Questions on fluids: pressure, upthrust and viscosity

Question 1

State the equation for the pressure due to a column of fluid, defining each symbol, and state one quantity the pressure does not depend on.

Accepted Answer

p = ρgh, where ρ is the density of the fluid, g the gravitational field strength and h the depth below the surface (1). It does not depend on the shape or width of the container, nor on the total volume of fluid (1).

Question 2

Calculate the pressure due to the water at the bottom of a swimming pool 2.5 m deep (density of water 1000 kg m⁻³).

Accepted Answer

p = ρgh = 1000 × 9.81 × 2.5 (1) p = 24525 Pa ≈ 2.5 × 10⁴ Pa (1)

Question 3

State what is meant by upthrust and give its size in terms of the fluid displaced.

Accepted Answer

Upthrust is the upward force a fluid exerts on a body in it, caused by pressure on the lower surface exceeding pressure on the upper surface (1). It equals the weight of fluid displaced by the body (1).

Question 4

State the condition, in terms of forces, for an object to float in a fluid, and the condition this places on the object's average density.

Accepted Answer

The upthrust must equal the object's weight, i.e. the object displaces its own weight of fluid before becoming fully submerged (1). This requires the object's average density to be less than (or equal to) the density of the fluid (1).

Question 5

Distinguish between laminar flow and turbulent flow.

Accepted Answer

Laminar flow is smooth flow in layers that do not mix (1). Turbulent flow is chaotic, with eddies and mixing between layers, and occurs above a certain flow speed (1).

Question 6

A concrete block of weight 750 N rests on level ground on a face of area 0.30 m². Calculate the pressure the block exerts on the ground.

Accepted Answer

p = F/A = 750/0.30 (1) = 2500 Pa, that is 2.5 × 10³ Pa (1).

Question 7

A stone of volume 3.0 × 10⁻⁴ m³ is fully submerged in water. Calculate the upthrust on it.

Accepted Answer

Upthrust = weight of water displaced (1) = ρVg = 1000 × 3.0 × 10⁻⁴ × 9.81 (1) = 2.9 N (1)

Question 8

A submarine cruises 30 m below the surface of sea water of density 1025 kg m⁻³. Calculate the pressure due to the water at this depth, and state what must be added to find the total pressure on the hull.

Accepted Answer

p = ρgh = 1025 × 9.81 × 30 (1) p = 3.02 × 10⁵ Pa (1) Atmospheric pressure acting on the sea surface must be added for the total (1)

Question 9

Explain, in terms of pressure, why the upthrust on a submerged object acts upwards.

Accepted Answer

Pressure in a fluid increases with depth, so the pressure on the object's lower surface is greater than on its upper surface (1). The upward push on the bottom exceeds the downward push on the top, leaving a net upward force (1).

Question 10

An iceberg of density 920 kg m⁻³ floats in fresh water of density 1000 kg m⁻³. Calculate the fraction of its volume below the surface.

Accepted Answer

Floating: ρ_iceV_wholeg = ρ_waterV_subg (1) V_sub/V_whole = 920/1000 = 0.92 (1) About 92% of the iceberg is submerged (1)

Question 11

State the conditions under which Stokes' law applies to the drag on a moving sphere.

Accepted Answer

The sphere must be small and moving slowly enough that the flow around it is laminar, with no turbulence (1), in a fluid of uniform viscosity (1).

Question 12

A cube of side 0.10 m is held fully submerged in fresh water (density 1000 kg m⁻³) with its top face horizontal at a depth of 0.50 m. Calculate the water pressure on the top face and on the bottom face, and hence the resultant force the water pressure exerts on the cube. (g = 9.81 m s⁻².)

Accepted Answer

Top face: p = ρgh = 1000 × 9.81 × 0.50 = 4905 Pa (1). Bottom face, at 0.60 m: p = 1000 × 9.81 × 0.60 = 5886 Pa (1). Resultant force = (pressure difference) × area = (5886 − 4905) × 0.10² (1) = 9.8 N upwards, which is exactly the upthrust, the weight of water displaced (1). (Equal pressures on the vertical faces cancel.)

Question 13

A flat rectangular pontoon of surface area 3.0 m² floats on a lake. When a person of weight 600 N steps aboard, the pontoon floats lower in the water. Calculate the extra depth by which it sinks. (Density of water 1000 kg m⁻³; g = 9.81 m s⁻².)

Accepted Answer

The extra upthrust must equal the person's weight: ρgAd = 600 (1). d = 600/(1000 × 9.81 × 3.0) (1) = 0.020 m ≈ 2.0 cm (1).

Question 14

A helium balloon has a total volume of 12 m³ and a total weight, including the helium and its load, of 90 N. The density of the surrounding air is 1.2 kg m⁻³. Calculate the upthrust on the balloon and the resultant force acting on it, and state what happens when it is released. (g = 9.81 m s⁻².)

Accepted Answer

Upthrust = weight of air displaced = ρVg = 1.2 × 12 × 9.81 = 141 N (1). Resultant force = 141 − 90 = 51 N upwards (1). The balloon accelerates upwards when released (1).

Question 15

The definition p = F/A serves a solid, a liquid and a gas alike. State what fixes the area used for a solid resting on a surface, and describe how the directions in which the pressure acts differ between a solid and a fluid.

Accepted Answer

For a solid the area is the contact patch between the solid and the surface it rests on, so for an unchanged force the geometry of that contact fixes the pressure; stiletto heels and snowshoes are the same weight over very different areas (1). A solid presses only on the surface it touches, and only along the direction in which the force is applied (1). A fluid presses at right angles on every surface it touches, and at any point inside it the pressure acts equally in all directions, sideways and upwards as much as downwards (1).

Question 16

A sphere of radius 1.5 mm moves at 0.12 m s⁻¹ through oil of viscosity 0.25 Pa s. Calculate the viscous drag on it.

Accepted Answer

F = 6πηrv (1) = 6π × 0.25 × 1.5 × 10⁻³ × 0.12 (1) = 8.48 × 10⁻⁴ N ≈ 8.5 × 10⁻⁴ N (1)

Question 17

A small sphere falls through glycerol of viscosity 0.85 Pa s. Its weight exceeds the upthrust on it by 1.8 × 10⁻⁴ N, and its radius is 1.0 mm. Calculate its terminal velocity.

Accepted Answer

At terminal velocity, drag = weight − upthrust = 1.8 × 10⁻⁴ N (1)
From Stokes' law, v = F/6πηr (1)
v = 1.8 × 10⁻⁴ / (6π × 0.85 × 1.0 × 10⁻³) (1)
v = 0.011 m s⁻¹ ≈ 1.1 × 10⁻² m s⁻¹ (1)

Question 18

Estimate the depth of fresh water that produces a pressure equal to one atmosphere (1.01 × 10⁵ Pa), and comment on what this means for a diver at 20 m.

Accepted Answer

h = p/ρg = 101000/(1000 × 9.81) (1)
h = 10.3 m ≈ 10 m (1)
Every 10 m of water adds about one atmosphere, so a diver at 20 m experiences roughly three atmospheres in total: two from the water plus one from the air above it (1)

Question 19

A student measures the viscosity of syrup by timing a ball bearing falling between two marks. Explain why the marks must be placed well below the surface, and why the temperature must be recorded.

Accepted Answer

The sphere must reach terminal velocity before the first mark, or the measured speed is not the steady Stokes' law speed (1); placing the marks deep allows the acceleration phase to finish (1). Viscosity falls steeply as temperature rises, so a result without its temperature cannot be compared with any other (1).

Question 20

A steel sphere of radius 0.60 mm (density 7800 kg m⁻³) falls at terminal velocity through oil of density 920 kg m⁻³ and viscosity 0.30 Pa s. Assume the flow around the sphere is laminar. Calculate the terminal velocity. (g = 9.81 m s⁻².)

Accepted Answer

V = (4/3)πr³ = (4/3)π × (0.60 × 10⁻³)³ = 9.05 × 10⁻¹⁰ m³ (1). Weight − upthrust = (ρ_steel − ρ_oil)Vg (1) = 6880 × 9.05 × 10⁻¹⁰ × 9.81 = 6.11 × 10⁻⁵ N (1). At terminal velocity the viscous drag equals this: 6πηrv = 6.11 × 10⁻⁵ N (1). v = 6.11 × 10⁻⁵/(6π × 0.30 × 0.60 × 10⁻³) = 0.018 m s⁻¹ (1).

Question 21

A food manufacturer requires its syrup to have a viscosity of at least 2.0 Pa s at 20 °C. In a quality-control test at 20 °C, a ball of radius 1.5 mm falls through a sample at a steady speed of 9.5 mm s⁻¹; its weight exceeds the upthrust on it by 5.7 × 10⁻⁴ N. Deduce whether the batch meets the requirement.

Accepted Answer

At steady (terminal) speed, viscous drag = weight − upthrust = 5.7 × 10⁻⁴ N (1). From Stokes' law, η = F/(6πrv) = 5.7 × 10⁻⁴/(6π × 1.5 × 10⁻³ × 9.5 × 10⁻³) (1) = 2.1 Pa s (1). 2.1 Pa s is above 2.0 Pa s, so the batch meets the requirement (1).

Question 22

A small steel sphere is released from rest at the surface of a deep column of glycerol. Describe and explain, in terms of the forces acting, the motion of the sphere from release until it falls at a steady speed.

Accepted Answer

Throughout the fall the sphere's weight acts downwards and a constant upthrust acts upwards; at release the sphere is at rest, so the viscous drag is zero (1). The resultant force (weight − upthrust) acts downwards, so the sphere accelerates downwards (1). As its speed increases, the viscous drag, given by Stokes' law F = 6πηrv, increases in proportion to the speed (1). The resultant force therefore decreases, so the acceleration decreases (1). Eventually weight = upthrust + drag, and the resultant force is zero (1). The acceleration is then zero and the sphere falls at a constant (terminal) velocity (1).