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Lesson 1 Questions: Density, Liquid Pressure and Pascal Principle

Physics

Lesson 1 Questions: Density, Liquid Pressure and Pascal Principle

مقدمة الدرس

مقدمة الدرس

  • This lesson covers density, the pressure that a liquid exerts at a point inside it, and Pascal’s principle.
  • The questions and worked examples below apply these ideas to numbers and to a practical device.
  • By the end of this lesson, you should be able to:
  • Recognize density and use the relation between mass, volume and density.
  • Calculate the pressure at a point inside a liquid, including the effect of atmospheric pressure.
  • Solve problems related to density and pressure.
  • Explain Pascal’s principle and apply it to the hydraulic press.
Lesson 1 Questions: Density, Liquid Pressure and Pascal Principle · 1 / 6
المفاهيم والمصطلحات الأساسية

المفاهيم والمصطلحات الأساسية

Comparison aspectMain ideaKeywords
Density (ρ)The mass of a substance per unit volume, ρ = m / V. Its SI unit is kg/m3.masssubstanceunitvolume
Relative density, The ratio of the density of a material to the density of water at 4 °C. The density of water at 4 °C is 1000 kg/m3. Relative density has no unit.ratiomaterialwater1000
Pressure at a point, The average force acting perpendicularly on a unit area around that point, P = F / A. Its unit is the newton per square metre (N/m2).averageforceactingperpendicularly
Pascal (Pa)The SI unit of pressure, where 1 Pa = 1 N/m2.unitpressure
Pressure in a liquid at depth hThe pressure due to the liquid alone is P = hρg, where ρ is the liquid density and g is the acceleration due to gravity. If the free surface of the liquid is exposed to the air, the total pressure is P = Pa + hρg, where Pa is atmospheric pressure.alonedensityaccelerationgravity
Comparison aspectMain ideaKeywords
Pascal’s principlePressure applied to a confined liquid is passed on undiminished to every part of the liquid and to the walls of the container that holds itPressureappliedconfinedliquid
Hydraulic press
  • A device with a small piston of cross-sectional area a and a large piston of cross-sectional area A
  • with a liquid between them. A force on the small piston produces a larger force on the large piston
devicesmallpistoncross
Lesson 1 Questions: Density, Liquid Pressure and Pascal Principle · 2 / 6
الشرح والتفصيل العلمي الشامل

الشرح والتفصيل العلمي الشامل

Multiple choice

1.→Which factor does not affect the pressure at the bottom of a container filled with liquid?

(a) height of the liquid in the vessel

(b) density of the liquid

(c) area of the vessel’s base

(d) atmospheric pressure

2.→The water pressure at the bottom of a high dam lake, which affects the dam structure, depends on:

(a) surface area of the water in the lake

(b) length of the dam structure

(c) depth of the water

(d) thickness of the dam wall

3.→In a hydraulic press, the ratio of the radii of the two cylindrical pistons is 2 : 9. The ratio of the forces on the two pistons is:

(a) 2 : 9

(b) 9 : 2

(c) 81 : 4

(d) 4→ 81

Definitions

4.Define pressure at a point.
5.Define Pascal’s principle.

Essay questions

Comparison aspectMain ideaKeywords
6.Prove that the pressure P at any point within a liquid at depth h is given by P = Pa + ρgh, where Pa is atmospheric pressure, ρ is the liquid density and g is the acceleration due to gravity.Provepressurepointwithin
7.Describe the manometer and explain how it works in measuring the pressure of a gas in a reservoirDescribemanometerexplainworks
8.What factors determine the type of liquid used in a manometer for measuring the pressure difference between a gas in a reservoir and atmospheric pressure?Whatfactorsdeterminetype
9.What is meant by Pascal’s principle? Explain one of its applicationsWhatmeantPascalprinciple

Problems

10.→The pressure on the base of a cylindrical container of oil is 1.5 × 104 N/m2. Calculate the total force acting on the base if the base diameter is 8 m.
Comparison aspectMain ideaKeywords
11.A fish tank has a base area of 1000 cm2 and contains water weighing 4000 N. Calculate the water pressure at the base of the tank.fishtankbasearea
12.
  • In a hydraulic press
  • the small and large pistons have diameters of 2 cm and 24 cm. A force of 2000 N acts on the large piston. Calculate the force acting on the small piston and the mechanical advantage of the press
hydraulicpresssmalllarge
13.The pressure at the water surface of a lake is 1 atm, and the pressure at its bottom is 3 atm. Find the depth of the lake. Take the density of water as 1000 kg/m3 and g as 9.8 m/s2.pressurewatersurfacelake
Lesson 1 Questions: Density, Liquid Pressure and Pascal Principle · 3 / 6
المقارنات وخريطة المفاهيم الجوهرية

المقارنات وخريطة المفاهيم الجوهرية

Comparison aspectMain ideaKeywords
•The answers to the numerical questions are given belowanswersnumericalquestionsgiven
Question 107.54 × 105 N
Question 114 × 104 N/m2
Question 12
  • The force on the small piston is 13.9 N
  • and the mechanical advantage of the press is 144
forcesmallpistonmechanical
Question 13The depth of the lake is 20.673 mdepthlake
Lesson 1 Questions: Density, Liquid Pressure and Pascal Principle · 4 / 6
أمثلة ومسائل محلولة بالخطوات

أمثلة ومسائل محلولة بالخطوات

  • Example A pressure of a solid block
  • A solid rectangular block measures 5 cm × 10 cm × 20 cm and has a density of 5000 kg/m3.
  • It rests on a horizontal surface.
  • Take g = 10 m/s2.
  • Find the maximum and minimum pressure it exerts on the surface.
  • Step 1: weight. The volume is 5 × 10 × 20 cm3 = 1000 cm3 = 1 × 10-3 m3. The force is the weight of the block: F = V × ρ × g = 1 × 10-3 × 5000 × 10 = 50 N.
  • Step 2: maximum pressure. The block stands on its smallest face, 5 cm × 10 cm = 5 × 10-3 m2. P = F / A = 50 / (5 × 10-3) = 1 × 104 N/m2.
  • Step 3: minimum pressure. The block lies on its largest face, 10 cm × 20 cm = 0.02 m2. P = 50 / 0.02 = 2500 N/m2.
  • Example B total pressure at the bottom of a tank
  • A tank contains salt water of density 1030 kg/m3 to a depth of 1 m.
  • The base area is 1000 cm2, the water surface is exposed to the air, atmospheric pressure is 1.013 × 105 N/m2, and g = 10 m/s2.
  • Find the total pressure and the total force on the base.
  • Total pressure P = Pa + ρgh = 1.013 × 105 + 1030 × 10 × 1 = 1.013 × 105 + 0.103 × 105 = 1.116 × 105 N/m2.
  • Total force The base area is 1000 cm2 = 0.1 m2. F = P × A = 1.116 × 105 × 0.1 = 1.116 × 104 N.
  • Example C hydraulic press
  • The small piston of a hydraulic press has a cross-sectional area of 10 cm2, and a force of 100 N acts on it.
  • The large piston has an area of 800 cm2.
  • Take g = 10 m/s2.
(A) Force on the large pistonF = f × A / a = 100 × 800 / 10 = 8 × 103 N.
(B) Mechanical advantageη = F / f = A / a = 800 / 10 = 80.
(C) Distance moved by the small pistonFor the large piston to rise 1 cm, the small piston must move y1 = F × y2 / f = 8000 × 1 / 100 = 80 cm.
Lesson 1 Questions: Density, Liquid Pressure and Pascal Principle · 5 / 6
تريكات الامتحان وكبسولة التفوق الوزاري

تريكات الامتحان وكبسولة التفوق الوزاري

  • Write every area in m2 before using P = F / A. One cm2 equals 1 × 10-4 m2, so 1000 cm2 = 0.1 m2.
  • Remember the pressure units 1 Pa = 1 N/m2, 1 bar = 105 Pa, 1 atm = 760 mm Hg = 1.013 × 105 Pa, and 1 Torr = 1 mm Hg.
  • If the liquid surface is exposed to the air, add atmospheric pressure: P = Pa + hρg. The term hρg alone is the pressure due to the liquid.
  • Points at the same horizontal level in the same liquid have the same pressure.
  • Pressure increases with depth and with the density of the liquid.
  • In connected vessels with a horizontal base, the liquid rises to the same level whatever the shape of the vessels.
  • Pressure depends on area as well as force. A pointed heel can exert more pressure than an elephant’s foot because its area is much smaller.
  • Relative density has no unit, because it is a ratio of two densities.
  • In a hydraulic press, use the cross-sectional area of each piston. The mechanical advantage is F / f = A / a, and the small piston moves a greater distance than the large piston.
Lesson 1 Questions: Density, Liquid Pressure and Pascal Principle · 6 / 6