9th Class Physics · Punjab Board · Complete learning guide
Chapter 6Mechanical Properties of Matter
Every topic, every MCQ and every exercise question, explained step by step in simple language. Written for the Punjab Board new-syllabus Physics book, English medium.
An external force that changes the size or shape of a body, for example squeezing a sponge or stretching a rubber band.
Definition
Elasticity
The property of a solid by which it returns to its original size and shape when the deforming force is removed.
Definition
Elastic limit
The greatest deforming force (or stretch) a body can take and still return to its original shape. Beyond it, the change becomes permanent.
Definition
Inelastic (plastic) materials
Materials, such as clay, dough and plasticine, that do not return to their original shape when the deforming force is removed.
Elastic
Inelastic (plastic)
Rubber band, spring, steel wire
Plasticine, clay, wet dough
Regains its shape
Keeps its new shape
Check your book
In physics "more elastic" means harder to deform and better at returning to shape. So steel is more elastic than rubber, and water (very hard to compress) is more elastic than air.
Topic 2
Hooke's law
Definition
Hooke's law
Within the elastic limit, the extension (or compression) of a spring is directly proportional to the force applied to it.
F = kx
F = force, x = extension, k = spring constant
Definition
Spring constant (k)
The force needed per unit extension of a spring: k = F ÷ x. SI unit: N m⁻¹. A stiff spring has a large k.
Force against extension is a straight line through the origin up to the elastic limit; its slope is k.
Worked exampleA spring stretches 20 mm under a 40 N load. Find k.
x = 20 mm = 0.02 m.
k = F ÷ x = 40 ÷ 0.02.
Answer2000 N m⁻¹
Remember
Always change mm or cm to metres before using k = F/x in N m⁻¹.
Topic 3
Applications of Hooke's law
In a spring balance the stretch of the spring, read by a pointer, is a measure of the weight.
Application
How the spring is used
Spring balance (spring scale)
The extension of the spring is proportional to the weight hung on it, so a pointer on a scale shows the weight
Compression balance (kitchen scale)
The load compresses the spring; the compression turns a pointer over a dial
Balance wheel of a mechanical clock
A fine spring makes the wheel swing back and forth at a steady rate, controlling the hands
Galvanometer
A tiny hair spring carries current to the coil and brings the pointer back to zero; the deflection is proportional to the current
Springs are also used in car suspensions, mattresses, pens and door closers.
Exam tip
Springs are made of steel, not iron: steel is more elastic, so it can be stretched further and still return to its shape.
Topic 4
Springs in series and in parallel
Suppose a load L stretches one spring by x, so k = L ÷ x.
One spring, two in series, and two in parallel, each carrying the same load L.
Arrangement
Load on each spring
Total extension
Spring constant
One spring
L
x
k
Two in series
L (each carries the full load)
x + x = 2x
L ÷ 2x = k/2
Two in parallel
L/2 (load shared)
x/2
L ÷ (x/2) = 2k
Remember
Series makes a softer spring (half the k); parallel makes a stiffer one (double the k).
Topic 5
Density
Definition
Density
The mass per unit volume of a substance.
ρ = m ÷ V
SI unit kg m⁻³; also g cm⁻³ (1 g cm⁻³ = 1000 kg m⁻³)
Substance
Density (kg m⁻³)
Air
1.3
Ice
920
Water
1000
Sea water
1030
Aluminium
2700
Iron
7800
Mercury
13 600
Gold
19 300
Worked exampleAn iron block measures 3 cm × 2 cm × 2 cm and has a mass of 94 g. Find the density of iron. (Solved Example 6.1)
Volume = 3 × 2 × 2 = 12 cm³ = 12 × 10⁻⁶ m³.
Mass = 94 g = 0.094 kg.
ρ = m ÷ V = 0.094 ÷ (12 × 10⁻⁶).
Answer≈ 7.83 × 10³ kg m⁻³
Remember
Density belongs to the material, not to the object: a small iron nail and a large iron gate have the same density.
Exam tip
Engineers estimate the load of a building from the densities and volumes of its materials: mass = density × volume, weight = mass × g.
Topic 6
Pressure
Definition
Pressure
The force acting normally (at right angles) on unit area of a surface. It is a scalar.
P = F ÷ A
SI unit pascal: 1 Pa = 1 N m⁻²
The same force gives a large pressure on a small area and a small pressure on a large area.
The same brick presses hardest on its smallest face and least on its largest face.
Small area: large pressure
Large area: small pressure
Sharp knife, needle, nail, drawing pin
Wide straps on school bags
Studs on football boots for grip
Broad tyres on tractors, caterpillar tracks
Deer's small hooves grip the ground when running
Elephant's large feet protect its bones
Walking barefoot on pebbles hurts
Skis and snowshoes stop sinking into snow
Topic 7
Atmospheric pressure
Definition
Atmospheric pressure
The pressure the air of the atmosphere exerts on the Earth's surface and on everything on it. It acts equally in all directions. At sea level it is about 1.013 × 10⁵ Pa.
Air molecules move randomly and keep hitting every surface. These collisions push on the surface; the force per unit area is the atmospheric pressure.
The collapsing-can experiment
A can full of steam is sealed and cooled: the steam condenses, the pressure inside falls, and the air outside crushes the can.
Boil a little water in a tin can until it is full of steam.
Take it off the burner and close its mouth with an airtight cap.
Pour cold water over it: the steam condenses and leaves very low pressure inside.
The much larger atmospheric pressure outside crushes the can.
Remember
Atmospheric pressure decreases with height, because there is less air above.
Topic 8
Measuring atmospheric pressure: the mercury barometer
Definition
Barometer
An instrument that measures atmospheric pressure.
Mercury barometer: atmospheric pressure on the dish holds up a column of mercury about 760 mm high, with a vacuum above it.
Construction
Fill a glass tube about 1 m long, closed at one end, with mercury.
Close the open end with a finger, turn the tube upside down, and dip it in a dish of mercury.
The mercury falls to about 76 cm above the dish, leaving a vacuum at the top.
A metre scale beside the tube gives the height h of the column.
Point A inside the tube and point B on the open mercury surface are at the same level, so the pressure is the same at both. At B it is atmospheric pressure; at A it is the pressure of the mercury column, ρgh. So atmospheric pressure = ρgh.
Worked exampleFind the pressure of a column of mercury 76 cm high (ρ = 13.6 × 10³ kg m⁻³). (Solved Example 6.2)
h = 76 cm = 0.76 m.
P = ρgh = 13 600 × 10 × 0.76.
Answer1.0336 × 10⁵ Pa
Exam tip
Mercury is used because it is very dense, so the column is short (76 cm). A water barometer would need a tube about 10.3 m tall.
Check your book
If air gets into the vacuum at the top, it pushes the mercury down and the reading is too low. If the top of the tube actually leaks, the column falls all the way to the dish level.
Topic 9
The manometer
Definition
Manometer
A U-shaped glass tube containing mercury (or water), used to measure the pressure of a gas compared with atmospheric pressure.
Gas connected to one arm: (a) gas pressure above atmospheric; (b) gas pressure below atmospheric.
With both ends open, atmospheric pressure acts on both sides, so the levels are equal.
Connect a gas supply to one arm and leave the other open to the air.
If the level in the gas arm is pushed down, the gas pressure is greater than atmospheric pressure by ρgh.
If the level in the gas arm rises, the gas pressure is less than atmospheric pressure by ρgh.
gas pressure = atmospheric pressure ± ρgh
Topic 10
Uses of atmospheric pressure
Use
How it works
Weather forecasting
A falling barometer reading often means rain or a storm; a rising one means fair weather
Finding altitude (altimeter)
Pressure falls with height, so measuring it tells pilots and climbers their height
Tyre pressure gauge
Measures the air pressure in car and bicycle tyres
Drinking through a straw
Sucking lowers the pressure in the straw; atmospheric pressure pushes the drink up
Syringe and dropper
Pulling the piston lowers the pressure inside; atmospheric pressure pushes liquid in
Readings from a hot-air balloon (Numerical 6.10): pressure falls steadily as the balloon rises.
Topic 11
Pressure in liquids
A liquid presses on the bottom and the sides of its container and on anything in it. The pressure increases with depth and acts equally in all directions at a given depth.
The column of liquid of height h above area A presses on A with its weight mg.
Derivation of P = ρgh
Take an area A at depth h. The liquid column above it has volume V = Ah.
Its mass is m = ρV = ρAh.
Its weight, the force on A, is F = mg = ρAhg.
Pressure P = F ÷ A = ρAhg ÷ A = ρgh.
P = ρgh
Liquid pressure depends on
the depth h,
the density ρ of the liquid,
and g.
It does not depend on the shape or width of the container or the area of the surface: water 1 m below the surface of a small pool and of a huge lake has the same pressure.
Worked exampleA tank 2 m deep sits on a building 20 m high. Find the water pressure at the ground floor when the tank is full (ρ = 1000 kg m⁻³). (Solved Example 6.3)
Depth from the water surface to the ground floor: h = 2 + 20 = 22 m.
P = ρgh = 1000 × 10 × 22.
Answer2.2 × 10⁵ Pa
Exam tip
This is why dams are thicker at the bottom, why divers wear special suits in deep water, and why a giraffe needs special valves and a strong heart: its brain is about 2 m above its heart.
Topic 12
Pascal's law and hydraulic machines
Definition
Pascal's law
Pressure applied at one point of an enclosed fluid is transmitted equally, without loss, to every part of the fluid and the walls of the container.
Hydraulic press: a force multiplier
A small force F₁ on the small piston (area A₁) gives a large force F₂ on the large piston (area A₂).
A force F₁ on the small piston of area A₁ produces a pressure P = F₁ ÷ A₁.
By Pascal's law this pressure reaches the large piston unchanged.
Force on the large piston: F₂ = P × A₂ = F₁ × (A₂ ÷ A₁).
Since A₂ > A₁, F₂ > F₁: a small force lifts a large load.
F₁ ÷ A₁ = F₂ ÷ A₂, so F₂ = F₁ × A₂ ÷ A₁
Worked exampleThe pistons of a hydraulic press have diameters 5 cm and 25 cm. A force of 160 N acts on the small piston. Find the pressure and the force on the large piston. (Solved Example 6.4)
A₁ = π(0.025)² = 1.96 × 10⁻³ m².
P = F₁ ÷ A₁ = 160 ÷ (1.96 × 10⁻³) ≈ 8.15 × 10⁴ Pa, and the same pressure acts on the large piston.
F₂ = F₁ × (A₂ ÷ A₁) = 160 × (25 ÷ 5)² = 160 × 25.
AnswerP ≈ 8.15 × 10⁴ Pa, F₂ = 4000 N
Uses
Hydraulic press: compresses cotton bales and shapes metal sheets.
Car lift: lifts cars at service stations.
Hydraulic brakes: see the diagram below.
Pressing the pedal pushes the small piston of the master cylinder; the pressure reaches the larger wheel cylinders, which press the brake pads on the disc.
Remember
The liquid must be almost incompressible, like oil. An air bubble in brake lines would squash instead of passing on the pressure.
The wording follows published solved exercises and every answer was re-checked by hand, so the wording or the order of options may differ slightly from your print.
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Set A: Textbook exercise 8 MCQs
The multiple-choice questions at the end of Chapter 6 in your book, in the book's option order.
Q6.1Deformation and Elasticity
A wire is stretched by a weight w. If the diameter of the wire is reduced to half of its previous value, the extension will become:
Why do heavy animals like an elephant have a large area of the foot?
Answer
An elephant's weight is very large. Spread over large flat feet, the pressure on the ground (and on its leg bones) is P = F ÷ A, which stays small. With small feet the pressure would be too much for its bones and it would sink into soft ground.
Short Q 6.2
Why do animals like deer, which run fast, have a small area of the foot?
Answer
A small hoof gives a large pressure on the ground for the same weight (P = F ÷ A). The hooves dig into the ground and grip it, so the deer can push off hard and run fast without slipping.
Short Q 6.3
Why is it painful to walk barefoot on pebbles?
Answer
The foot touches the pebbles only at a few small points, so the contact area is small. The same body weight then gives a very large pressure (P = F ÷ A) at those points, which hurts.
Short Q 6.4
State Pascal's law. Give an application of Pascal's law.
Answer
Definition
Pascal's law
Pressure applied at one point of an enclosed fluid is transmitted equally, without loss, to all parts of the fluid.
Applications: hydraulic press, car lift at service stations, hydraulic brakes of vehicles.
Short Q 6.5
State what you mean by elasticity of a solid.
Answer
Definition
Elasticity
The property of a solid by which it regains its original size and shape when the deforming force is removed, for example a rubber band or a spring.
Short Q 6.6
What is Hooke's law? Does an object remain elastic beyond its elastic limit? Give a reason.
Answer
Definition
Hooke's law
Within the elastic limit, the extension of a spring is directly proportional to the applied force: F = kx.
No. Beyond the elastic limit the deformation becomes permanent: the object does not return to its original shape when the force is removed, and F is no longer proportional to x.
Short Q 6.7
Distinguish between force and pressure.
Answer
Force
Pressure
A push or pull that changes, or tends to change, the motion or shape of a body
Force acting normally per unit area
Vector
Scalar
Unit: newton (N)
Unit: pascal (Pa = N m⁻²)
F = ma
P = F ÷ A
Short Q 6.8
What is the relationship between liquid pressure and the depth of the liquid?
Answer
Liquid pressure is directly proportional to the depth: P = ρgh. Twice as deep means twice the pressure due to the liquid.
Short Q 6.9
What is the basic principle used to measure atmospheric pressure with a simple mercury barometer?
Answer
Atmospheric pressure pushing on the mercury in the dish holds up a column of mercury in the tube. Points at the same level in the mercury have the same pressure, so the pressure of the column, ρgh, equals atmospheric pressure. The height h of the column is therefore a measure of atmospheric pressure (about 760 mm at sea level).
Simple mercury barometer.
Short Q 6.10
State the basic principle used in the hydraulic brake system of automobiles.
Answer
Pascal's law. Pressing the brake pedal pushes the piston of the master cylinder and raises the pressure in the brake oil. This pressure is passed on equally to the larger pistons in the wheel cylinders, which push the brake pads against the wheels with a large force.
These test whether you really understand. Give a reason every time.
CRQ 6.1
A spring of spring constant k hangs vertically from a fixed point. A load of weight L hung from it causes an extension x (within the elastic limit). Identical springs, each of spring constant k, are arranged in series and in parallel. For each arrangement find (i) the total extension in terms of x (ii) the spring constant in terms of k.
Answer
One spring, two in series, two in parallel; each carries the load L.
Two springs in series
Each spring carries the full load L, so each stretches by x.
Total extension = x + x = 2x.
Spring constant = L ÷ 2x = ½(L ÷ x) = k/2.
Two springs in parallel
The load is shared: each spring carries L/2, so each stretches by x/2.
Total extension = x/2.
Spring constant = L ÷ (x/2) = 2(L ÷ x) = 2k.
Arrangement
Total extension
Spring constant
Series
2x
k/2
Parallel
x/2
2k
CRQ 6.2
Springs are made of steel instead of iron. Why?
Answer
Steel is more elastic than iron. It has a larger elastic limit, so it can be stretched or squeezed further and still return exactly to its shape, again and again. Iron would be permanently bent much sooner.
CRQ 6.3
Which of the following materials is more elastic? (a) iron or rubber (b) air or water
Answer
(a) Iron. A much larger force is needed to change the length of iron than rubber by the same amount, and iron returns fully to its shape. In physics, "more elastic" means harder to deform and better at recovering.
(b) Water. Water is far harder to compress than air, so it is more elastic.
CRQ 6.4
How does water pressure one metre below the surface of a swimming pool compare with water pressure one metre below the surface of a very large and deep lake?
Answer
They are the same. Liquid pressure P = ρgh depends only on the depth and the density of the liquid, not on how wide or how deep the whole body of water is. At 1 m: P = 1000 × 10 × 1 = 1.0 × 10⁴ Pa (plus atmospheric pressure) in both.
CRQ 6.5
What happens to the pressure in all parts of a confined liquid if the pressure is increased in one part? Give an example from daily life where this principle is applied.
Answer
By Pascal's law, the increase in pressure is passed on equally to every part of the liquid and the walls of its container.
Example: hydraulic brakes. Pressing the pedal raises the pressure in the brake oil, and the same increase reaches the cylinders at all four wheels at once. Other examples: a car lift and a hydraulic jack.
CRQ 6.6
If some air remains trapped at the top of the mercury column of a barometer, which is supposed to be a vacuum, how would it affect the height of the mercury column?
Answer
The trapped air presses down on the mercury. Now atmospheric pressure = pressure of the trapped air + pressure of the column, so the column is shorter than it should be and the barometer reads too low.
CRQ 6.7
Why is a long neck not a problem for a giraffe when it raises its neck suddenly?
Answer
A giraffe's brain is about 2 m above its heart, so blood must be pushed up against a pressure of ρgh ≈ 1050 × 10 × 2 ≈ 2 × 10⁴ Pa. Its heart is very large and strong and keeps a high blood pressure.
When it raises its head quickly, valves in the neck veins and a special network of small blood vessels near the brain keep the blood flow to the brain steady, so the blood pressure there does not suddenly drop and the giraffe does not faint (and when it lowers its head, they stop blood rushing in).
CRQ 6.8
The end of the glass tube used in a simple barometer is not properly sealed; some leak is present. What will be its effect?
Answer
Air enters through the leak into the space above the mercury. This air presses down on the column, so the reading falls. As more air leaks in, the pressure above the mercury becomes equal to atmospheric pressure, and the mercury falls to the same level as in the dish. The barometer then gives no reading at all.
Check your book
Some solved copies say only "it gives a lower reading". That is the start; a real leak empties the column completely.
CRQ 6.9
Comment on the statement: "Density is a property of a material, not the property of an object made of that material."
Answer
The statement is correct. Density = mass ÷ volume. A bigger object of the same material has more mass but also more volume in the same proportion, so the ratio stays the same. A 1 g iron pin and a 100 kg iron gate both have a density of about 7800 kg m⁻³. Density depends on what the object is made of, not on its size or shape.
CRQ 6.10
How is the load of a large structure estimated by an engineer?
Answer
From the drawings, the engineer finds the volume of each material: concrete, steel, bricks, glass.
Multiplying each volume by the density of that material gives its mass: m = ρV.
Multiplying by g gives the weight, the load: W = mg.
Adding all of these gives the total load, which tells the engineer how strong the foundations and pillars must be.
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
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.
Spring balance: extension ∝ weight.
Worked exampleA spring stretches 20 mm under 40 N
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.
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.
(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
Small piston A₁, large piston A₂, joined by a pipe full of liquid.
A force F₁ on the small piston gives a pressure P = F₁ ÷ A₁.
This pressure reaches the large piston unchanged, so F₂ = P × A₂ = F₁ × (A₂ ÷ A₁).
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
Brake pedal, master cylinder, brake line and wheel cylinder.
The brake pedal pushes the small piston of the master cylinder, which is full of brake oil.
The pressure travels through the brake lines to the larger pistons of the wheel cylinders.
These push the brake pads against the discs (or drums); friction slows the wheels.
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
Liquid column of height h on area A.
Derivation
Consider an area A at depth h. The liquid above it forms a column of volume V = Ah.
Mass of the column: m = ρV = ρAh.
Force on A = weight of the column: F = mg = ρAhg.
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
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
Heating, sealing and cooling a can.
Boil a little water in a tin can so that steam drives out the air.
Close the can tightly and cool it with cold water.
The steam condenses, so the pressure inside becomes very low.
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
Measure the mass on a balance.
Measure the length, breadth and height with a ruler or vernier callipers; volume = l × b × h.
ρ = m ÷ V.
(b) Irregular solid
Measure the mass on a balance.
Put water in a measuring cylinder and note the level V₁.
Lower the solid in on a thread and note the new level V₂.
Volume = V₂ − V₁; ρ = m ÷ (V₂ − V₁).
(c) Liquid
Find the mass of an empty beaker.
Pour in a measured volume of liquid from a measuring cylinder and find the new mass.
Mass of liquid = difference; ρ = mass ÷ volume.
Worked exampleA 60 g stone raises the water from 40 cm³ to 44 cm³
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.
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
Smallest face: 50 ÷ (50 × 10⁻⁴) = 1.0 × 10⁴ Pa.
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
Note the position of the pointer with no load.
Add 1 N and note the new reading; the difference is the extension x.
Add weights one by one, noting x each time.
Remove the weights one by one and check the pointer returns to the same readings (the elastic limit has not been passed).
Plot force F against extension x.
Result
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
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
The driver presses the pedal; the master cylinder piston pushes on the oil.
The pressure is transmitted equally through the lines to every wheel cylinder.
The larger pistons push the brake pads against the discs with a large force.
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
ρ(water) × g × h(water) = ρ(mercury) × g × h(mercury).
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.
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
One spring, series and parallel arrangements.
Series
Each spring carries the whole load L, so each stretches by x.
Total extension = 2x.
k(series) = L ÷ 2x = k/2.
Parallel
The load is shared equally, L/2 on each.
Each, and so the pair, stretches by x/2.
k(parallel) = L ÷ (x/2) = 2k.
Extension
Spring constant
Behaviour
Series
2x
k/2
Softer
Parallel
x/2
2k
Stiffer
Conclusion
Car suspensions use springs side by side (parallel) to carry heavy loads with little stretch.
A spring is stretched 20 mm by a load of 40 N. Calculate the spring constant. If an object causes an extension of 16 mm, what will be its weight?
Given Data
x₁ = 20 mm = 0.02 m
F₁ = 40 N
x₂ = 16 mm = 0.016 m
To Find
Spring constant k
Weight F₂ for 16 mm
Formulas
k = F ÷ x
F = kx
Calculation
k = F₁ ÷ x₁ = 40 ÷ 0.02 = 2000 N m⁻¹.
F₂ = k × x₂ = 2000 × 0.016 = 32 N.
Result
k = 2000 N m⁻¹; weight = 32 N
Numerical 6.2
The mass of 5 litres of milk is 4.5 kg. Find its density in SI units.
Given Data
m = 4.5 kg
V = 5 litres = 5 × 10⁻³ m³
To Find
Density ρ
Formula
ρ = m ÷ V
Calculation
1 litre = 10⁻³ m³, so V = 5 × 10⁻³ m³.
ρ = 4.5 ÷ (5 × 10⁻³) = 900 kg m⁻³.
Result
ρ = 900 kg m⁻³
Note: 1000 litres = 1 m³.
Numerical 6.3
When a solid of mass 60 g is lowered into a measuring cylinder, the level of water rises from 40 cm³ to 44 cm³. Calculate the density of the solid.
Given Data
m = 60 g = 0.06 kg
V₁ = 40 cm³
V₂ = 44 cm³
To Find
Density ρ
Formulas
V = V₂ − V₁
ρ = m ÷ V
Calculation
V = 44 − 40 = 4 cm³ = 4 × 10⁻⁶ m³.
ρ = 0.06 ÷ (4 × 10⁻⁶) = 15 000 kg m⁻³.
Result
ρ = 1.5 × 10⁴ kg m⁻³ (15 g cm⁻³)
Numerical 6.4
A block of density 8 × 10³ kg m⁻³ has a volume of 60 cm³. Find its mass.
Given Data
ρ = 8 × 10³ kg m⁻³
V = 60 cm³ = 60 × 10⁻⁶ m³
To Find
Mass m
Formula
m = ρV
Calculation
m = 8 × 10³ × 60 × 10⁻⁶ = 0.48 kg.
Result
m = 0.48 kg (480 g)
Numerical 6.5
A brick measures 5 cm × 10 cm × 20 cm. If its mass is 5 kg, calculate the maximum and minimum pressure the brick can exert on a horizontal surface. (g = 10 m s⁻²)
Smallest face 5 cm × 10 cm, largest face 10 cm × 20 cm.
Given Data
m = 5 kg, so F = mg = 50 N
smallest face = 5 × 10 = 50 cm² = 50 × 10⁻⁴ m²
largest face = 10 × 20 = 200 cm² = 200 × 10⁻⁴ m²
To Find
Maximum pressure
Minimum pressure
Formula
P = F ÷ A
Calculation
Maximum pressure, on the smallest face: P = 50 ÷ (50 × 10⁻⁴) = 1.0 × 10⁴ Pa.
Minimum pressure, on the largest face: P = 50 ÷ (200 × 10⁻⁴) = 2.5 × 10³ Pa.
Result
Maximum 1.0 × 10⁴ Pa; minimum 2.5 × 10³ Pa
Numerical 6.6
What will be the height of the column in a barometer at sea level if mercury is replaced by water of density 1000 kg m⁻³? (Density of mercury = 13.6 × 10³ kg m⁻³; the mercury column is 0.76 m.)
Given Data
ρ(water) = 1000 kg m⁻³
ρ(mercury) = 13.6 × 10³ kg m⁻³
h(mercury) = 0.76 m
To Find
Height of the water column
Formula
ρ₁gh₁ = ρ₂gh₂
Calculation
The same atmospheric pressure holds up both columns: 1000 × g × h = 13 600 × g × 0.76.
h = 13 600 × 0.76 ÷ 1000 = 10.34 m.
Result
h ≈ 10.3 m
Numerical 6.7
In the hydraulic brake system of a car, the force exerted normally on its piston of cross-sectional area 5 cm² is 500 N. What will be the pressure transferred to the brake oil? What will be the force on the second piston of cross-sectional area 20 cm²?
The same pressure acts on both pistons (Pascal's law).
Given Data
A₁ = 5 cm² = 5 × 10⁻⁴ m²
F₁ = 500 N
A₂ = 20 cm² = 20 × 10⁻⁴ m²
To Find
Pressure P in the oil
Force F₂ on the second piston
Formulas
P = F₁ ÷ A₁
F₂ = P × A₂
Calculation
P = 500 ÷ (5 × 10⁻⁴) = 1.0 × 10⁶ Pa.
F₂ = 1.0 × 10⁶ × 20 × 10⁻⁴ = 2000 N.
Result
P = 1.0 × 10⁶ Pa; F₂ = 2000 N
Numerical 6.8
Find the water pressure on a deep-sea diver at a depth of 10 m, where the density of sea water is 1030 kg m⁻³. (g = 10 m s⁻²)
Given Data
h = 10 m
ρ = 1030 kg m⁻³
g = 10 m s⁻²
To Find
Water pressure P
Formula
P = ρgh
Calculation
P = 1030 × 10 × 10 = 1.03 × 10⁵ Pa.
Result
P = 1.03 × 10⁵ Pa
Note: That is about one more atmosphere; the total pressure on the diver is about twice atmospheric pressure.
Numerical 6.9
The cross-sectional areas of the small and large pistons of a hydraulic press are 10 cm² and 100 cm². What force must be exerted on the small piston to lift a car of weight 4000 N?
F₁ on the small piston, the 4000 N car on the large one.
Given Data
A₁ = 10 cm²
A₂ = 100 cm²
F₂ = 4000 N
To Find
Force F₁ on the small piston
Formula
F₁ ÷ A₁ = F₂ ÷ A₂
Calculation
F₁ = F₂ × A₁ ÷ A₂ = 4000 × 10 ÷ 100.
F₁ = 400 N.
Result
F₁ = 400 N
Note: Only the ratio of the areas matters, so cm² need not be changed to m² here.
Numerical 6.10
In a hot-air balloon the following readings were taken: altitude 150, 500, 800, 1140, 1300, 1500 m; pressure 99.5, 95.7, 92.4, 88.9, 87.2, 85.3 kPa. Draw a graph of pressure against altitude and find (a) the air pressure at sea level (b) the height at which the air pressure is 90 kPa.
The points lie close to a straight line; extend it back to 0 m, and read across from 90 kPa.
Plot the points and draw the best straight line through them; the pressure falls about 1.06 kPa for every 100 m.
(a) Extend the line back to altitude 0: it meets the pressure axis at about 101 kPa.
(b) Read across from 90 kPa to the line and down: about 1030–1040 m (between the 800 m and 1140 m readings).
Result
(a) about 101 kPa (b) about 1030–1040 m
Note: One solved copy answers (b) with "88.9 kPa at 1140 m", which is just a row of the table. The question asks for 90 kPa.
Numerical 6.11
If the pressure in a hydraulic press is increased by an additional 10 N cm⁻², how much extra load will the output platform support if its cross-sectional area is 50 cm²?
Given Data
extra pressure P = 10 N cm⁻²
A = 50 cm²
To Find
Extra load F
Formula
F = P × A
Calculation
F = 10 N cm⁻² × 50 cm² = 500 N.
Result
Extra load = 500 N
Note: Both are in cm units, so they can be multiplied directly (10 N cm⁻² = 1.0 × 10⁵ Pa and 50 cm² = 5 × 10⁻³ m² give the same 500 N).
An external force that changes the size or shape of a body, for example squeezing a sponge or stretching a rubber band.
Elasticity
The property of a solid by which it returns to its original size and shape when the deforming force is removed.
Elastic limit
The greatest deforming force (or stretch) a body can take and still return to its original shape. Beyond it, the change becomes permanent.
Inelastic (plastic) materials
Materials, such as clay, dough and plasticine, that do not return to their original shape when the deforming force is removed.
Hooke's law
Within the elastic limit, the extension (or compression) of a spring is directly proportional to the force applied to it.
Spring constant (k)
The force needed per unit extension of a spring: k = F ÷ x. SI unit: N m⁻¹. A stiff spring has a large k.
Density
The mass per unit volume of a substance.
Pressure
The force acting normally (at right angles) on unit area of a surface. It is a scalar.
Atmospheric pressure
The pressure the air of the atmosphere exerts on the Earth's surface and on everything on it. It acts equally in all directions. At sea level it is about 1.013 × 10⁵ Pa.
Barometer
An instrument that measures atmospheric pressure.
Manometer
A U-shaped glass tube containing mercury (or water), used to measure the pressure of a gas compared with atmospheric pressure.
Pascal's law
Pressure applied at one point of an enclosed fluid is transmitted equally, without loss, to every part of the fluid and the walls of the container.
Common mistakes to avoid
Using mm or cm for the extension when k is in N m⁻¹.
Converting cm² to m² by dividing by 100 instead of 10⁴ (and cm³ to m³ by 10⁶).
Thinking liquid pressure depends on the area or shape of the container: it depends on depth only.
Forgetting to add the 20 m of pipe height when the tank is on a building.
Using the diameter instead of the radius in A = πr².
Saying rubber is more elastic than steel: in physics steel is more elastic.
Saying a barometer measures gas pressure: that is a manometer.
Mixing up force (N, vector) and pressure (Pa, scalar).
Extra practice tap a question to see its answer
A 30 N load stretches a spring by 0.06 m. Find k.Show answer
30 ÷ 0.06 = 500 N m⁻¹
A spring with k = 400 N m⁻¹ is stretched 5 cm. Find the force.Show answer
400 × 0.05 = 20 N
Two springs of k = 300 N m⁻¹ are joined in parallel. Find the combined k.Show answer
600 N m⁻¹
A 2 kg block has a volume of 0.001 m³. Find its density.Show answer
2 ÷ 0.001 = 2000 kg m⁻³
Find the mass of 2 m³ of water.Show answer
1000 × 2 = 2000 kg
A 600 N man stands on 0.03 m² of shoe sole. Find the pressure.Show answer
600 ÷ 0.03 = 2 × 10⁴ Pa
A 20 N force acts on 4 cm². Find the pressure in pascals.Show answer
20 ÷ (4 × 10⁻⁴) = 5 × 10⁴ Pa
Find the water pressure at the bottom of a 3 m deep tank.Show answer
1000 × 10 × 3 = 3 × 10⁴ Pa
Find the pressure of a 50 cm column of mercury.Show answer
13 600 × 10 × 0.5 = 6.8 × 10⁴ Pa
A press has pistons of 2 cm² and 200 cm². What load does a 50 N force lift?Show answer
50 × 200 ÷ 2 = 5000 N
Atmospheric pressure is 1.0 × 10⁵ Pa. Find the force on a 2 m² table top.Show answer
2 × 10⁵ N
Why is a mercury barometer used rather than a water one?Show answer
Mercury is 13.6 times denser, so the column is 76 cm instead of about 10 m
Name the law used in hydraulic brakes.Show answer
Pascal's law
What happens to a spring loaded beyond its elastic limit?Show answer
It is permanently stretched and no longer obeys Hooke's law
Final self-test
Final self-test: all 24 MCQs, no answers
Give yourself 25 minutes. Pick one option for each question, then submit to see your score and which questions to revise. Any question you get wrong tells you which topic to read again.
