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

Answer:

A) $V = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}$

B) $d = 4181.49 kg/m^{3} = 4.18 g/cm^{3}$

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:

$m = \frac{W_{d}}{g} = \frac{4.8 N}{9.81 m/s^{2}} = 0.49 kg$

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

$V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}$

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

$d = \frac{W_{a}}{g*V} = \frac{20.1 N}{9.81 m/s^{2}*4.92 \cdot 10^{-4} m^{3}} = 4181.49 kg/m^{3} = 4.18 g/cm^{3}$

I hope it helps you!