When the rock is suspended in the air, the net force on it is
∑ <em>F₁</em> = <em>T₁</em> - <em>m₁g</em> = 0
where <em>T₁</em> is the magnitude of tension in the string and <em>m₁g</em> is the rock's weight. So
<em>T₁</em> = <em>m₁g</em> = 37.8 N
When immersed in water, the tension reduces to <em>T₂</em> = 32.0 N. The net force on the rock is then
∑ <em>F₂</em> = <em>T₂</em> + <em>B₂</em> - <em>m₁g</em> = 0
where <em>B₂</em> is the magnitude of the buoyant force. Then
<em>B₂</em> = <em>m₁g</em> - <em>T₂</em> = 37.8 N - 32.0 N = 5.8 N
<em>B₂</em> is also the weight of the water that was displaced by submerging the rock. Let <em>m₂</em> be the mass of the displaced water; then
5.8 N = <em>m₂g</em> ==> <em>m₂</em> ≈ 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 <em>V</em> that was displaced was
1.00 × 10³ kg/m³ = <em>m₂</em>/<em>V</em> ==> <em>V</em> ≈ 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 <em>T₃</em> = 20.2 N, and so the net force on the rock is
∑ <em>F₃</em> = <em>T₃</em> + <em>B₃</em> - <em>m₁g</em> = 0
which means the buoyant force is
<em>B₃</em> = <em>m₁g</em> - <em>T₃</em> = 37.8 N - 20.2 N = 17.6 N
The mass <em>m₃</em> of the liquid displaced is then
17.6 N = <em>m₃g</em> ==> <em>m₃</em> ≈ 1.80 kg
Then the density <em>ρ</em> of the unknown liquid is
<em>ρ</em> = <em>m₃</em>/<em>V</em> ≈ (1.80 kg)/(0.000592 m³) ≈ 3040 kg/m³ = 3.04 g/cm³
