State what is meant by the density of a material.
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The mass per unit volume of the material.
Density Upthrust And Viscosity
A block of aluminium has dimensions 0.20 m × 0.15 m × 0.10 m and a mass of 8.1 kg. Calculate the density of aluminium.
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Finding the volume $$V=0.20\times 0.15\times 0.10=3.0\times {10}^{-3}{\text{ m}}^{3}$$ Finding the density $$\rho =\frac{m}{V}=\frac{8.1}{3.0\times {10}^{-3}}$$ $$\rho =2700{\text{ kg m}}^{-3}$$ The density of aluminium is 2700 kg m⁻³.
A copper wire has density 8960 kg m⁻³ and volume 2.5 × 10⁻⁶ m³. Determine the mass of the wire.
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$$m=\rho V=8960\times 2.5\times {10}^{-6}$$ $$m=0.0224\text{ kg}=22.4\text{ g}$$ The mass of the wire is 0.0224 kg (22.4 g).
State the relationship between upthrust and the fluid surrounding a submerged object.
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Upthrust equals the weight of fluid displaced by the object.
A cube of side 0.050 m is fully submerged in water of density 1000 kg m⁻³. Calculate the upthrust on the cube. Take g = 9.81 N kg⁻¹.
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Finding the volume displaced $$V=(0.050{)}^{3}=1.25\times {10}^{-4}{\text{ m}}^{3}$$ Finding the upthrust $$U=\rho Vg=1000\times 1.25\times {10}^{-4}\times 9.81$$ $$U=1.23\text{ N}$$ The upthrust on the cube is 1.23 N.
Explain why a steel ship can float on water even though steel is denser than water.
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The ship’s hull encloses a large volume of air, so the average density of the ship is less than that of water; the ship displaces a weight of water equal to its own weight before it is fully submerged, so upthrust balances weight and the ship floats.
State the three conditions that must be satisfied for Stokes’ Law to apply to a moving object.
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The object is small and spherical; it moves at a low speed through the fluid; the flow around it is laminar / non-turbulent.
Describe how the viscosity of a liquid changes with temperature, and give a molecular reason for this change.
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Viscosity decreases as temperature increases; molecules gain kinetic energy and move past one another more easily, reducing internal resistance to flow.
A small oil droplet of radius 2.0 × 10⁻⁶ m moves through air of viscosity 1.8 × 10⁻⁵ Pa s at a speed of 1.2 × 10⁻⁴ m s⁻¹. Calculate the viscous drag force on the droplet.
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$$F=6\pi \eta rv$$ $$F=6\pi \times 1.8\times {10}^{-5}\times 2.0\times {10}^{-6}\times 1.2\times {10}^{-4}$$ $$F=8.1\times {10}^{-14}\text{ N}$$ The viscous drag force on the droplet is 8.1 × 10⁻¹⁴ N.
Predict what happens to the viscous drag on a sphere if its speed doubles while all other quantities stay the same.
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Drag force is directly proportional to speed; therefore the drag force also doubles.
In the falling-ball experiment, explain why the upper marker is positioned below the surface of the liquid rather than at the surface.
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The ball must have reached terminal velocity before timing begins; placing the marker below the surface allows the ball to accelerate and reach a constant speed before it is timed.
A steel ball of radius 1.20 mm and density 7800 kg m⁻³ falls through glycerol of density 1260 kg m⁻³. The terminal velocity is measured as 0.052 m s⁻¹. Take g = 9.81 N kg⁻¹. Show that the viscosity of the glycerol is approximately 0.39 Pa s.
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Converting the radius $$r=1.20\times {10}^{-3}\text{ m}$$ Applying the terminal-velocity condition (weight = upthrust + drag) $$\eta =\frac{2{r}^{2}g({\rho }_{\text{ball}}-{\rho }_{\text{fluid}})}{9v}$$ $$\eta =\frac{2\times (1.20\times {10}^{-3}{)}^{2}\times 9.81\times (7800-1260)}{9\times 0.052}$$ $$\eta =0.39\text{ Pa s}$$ The viscosity of the glycerol is approximately 0.39 Pa s.
Suggest two improvements that would reduce the uncertainty in the calculated viscosity.
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Increase the distance d between markers to reduce percentage uncertainty in time; repeat each measurement and take a mean / use video recording with frame-by-frame playback to remove reaction-time error / measure radius using a micrometer at three positions and average.
Discuss the limitations of using the falling-ball method to determine viscosity, and explain how each limitation affects the result. In your answer you should: • identify at least three sources of error; • explain how each affects the measured viscosity; • suggest one method to reduce the overall uncertainty.
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Reaction-time error in starting and stopping the stopwatch increases random uncertainty in v, which propagates into η; if the ball has not reached terminal velocity at the upper marker, the measured v is too low, so η is overestimated; if the ball is too large or moves too fast, flow becomes turbulent and Stokes’ Law no longer applies, so the calculated η is invalid. The cylinder walls exert an additional drag if the ball is close to them, increasing the apparent viscosity; temperature changes alter η during the experiment because viscosity is temperature dependent. To reduce uncertainty, repeat the experiment at constant temperature with a small ball and a long fall distance, averaging multiple drops, and ensure the ball is released centrally in the cylinder.