Lesson 4.1.2
4.1.2 Viscous drag and Stokes' law Quiz: Pearson Edexcel Physics, Unit 4
20 questions
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Lesson 4.1.2, Viscous drag and Stokes' law: 20 multiple choice questions for the Pearson Edexcel Physics (9PH0), Unit 4: Materials, written with Revision Ninja.
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The 20 questions
-
Which equation is Stokes' law for the viscous drag on a small sphere?
- F = 6 pi eta r v
- F = 3 pi eta r v^2
- F = 6 pi eta r^2 v
- F = 4 pi eta r v^2
-
Under which conditions does Stokes' law apply?
- Objects of any shape moving at any speed
- Small spherical objects moving at low speeds in laminar flow
- Large spherical objects moving at high speeds in turbulent flow
- Objects moving at speeds above the speed of sound in the fluid
-
What is meant by the viscosity of a liquid?
- Its ability to conduct electric current
- The upthrust acting on a body placed in it
- Its resistance to flow, arising from internal friction between layers of the liquid
- Its density per unit volume
-
How does the viscosity of a liquid change as its temperature increases?
- It stays the same
- It decreases
- It increases in direct proportion to the absolute temperature
- It increases
-
A sphere falling through a viscous fluid reaches terminal velocity. What is true at that point?
- Its acceleration is at a maximum
- The net force on it is zero, so drag plus upthrust balances its weight
- The drag force on it is zero
- Its weight is zero
-
What is the SI unit of viscosity?
- N m
- Pa s
- kg m s^-1
- m s^-2
-
A small sphere of radius r and density rho_s falls at terminal velocity v through a fluid of density rho_f and viscosity eta. Which expression gives v?
- v = 2 r^2 (rho_s - rho_f) g/(9 eta)
- v = 9 r^2 (rho_s - rho_f) g/(2 eta)
- v = 2 r (rho_s - rho_f) g/(9 eta)
- v = 2 r^2 (rho_s + rho_f) g/(9 eta)
-
A sphere of radius 2.0 mm moves at 0.050 m/s through a liquid of viscosity 0.80 Pa s. What is the viscous drag on it?
- 3.0 x 10^-2 N
- 1.5 x 10^-3 N
- 6.0 x 10^-4 N
- 3.0 x 10^-3 N
-
A steel sphere of radius 1.0 mm (density 2700 kg/m^3) falls through a liquid of density 1000 kg/m^3 and viscosity 1.0 Pa s. What is its terminal velocity? Take g = 9.8 N/kg.
- 7.4 x 10^-3 m/s
- 3.7 x 10^-1 m/s
- 1.9 x 10^-2 m/s
- 3.7 x 10^-3 m/s
-
If the radius of a sphere falling at terminal velocity in a fixed fluid is doubled, by what factor does its terminal velocity change?
- 4 times greater
- 2 times greater
- 8 times greater
- Unchanged
-
At terminal velocity a sphere has weight 0.050 N and upthrust 0.020 N. What is the drag force on it?
- 0.030 N
- 0.070 N
- 0.020 N
- 0.050 N
-
A sphere falls at a constant terminal velocity of 0.20 m/s. How long does it take to fall 0.50 m at this speed?
- 0.40 s
- 0.10 s
- 10 s
- 2.5 s
-
In a falling-ball experiment, a sphere of radius 0.50 mm, with density difference 1500 kg/m^3 from the liquid, falls at terminal velocity 0.0040 m/s. What is the viscosity of the liquid? Take g = 9.8 N/kg.
- 0.020 Pa s
- 0.20 Pa s
- 0.0020 Pa s
- 2.0 Pa s
-
The drag on a sphere moving at 0.10 m/s in a fluid of viscosity 0.10 Pa s is 0.00377 N. What is the radius of the sphere?
- 2.0 cm
- 20 m
- 0.20 cm
- 2.0 mm
-
A sphere moving slowly through a fluid has drag 0.0030 N at speed 0.10 m/s. If its speed doubles to 0.20 m/s, staying in laminar flow, what is the new drag?
- 0.0030 N
- 0.0120 N
- 0.0060 N
- 0.0015 N
-
A student uses Stokes' law to calculate drag on a large raindrop falling at 10 m/s through air. What is the best evaluation?
- Stokes' law applies, because air is a viscous fluid at any speed
- Stokes' law applies, but only with a different constant for air
- Stokes' law does not apply, because the drop is not small and the flow at this speed is turbulent
- Stokes' law applies, because raindrops are spherical
-
A steel ball of radius 1.0 mm falls at constant speed 0.40 m/s in a liquid. Its density difference from the liquid is 6600 kg/m^3. What is the viscosity of the liquid? Take g = 9.8 N/kg.
- 3.6 Pa s
- 0.0036 Pa s
- 0.36 Pa s
- 0.036 Pa s
-
A student says that a falling sphere's terminal velocity depends only on its mass. Which evaluation is correct?
- The claim is right, because heavier objects always fall faster through any fluid
- The claim is wrong, because terminal velocity does not depend on the fluid at all
- The claim is right, because drag depends only on the mass of the sphere
- The claim is wrong, because terminal velocity depends on radius, density difference and the fluid's viscosity
-
A sphere of radius 0.50 mm and density 2500 kg/m^3 falls through oil of density 900 kg/m^3 and viscosity 0.50 Pa s. What is its terminal velocity? Take g = 9.8 N/kg.
- 0.17 mm/s
- 17 mm/s
- 3.5 mm/s
- 1.7 mm/s
-
Why must temperature be controlled in a falling-ball viscosity experiment?
- Stokes' law is valid only at one fixed temperature for every fluid
- Temperature changes the mass of the ball, so its weight changes during the fall
- The viscosity of a liquid depends strongly on temperature, so the terminal velocity and calculated viscosity would change
- Temperature changes the value of g in the laboratory
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