Physics
What is density, and what is its SI unit?
Mass per unit volume. The SI unit is kg m⁻³.
What is upthrust?
The upward force that a fluid exerts on an object partly or fully immersed in it.
An object floats at rest on water. How does its upthrust compare with its weight?
Upthrust equals weight, so the resultant force is zero.
What is a fluid? Give one liquid example and one gas example.
A substance that can flow. Examples: water and air.
State Stokes’ law for the viscous drag on a sphere.
F = 6πηrv, where η is dynamic viscosity, r is sphere radius and v is its speed relative to the fluid
Write the equation linking density ρ, mass m and volume V. Rearrange it to make V the subject.
ρ = m/V, so V = m/ρ.
State Archimedes’ principle.
The upthrust on an object equals the weight of the fluid it displaces.
A solid object has density 1200 kg m⁻³. It is released from rest in water of density 1000 kg m⁻³. Does it initially rise or sink? Explain using forces.
It sinks. Its weight exceeds the upthrust because its density is greater than the water’s density.
Describe the difference between laminar flow and turbulent flow.
In laminar flow, fluid moves in smooth layers with little mixing between them. Turbulent flow involves irregular fluctuations, eddies and mixing.
State the conditions under which the Stokes’ law model used in this topic applies.
A small spherical object moves slowly relative to the fluid, with laminar flow around it. Accept “creeping flow” or “very low Reynolds number” as a more precise flow condition.
An oil sample has mass 0.36 kg and volume 4.0 × 10⁻⁴ m³. Calculate its density.
ρ = 0.36/(4.0 × 10⁻⁴) = 900 kg m⁻³.
A balloon displaces 2.0 m³ of air of density 1.2 kg m⁻³. Calculate the upthrust.
U = ρVg = 1.2 × 2.0 × 9.81 = 23.5 N, approximately 24 N.
A uniform cylinder of volume 1.6 m³ floats vertically in water, with one-third of its height above the surface. Calculate the upthrust.
V submerged = (2/3) × 1.6 = 1.067 m³. U = 1000 × 1.067 × 9.81 = 1.0464 × 10⁴ N, approximately 1.05 × 10⁴ N.
What does viscosity describe? Give its symbol and SI unit.
A fluid’s resistance to flow or relative motion between its layers. Symbol η (eta). Unit Pa s, also expressible as kg m⁻¹ s⁻¹.
A sphere has radius 1.0 × 10⁻³ m and moves at 0.020 m s⁻¹ through a fluid of viscosity 0.50 Pa s. Calculate the drag, assuming Stokes’ law applies.
F = 6π × 0.50 × 1.0 × 10⁻³ × 0.020 = 1.88 × 10⁻⁴ N.
Convert the density of gold, 19.3 g cm⁻³, into kg m⁻³.
1.93 × 10⁴ kg m⁻³. Multiply the value in g cm⁻³ by 1000.
A fully submerged object displaces 3.0 × 10⁻³ m³ of water. What upthrust does it experience? If the same object is fully submerged in oil of density 800 kg m⁻³, what is the new upthrust?
Water: U = 1000 × 3.0 × 10⁻³ × 9.81 = 29.4 N. Oil: U = 800 × 3.0 × 10⁻³ × 9.81 = 23.5 N.
A solid sphere of volume 0.080 m³ and density 1200 kg m⁻³ rests fully submerged on the base of a water tank. Water acts over the submerged surface, so take U = ρwaterVg. Calculate the normal reaction from the base.
R + U = W. R = (1200 − 1000) × 0.080 × 9.81 = 157 N.
How does increasing temperature usually affect the viscosity of a liquid and a gas?
Liquid viscosity generally decreases, while gas viscosity generally increases. A decrease does not by itself establish inverse proportionality.
A sphere experiences viscous drag F while moving through a fluid. Its radius doubles and its speed triples. The fluid’s viscosity stays constant and Stokes’ law remains valid. Express the new drag in terms of F.
6F, because drag is proportional to rv and 2 × 3 = 6.
A sphere moves through a liquid at constant speed. The liquid is heated so its viscosity falls to half its original value. If the sphere’s speed and radius stay unchanged and Stokes’ law remains valid, what happens to the viscous drag? Explain.
The drag halves. F = 6πηrv, so drag is directly proportional to viscosity when radius and speed are constant.
A sphere of radius 2.0 × 10⁻³ m moves at 0.030 m s⁻¹ through a fluid and experiences drag 1.13 × 10⁻³ N. Calculate the viscosity, assuming Stokes’ law applies.
η = F/(6πrv) = (1.13 × 10⁻³)/(6π × 2.0 × 10⁻³ × 0.030) = 1.00 Pa s.
A small sphere of radius 1.0 × 10⁻³ m and density 2500 kg m⁻³ falls through a fluid of density 1000 kg m⁻³ and viscosity 0.50 Pa s. Using Stokes’ law and sphere volume V = (4/3)πr³, calculate its terminal speed.
At terminal speed, (ρsphere − ρfluid)Vg = 6πηrv. Hence v = 2r²g(ρsphere − ρfluid)/(9η) = [2 × (1.0 × 10⁻³)² × 9.81 × 1500]/(9 × 0.50) = 6.54 × 10⁻³ m s⁻¹ downwards.
A ball of mass 10 kg is released from rest while fully submerged in water. Its initial acceleration is 4.0 m s⁻² downwards. Neglect drag at release. Calculate its volume.
mg − U = ma, so U = 10 × (9.81 − 4.0) = 58.1 N. V = U/(ρg) = 58.1/(1000 × 9.81) = 5.92 × 10⁻³ m³.
A cylinder of volume 1.2 m³ and density 650 kg m⁻³ is held fully submerged in water by a light string attached to the tank floor. Calculate the string tension and the cylinder’s initial acceleration if the string snaps. Neglect drag at the instant it snaps.
T = U − W = (1000 − 650) × 1.2 × 9.81 = 4120 N. Mass = 650 × 1.2 = 780 kg. After the string snaps, a = (U − W)/m = 4120.2/780 = 5.28 m s⁻² upwards.