Question

A rock is suspended by a light string. When the rock is in air, the tension in the string is 37.8 N. When the rock is totally immersed in water, the tension is 32.0 N. When the rock is totally immersed in an unknown liquid, the tension is 20.2 N. What is the Density of the unknown liquid?

Answer 1

When the rock is suspended in the air, the net force on it is

∑ F₁ = T₁ - m₁g = 0

where T₁ is the magnitude of tension in the string and m₁g is the rock's weight. So

T₁ = m₁g = 37.8 N

When immersed in water, the tension reduces to T₂ = 32.0 N. The net force on the rock is then

∑ F₂ = T₂ + B₂ - m₁g = 0

where B₂ is the magnitude of the buoyant force. Then

B₂ = m₁g - T₂ = 37.8 N - 32.0 N = 5.8 N

B₂ is also the weight of the water that was displaced by submerging the rock. Let m₂ be the mass of the displaced water; then

5.8 N = m₂g   ==>   m₂ ≈ 0.592 kg

If one takes the density of water to be 1.00 g/cm³ = 1.00 × 10³ kg/m³, then the volume of water V that was displaced was

1.00 × 10³ kg/m³ = m₂/V   ==>   V ≈ 0.000592 m³ = 592 cm³

and this is also the volume of the rock.

When immersed in the unknown liquid, the tension reduces further to T₃ = 20.2 N, and so the net force on the rock is

∑ F₃ = T₃ + B₃ - m₁g = 0

which means the buoyant force is

B₃ = m₁g - T₃ = 37.8 N - 20.2 N = 17.6 N

The mass m₃ of the liquid displaced is then

17.6 N = m₃g   ==>   m₃ ≈ 1.80 kg

Then the density ρ of the unknown liquid is

ρ = m₃/V ≈ (1.80 kg)/(0.000592 m³) ≈ 3040 kg/m³ = 3.04 g/cm³