Unit notes and practice questions
Everything here is for unit 3 of the Grade 12 Physics textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.
The cards filed under this chapter on the Grade 12 Physics page. Open one at a time.
When you use it
Use to convert between pressure gauge readings and the true absolute pressure relative to complete vacuum.
Watch out
Pressure measuring gauges read zero at atmospheric pressure. Never substitute gauge pressure directly into the ideal gas equation without adding atmospheric pressure.
Drafted from Grade 12 Physics, pages 69-117, then checked twice before it went up
When you use it
Use to determine the upward buoyant force acting on any partially or completely submerged object in a fluid.
Watch out
Use the density of the fluid, not the density of the object, to calculate the buoyant force.
Drafted from Grade 12 Physics, pages 69-117, then checked twice before it went up
When you use it
Use to calculate the fraction or percentage of a floating object submerged beneath the surface of a fluid.
Watch out
This equation applies only to floating objects in static equilibrium, where the object density is less than the fluid density.
Drafted from Grade 12 Physics, pages 69-117, then checked twice before it went up
When you use it
Use when finding the force or piston area in hydraulic lifts, brakes, and presses where confined fluid transmits pressure equally.
Watch out
Make sure the areas A_1 and A_2 are in the same units before calculating. If diameters are given, use area proportional to diameter squared.
Drafted from Grade 12 Physics, pages 69-117, then checked twice before it went up
When you use it
Use to calculate total hydrostatic pressure at any depth below the open surface of a static liquid.
Watch out
Always add atmospheric pressure P_0 when asked for absolute pressure. If only gauge pressure is required, omit P_0.
Drafted from Grade 12 Physics, pages 69-117, then checked twice before it went up
10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.
Question 1
What is the SI unit of pressure?
Answer: A Pascal (Pa)
Why
In the SI system, pressure is measured in newtons per square meter (N/m2), which is defined as the pascal (Pa).
Question 2
For a fluid at rest in fluid statics, what is the value of the shear (tangential) stress?
Answer: B It is zero
Why
In a fluid at rest, there is no relative motion between adjacent fluid layers. Therefore, the shear stress is zero, and the only existing stress is the normal stress (pressure).
Question 3
According to Pascal's principle, what happens when an external pressure is applied to an enclosed static fluid?
Answer: B The pressure change is transmitted undiminished to every point in the fluid
Why
Pascal's principle states that a change in pressure applied to a static enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of its container.
Question 4
According to Archimedes' principle, the upward buoyant force acting on an object placed in a fluid equals the:
Answer: A Weight of the fluid displaced by the object
Why
Archimedes' principle states that the buoyant force on an object equals the weight of the fluid it displaces (FB = Wfluid).
Question 5
A mechanic in Addis Ababa measures the tire gauge pressure of a minibus to be 32.0 psi. If the local atmospheric pressure is 14.3 psi, what is the absolute pressure inside the tire?
Answer: C 46.3 psi
Why
Absolute pressure is found by adding atmospheric pressure to gauge pressure: Pabs = Patm + Pgage = 14.3 psi + 32.0 psi = 46.3 psi.
Question 6
A hydraulic press in a garage in Hawassa has a small piston with an area of 0.0020 m2 and a large piston with an area of 0.20 m2. If a total downward force of 18,000 N acts on the large piston, what minimum force must be applied to the small piston to lift the load?
Answer: A 180 N
Why
From Pascal's principle, F1 = (A1 / A2) × F2 = (0.0020 m2 / 0.20 m2) × 18,000 N = 0.01 × 18,000 N = 180 N.
Question 7
What is the gauge pressure in water at a depth of 25.0 m? (Take the density of water as 1.00 × 103 kg/m3 and g = 9.8 m/s2).
Answer: A 2.45 × 10<sup>5</sup> Pa
Why
The gauge pressure due to depth is P = ρgh = (1.00 × 103 kg/m3)(9.8 m/s2)(25.0 m) = 245,000 Pa = 2.45 × 105 Pa.
Question 8
Why does atmospheric pressure decrease as a person climbs to higher altitudes on a mountain?
Answer: B The depth and density of the air column above both decrease
Why
Atmospheric pressure is caused by the weight of the air column above. At higher altitudes, there is less depth of atmosphere above you and the air is less dense, so the atmospheric pressure decreases.
Question 9
A rectangular wooden block floats in Lake Tana with 75% of its total volume submerged in fresh water (ρ = 1000 kg/m3). What is the density of the wooden block?
Answer: B 750 kg/m<sup>3</sup>
Why
For a floating object, Vdisp / Vobj = ρobj / ρfluid. Therefore, ρobj = 0.75 × 1000 kg/m3 = 750 kg/m3.
Question 10
An open-end manometer is connected to a gas container. The fluid has a density of 850 kg/m3, and the liquid column in the open arm is 0.55 m higher than the arm connected to the gas. If atmospheric pressure is 96.0 kPa and g = 9.81 m/s2, what is the absolute pressure of the gas?
Answer: C 100.6 kPa
Why
When the open arm is higher, Pgas = Patm + ρgh. First, ρgh = (850 kg/m3)(9.81 m/s2)(0.55 m) = 4586 Pa = 4.6 kPa. Thus, Pgas = 96.0 kPa + 4.6 kPa = 100.6 kPa.
The app holds 12 questions on this unit, marked as you go. These are the first 10.
Unit 3 runs from page 69 to page 117 of the Ministry of Education edition.
Ministry of Education edition, English, 186 pages
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