A solid cylinder (radius = 0.150 m, height = 0.120 m) has a mass of 7.00 kg. This cylinder is fl oating in water. Then oil (ρ = 725 kg/m3 ) is poured on top of the water until the situation shown in the drawing results. How much of the height of the cylinder is in the oil?
The correct answer and explanation is :
To determine the height of the cylinder submerged in oil, we apply Archimedes’ principle, which states that the buoyant force on an object equals the weight of the fluid it displaces. For an object floating at equilibrium, the sum of the buoyant forces from both fluids equals the object’s weight.
Given:
- Radius of the cylinder, ( r = 0.150 \, \text{m} )
- Height of the cylinder, ( h = 0.120 \, \text{m} )
- Mass of the cylinder, ( m = 7.00 \, \text{kg} )
- Density of oil, ( \rho_{\text{oil}} = 725 \, \text{kg/m}^3 )
- Density of water, ( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 )
Calculations:
- Volume of the Cylinder: [ V_{\text{cylinder}} = \pi r^2 h = \pi (0.150 \, \text{m})^2 (0.120 \, \text{m}) \approx 8.482 \times 10^{-3} \, \text{m}^3 ]
- Weight of the Cylinder: [ W_{\text{cylinder}} = m \times g = 7.00 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 68.67 \, \text{N} ]
- Buoyant Forces: Let ( h_{\text{oil}} ) be the height of the cylinder submerged in oil, and ( h_{\text{water}} ) be the height submerged in water. Thus: [ h_{\text{oil}} + h_{\text{water}} = h = 0.120 \, \text{m} ] The buoyant force from oil: [ F_{\text{buoy, oil}} = \rho_{\text{oil}} \times g \times V_{\text{displaced, oil}} = \rho_{\text{oil}} \times g \times (\pi r^2 h_{\text{oil}}) ] The buoyant force from water: [ F_{\text{buoy, water}} = \rho_{\text{water}} \times g \times V_{\text{displaced, water}} = \rho_{\text{water}} \times g \times (\pi r^2 h_{\text{water}}) ] At equilibrium: [ F_{\text{buoy, oil}} + F_{\text{buoy, water}} = W_{\text{cylinder}} ] Substituting the expressions: [ \rho_{\text{oil}} \times g \times (\pi r^2 h_{\text{oil}}) + \rho_{\text{water}} \times g \times (\pi r^2 h_{\text{water}}) = m \times g ] Simplifying: [ \rho_{\text{oil}} \times h_{\text{oil}} + \rho_{\text{water}} \times h_{\text{water}} = m / (\pi r^2) ] [ 725 \times h_{\text{oil}} + 1000 \times (0.120 – h_{\text{oil}}) = \frac{7.00}{\pi \times (0.150)^2} ] [ 725 \times h_{\text{oil}} + 1000 \times 0.120 – 1000 \times h_{\text{oil}} = \frac{7.00}{\pi \times 0.0225} ] [ 725 \times h_{\text{oil}} + 120 – 1000 \times h_{\text{oil}} = 98.98 ] [ -275 \times h_{\text{oil}} = 98.98 – 120 ] [ -275 \times h_{\text{oil}} = -21.02 ] [ h_{\text{oil}} = \frac{21.02}{275} \approx 0.0764 \, \text{m} ] [ h_{\text{oil}} \approx 7.64 \, \text{cm} ]
Conclusion:
Approximately 7.64 cm of the cylinder’s height is submerged in oil. This calculation demonstrates how Archimedes’ principle can be applied to determine the extent of submersion of an object in multiple fluids based on their densities and the object’s dimensions.