of copper + Weight of water = Density of copper x (V - Vh) x g

where g is the acceleration due to gravity.

Subtracting the second equation from the first, we get:

Weight of air = Density of copper x Vh x g

Solving for Vh, we get:

Vh = Weight of air / (Density of copper x g)

Substituting the given values, we get:

Vh = (264 - 221) / (8.8 x 9.81) = 0.425 cm³