ParhaiKrlo
9th Class Physics · Punjab Board · Chapter 6

Chapter 6: Mechanical Properties of MatterLong Questions with Answers

11 long (comprehensive) questions of 9th Class Physics Chapter 6, Mechanical Properties of Matter, answered in full with headings, diagrams and tables, including important board-paper questions beyond the book. Punjab Board new syllabus.

Chapter 6· Mechanical Properties of Matter

Long Questions with Answers

Long answers, explained in full with headings. Learn the structure: definition, explanation, diagram or table, example.

Set A: Textbook comprehensive questions 5 questions

The long questions at the end of the chapter in your book.

Long Q 6.1

What is Hooke's law? Give three applications of this law.

Answer

Statement

Definition
Hooke's law

Within the elastic limit, the extension (or compression) of a spring is directly proportional to the applied force.

F ∝ x, so F = kx
Definition
Spring constant

k = F ÷ x, the force per unit extension; SI unit N m⁻¹.

Graph

extension xforce Felastic limitF = kx (slope = k)beyond: permanent stretchΔxΔF
F against x is a straight line up to the elastic limit.

Applications

  • Spring balance: the extension of the spring is a measure of the weight hung on it, read by a pointer on a scale. In a compression balance the load squeezes the spring instead.
  • Balance wheel of a mechanical clock: a fine spring makes the wheel swing back and forth regularly, controlling the speed of the hands.
  • Galvanometer: a hair spring brings the pointer back to zero; the deflection is proportional to the current.
0 N1 N2 N3 N4 N5 Nloadpointer reads 3 Nextension ∝ weight (Hooke's law)spring
Spring balance: extension ∝ weight.
Worked exampleA spring stretches 20 mm under 40 N
  1. k = 40 ÷ 0.02 = 2000 N m⁻¹.

Answerso a 16 mm stretch means 2000 × 0.016 = 32 N

Long Q 6.2

Describe the working and applications of a simple mercury barometer and a manometer.

Answer

Mercury barometer: construction

A glass tube about 1 m long, closed at one end, is filled with mercury and turned upside down in a dish of mercury. The mercury falls until its column is about 76 cm high, leaving a vacuum above it. A scale beside the tube measures the height.

vacuumglass tubemercuryh = 76 cmABair pressurepressure at A = pressure at B,so atmospheric pressure = ρgh
Simple mercury barometer.

Working

Point A in the tube and point B on the open surface are at the same level, so they have the same pressure. At B the pressure is atmospheric; at A it is the pressure of the mercury column. So atmospheric pressure = ρgh. At sea level, with P = 1.013 × 10⁵ Pa and ρ = 13.6 × 10³ kg m⁻³, h is about 760 mm.

Uses of the barometer

  • Measuring atmospheric pressure, for weather forecasting.
  • Finding altitude, because pressure falls with height.

Manometer

A U-tube containing mercury. With both ends open, the levels are equal. A gas supply is connected to one arm and the other is left open.

gasair pressureh(a) gas pressure = atmospheric + ρghgasair pressureh(b) gas pressure = atmospheric − ρgh
(a) Gas pressure more than atmospheric. (b) Gas pressure less than atmospheric.
  • If the mercury in the gas arm is lower (fig. a), the gas pressure is more than atmospheric by ρgh.
  • If it is higher (fig. b), the gas pressure is less than atmospheric by ρgh.
Conclusion

A barometer measures atmospheric pressure; a manometer measures the pressure of a gas compared with it.

Long Q 6.3

Describe Pascal's law. State its applications with examples.

Answer

Statement

Definition
Pascal's law

Pressure applied at one point of an enclosed fluid is transmitted equally, without loss, to all parts of the fluid.

Hydraulic press

F₁loadF₂A₁A₂small piston, area A₁large piston, area A₂liquid (oil)F₁ ÷ A₁ = F₂ ÷ A₂
Small piston A₁, large piston A₂, joined by a pipe full of liquid.
  1. A force F₁ on the small piston gives a pressure P = F₁ ÷ A₁.
  2. This pressure reaches the large piston unchanged, so F₂ = P × A₂ = F₁ × (A₂ ÷ A₁).
  3. Since A₂ > A₁, F₂ > F₁: the press is a force multiplier.
F₂ = F₁ × A₂ ÷ A₁

Cotton bales or metal sheets placed on the large piston are pressed against a fixed plate. The same idea lifts cars at service stations.

Hydraulic brakes

disclarger (wheel) cylinderbrake pads press the discbrake line (oil)brake fluidmaster cylinder(smaller piston)brake pedal
Brake pedal, master cylinder, brake line and wheel cylinder.
  1. The brake pedal pushes the small piston of the master cylinder, which is full of brake oil.
  2. The pressure travels through the brake lines to the larger pistons of the wheel cylinders.
  3. These push the brake pads against the discs (or drums); friction slows the wheels.
  4. When the pedal is released, springs pull the pads back.
Conclusion

Because the pressure is the same everywhere, all four wheels are braked equally.

Long Q 6.4

On what factors does the pressure of a liquid in a container depend? How is it determined?

Answer

Liquid pressure

A liquid presses on everything in it, in all directions, and the pressure increases with depth.

Diagram

Amghliquid surfacedensity ρmass of column m = ρAh, so P = mg ÷ A = ρgh
Liquid column of height h on area A.

Derivation

  1. Consider an area A at depth h. The liquid above it forms a column of volume V = Ah.
  2. Mass of the column: m = ρV = ρAh.
  3. Force on A = weight of the column: F = mg = ρAhg.
  4. Pressure: P = F ÷ A = ρAhg ÷ A.
P = ρgh

Factors

  • Depth h: the deeper the point, the greater the pressure.
  • Density ρ: a denser liquid gives more pressure at the same depth.
  • g: the gravitational field strength.

It does not depend on the shape of the container or the area of the surface.

Worked exampleSea water (1030 kg m⁻³) at 10 m
  1. P = 1030 × 10 × 10.

Answer1.03 × 10⁵ Pa

Long Q 6.5

Explain that the atmosphere exerts pressure. What are its applications? Give at least three examples.

Answer

Atmospheric pressure

The Earth is covered by a layer of air. Air molecules move randomly and keep striking every surface; these collisions produce a force on each unit of area, which is atmospheric pressure. Because the motion is random, it acts equally in all directions. At sea level it is about 1.013 × 10⁵ Pa.

Experiment: the collapsing can

1. heat: steam drives out the airseal it andpour cold water2. steam condenses: low pressure inside,atmospheric pressure crushes the can
Heating, sealing and cooling a can.
  1. Boil a little water in a tin can so that steam drives out the air.
  2. Close the can tightly and cool it with cold water.
  3. The steam condenses, so the pressure inside becomes very low.
  4. The atmospheric pressure outside, now much larger, crushes the can.

Applications

  • Barometer: measures atmospheric pressure by the height of mercury it holds up.
  • Weather forecasting: changes in pressure signal changes in the weather.
  • Altimeter: pressure falls with height, so it tells the altitude.
  • Tyre pressure gauge: measures the air pressure in tyres.
  • Straws and syringes: atmospheric pressure pushes the liquid in when the pressure inside is lowered.
Conclusion

We do not feel the atmosphere crushing us because the pressure of the fluids inside our bodies balances it.

Set B: More important long questions 6 questions

Long questions often set in board papers from this chapter, answered in the same full style.

Imp. LQ 1

Define density. How would you find the density of (a) a regular solid (b) an irregular solid (c) a liquid?

Answer

Definition

Definition
Density

Mass per unit volume: ρ = m ÷ V. SI unit kg m⁻³.

(a) Regular solid

  1. Measure the mass on a balance.
  2. Measure the length, breadth and height with a ruler or vernier callipers; volume = l × b × h.
  3. ρ = m ÷ V.

(b) Irregular solid

  1. Measure the mass on a balance.
  2. Put water in a measuring cylinder and note the level V₁.
  3. Lower the solid in on a thread and note the new level V₂.
  4. Volume = V₂ − V₁; ρ = m ÷ (V₂ − V₁).

(c) Liquid

  1. Find the mass of an empty beaker.
  2. Pour in a measured volume of liquid from a measuring cylinder and find the new mass.
  3. Mass of liquid = difference; ρ = mass ÷ volume.
Worked exampleA 60 g stone raises the water from 40 cm³ to 44 cm³
  1. V = 4 cm³; ρ = 60 ÷ 4 = 15 g cm⁻³.

Answer15 × 10³ kg m⁻³

Imp. LQ 2

What is pressure? Explain how pressure depends on area, with examples where it is increased and where it is decreased.

Answer

Definition

Definition
Pressure

The force acting normally per unit area: P = F ÷ A. SI unit: pascal (1 Pa = 1 N m⁻²). It is a scalar.

Pressure and area

For the same force, P ∝ 1/A: a smaller area gives a larger pressure.

on 5 × 10 cm faceP = 1.0 × 10⁴ Pa (largest)on 5 × 20 cm faceP = 5.0 × 10³ Paon 10 × 20 cm faceP = 2.5 × 10³ Pa (smallest)
The same brick on its three faces.

Where a large pressure is wanted

  • Knives, needles and nails are sharp.
  • Football boots have studs; deer have small hooves.

Where a small pressure is wanted

  • School bags have wide straps.
  • Tractors have broad tyres; tanks have tracks.
  • Elephants have large feet; foundations of buildings are wide.
Worked exampleA 50 N brick on faces of 50 cm² and 200 cm²
  1. Smallest face: 50 ÷ (50 × 10⁻⁴) = 1.0 × 10⁴ Pa.
  2. Largest face: 50 ÷ (200 × 10⁻⁴) = 2.5 × 10³ Pa.

Answerthe same brick gives four times the pressure on its smallest face

Imp. LQ 3

Describe an experiment to verify Hooke's law and to find the spring constant of a spring.

Answer

Apparatus

  • A spring hung from a stand
  • A metre rule fixed beside it
  • A pointer at the bottom of the spring
  • Slotted weights of 1 N each

Method

  1. Note the position of the pointer with no load.
  2. Add 1 N and note the new reading; the difference is the extension x.
  3. Add weights one by one, noting x each time.
  4. Remove the weights one by one and check the pointer returns to the same readings (the elastic limit has not been passed).
  5. Plot force F against extension x.

Result

extension xforce Felastic limitF = kx (slope = k)beyond: permanent stretchΔxΔF
A straight line through the origin verifies Hooke's law.

The graph is a straight line through the origin, so F ∝ x: Hooke's law is verified. Its slope gives k = F ÷ x.

Precautions

  • Do not overload the spring beyond its elastic limit.
  • Read the scale at eye level to avoid parallax.
  • Let the spring stop moving before taking a reading.
Imp. LQ 4

Explain the hydraulic brake system of a car with a labelled diagram.

Answer

Principle

Hydraulic brakes work on Pascal's law: pressure applied to an enclosed liquid is transmitted equally to all parts of it.

Diagram

disclarger (wheel) cylinderbrake pads press the discbrake line (oil)brake fluidmaster cylinder(smaller piston)brake pedal
Pedal, master cylinder, brake line, wheel cylinder and brake pads.

Parts

  • Brake pedal connected to the small piston of the master cylinder.
  • Brake lines (pipes) full of brake oil.
  • Wheel cylinders with larger pistons, attached to the brake pads.

Working

  1. The driver presses the pedal; the master cylinder piston pushes on the oil.
  2. The pressure is transmitted equally through the lines to every wheel cylinder.
  3. The larger pistons push the brake pads against the discs with a large force.
  4. Friction slows and stops the wheels. Releasing the pedal lets springs pull the pads back.

Advantages

  • All wheels are braked equally at the same moment.
  • A small push on the pedal gives a large braking force.
Conclusion

Air in the brake lines is dangerous: air is compressible and would squash instead of passing on the pressure.

Imp. LQ 5

Why is mercury used in a barometer? Find the height of a water barometer, and explain the effect of trapped air or a leak.

Answer

Why mercury

  • It is very dense (13.6 × 10³ kg m⁻³), so the column is only about 76 cm tall.
  • It does not stick to glass and is easy to see.
  • It hardly evaporates, so the space above stays a near vacuum.

A water barometer

Worked exampleSame atmospheric pressure, water instead of mercury
  1. ρ(water) × g × h(water) = ρ(mercury) × g × h(mercury).
  2. h(water) = 13 600 × 0.76 ÷ 1000.

Answerabout 10.3 m, far too tall to be practical

Trapped air

Air above the mercury pushes down on it, so the column is shorter and the reading is too low.

A leak

Air keeps entering until the pressure above the mercury equals atmospheric pressure; the mercury then falls to the dish level and the barometer stops working.

vacuumglass tubemercuryh = 76 cmABair pressurepressure at A = pressure at B,so atmospheric pressure = ρgh
A correct barometer has a vacuum above the mercury.
Imp. LQ 6

Explain springs joined in series and in parallel. Derive their spring constants.

Answer

Single spring

A load L stretches one spring by x, so its spring constant is k = L ÷ x.

Diagram

Lone springextension x, constant kLseriesextension 2x, constant k/2Lparallelextension x/2, constant 2k
One spring, series and parallel arrangements.

Series

  1. Each spring carries the whole load L, so each stretches by x.
  2. Total extension = 2x.
  3. k(series) = L ÷ 2x = k/2.

Parallel

  1. The load is shared equally, L/2 on each.
  2. Each, and so the pair, stretches by x/2.
  3. k(parallel) = L ÷ (x/2) = 2k.
ExtensionSpring constantBehaviour
Series2xk/2Softer
Parallelx/22kStiffer
Conclusion

Car suspensions use springs side by side (parallel) to carry heavy loads with little stretch.

End of Chapter 6 Long QuestionsNext: Chapter 6 Numericals →
6Chapter
Full guide: Mechanical Properties of Matter (notes, diagrams, all questions, self-test)
›
1Chapter
Physical Quantities and Measurements Long Questions
›
2Chapter
Kinematics Long Questions
›
3Chapter
Dynamics Long Questions
›
4Chapter
Turning Effect of Forces Long Questions
›
5Chapter
Work, Energy and Power Long Questions
›
7Chapter
Thermal Properties of Matter Long Questions
›
8Chapter
Magnetism Long Questions
›
9Chapter
Nature of Science Long Questions
›