Practise › Questions › Fluids: pressure, upthrust and viscosity
Fluids: pressure, upthrust and viscosity questions
Liquids and gases exert pressure on everything within them, and the pressure grows with depth. That explains flotation through Archimedes' principle, and a second property, viscosity, explains why objects fall slowly through a fluid. Between them they determine what floats, what sinks, and how fast a sphere falls through a liquid.
20 original questions · 61 marks · the fluids: pressure, upthrust and viscosity notes · Materials
On this topic the library is tagged for CIE, Edexcel, OCR A.
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening a mark scheme: the schemes award marks point by point, and the marks are easier to see when you have something of your own to compare against.
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.
Mark scheme
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).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).
Mark scheme
p = ρgh = 1000 × 9.81 × 2.5 (1)
p = 24525 Pa ≈ 2.5 × 104 Pa (1)State what is meant by upthrust and give its size in terms of the fluid displaced.
Mark scheme
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).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.
Mark scheme
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).Distinguish between laminar flow and turbulent flow.
Mark scheme
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).A stone of volume 3.0 × 10⁻⁴ m³ is fully submerged in water. Calculate the upthrust on it.
Mark scheme
Upthrust = weight of water displaced (1)
= ρVg = 1000 × 3.0 × 10−4 × 9.81 (1)
= 2.9 N (1)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.
Mark scheme
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)Explain, in terms of pressure, why the upthrust on a submerged object acts upwards.
Mark scheme
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).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.
State the conditions under which Stokes' law applies to the drag on a moving sphere.
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.)
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.)
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.)
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.
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.
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.
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.
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.)
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.
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.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise fluids: pressure, upthrust and viscosity one question at a time
The player marks nothing for you. It shows one question, waits, then shows the scheme so you can mark yourself, and brings a question back sooner when it went badly.