Question

A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 20.1 N; when it is completely immersed in water, the scale reads 15.3 N.

A) What is the volume of the block?
B) What is the density of the block?

Answer 1

A)
`V = 4.92 \cdot 10^(-4) m^(3) = 492 cm^(3)`

B)
`d = 4181.49 kg/m^(3) = 4.18 g/cm^(3)`

Step-by-step explanation:

A) Using the Archimedes' force we can find the weight of water displaced:

`W_(d) = W_(a) - W_(w)`

Where:

`W_(a)`: is the weight of the block in the air = 20.1 N

`W_(w)`: is the weight of the block in the water = 15.3 N

`W_(d) = W_(a) - W_(w) = 20.1 N - 15.3 N = 4.8 N`

Now, the mass of the water displaced is:

`m = (W_(d))/(g) = (4.8 N)/(9.81 m/s^(2)) = 0.49 kg`

The volume of the block can be found using the mass of water displaced and the density of the water:

`V = (m)/(d) = (0.49 kg)/(997 kg/m^(3)) = 4.92 \cdot 10^(-4) m^(3) = 492 cm^(3)`

B) The density of the block can be found as follows:

`d = (W_(a))/(g*V) = (20.1 N)/(9.81 m/s^(2)*4.92 \cdot 10^(-4) m^(3)) = 4181.49 kg/m^(3) = 4.18 g/cm^(3)`

I hope it helps you!