Class 9 Physics Notes – Chapter 6: Mechanical Properties of Matter

Punjab Board | 2026

A. Multiple Choice Questions

6.1. 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:

(a) One half
(b) Double
(c) One fourth
(d) Four times

Correct Answer: (d) Four times

Explanation:

The extension of a wire is given by:

\[ \Delta L=\frac{FL}{AY}=\frac{FL}{\pi r^2Y} \]

Since the diameter is halved, the radius \(r\) is also halved. Therefore, the cross-sectional area becomes one-fourth of its original value, and the extension increases four times.

6.2. Four wires of the same material are stretched by the same load. Their dimensions are given below. Which of them will elongate most?

(a) Length 1 m, diameter 1 mm
(b) Length 2 m, diameter 2 mm
(c) Length 3 m, diameter 3 mm
(d) Length 4 m, diameter 0.5 mm

Correct Answer: (d) Length 4 m, diameter 0.5 mm

Explanation:

Elongation is proportional to the length and inversely proportional to the square of the diameter:

\[ \Delta L\propto\frac{L}{d^2} \]

Wire (d) has the greatest length and the smallest diameter, giving the maximum value of \(\frac{L}{d^2}\).

6.3. Two metal plates of areas \(2\,\text{m}^2\) and \(3\,\text{m}^2\) are placed in a liquid at the same depth. The ratio of pressures on the two plates is:

(a) 1:1
(b) \(\sqrt{2}:\sqrt{3}\)
(c) 2:3
(d) 4:9

Correct Answer: (a) 1:1

Explanation:

Liquid pressure depends on depth, density and gravitational acceleration, not on the surface area of the submerged plate.

\[ P=\rho gh \]

At the same depth in the same liquid, the pressures are equal.

6.4. The pressure at any point in a liquid is proportional to:

(a) Density of the liquid
(b) Depth of the point below the surface
(c) Acceleration due to gravity
(d) All of the above

Correct Answer: (d) All of the above

Explanation:

Liquid pressure is given by:

\[ P=\rho gh \]

It depends on the liquid’s density \(\rho\), gravitational acceleration \(g\), and depth \(h\).

6.5. Pressure applied to an enclosed fluid is:

(a) Increased and applied to every part
(b) Diminished and transmitted to the walls
(c) Increased proportional to mass
(d) Transmitted unchanged to every portion of the fluid and the walls of the containing vessel

Correct Answer: (d) Transmitted unchanged to every portion of the fluid and the walls of the containing vessel

Explanation:

This is Pascal’s principle, which states that pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid and to the walls of its container.

6.6. The principle of a hydraulic press is based on:

(a) Hooke’s law
(b) Pascal’s law
(c) Conservation of energy
(d) Conservation of momentum

Correct Answer: (b) Pascal’s law

Explanation:

Hydraulic systems operate by transmitting pressure equally throughout an enclosed fluid according to Pascal’s law.

6.7. When a spring is compressed, what form of energy does it possess?

(a) Kinetic
(b) Potential
(c) Internal
(d) Heat

Correct Answer: (b) Potential

Explanation:

A compressed or stretched spring stores elastic potential energy.

6.8. What is the force exerted by the atmosphere on a rectangular block surface of length 50 cm and breadth 40 cm? The atmospheric pressure is 100 kPa.

(a) 20 kN
(b) 100 kN
(c) 200 kN
(d) 500 kN

Correct Answer: (a) 20 kN

Explanation:

Area:

\[ A=0.5\times0.4=0.2\,\text{m}^2 \]

Force:

\[ F=PA \]

\[ F=100000\times0.2=20000\,\text{N} \]

Therefore,

\[ \boxed{F=20\,\text{kN}} \]

B. Short Answer Questions

6.1. Why do heavy animals like elephants have a large area of the foot?

Answer:

A large foot area distributes their massive weight over a greater surface area, reducing pressure.

\[ P=\frac{F}{A} \]

This helps prevent them from sinking deeply into soft mud or ground.

6.2. Why do animals like deer that run fast have a small area of the foot?

Answer:

A small hoof area allows concentrated grip and traction on the ground, helping with rapid acceleration and agile running.

6.3. Why is it painful to walk barefoot on pebbles?

Answer:

Pebbles have a very small contact area with the sole of the foot. The body’s weight is concentrated over tiny points, creating high local pressure that stimulates pain receptors.

6.4. State Pascal’s law. Give an application of Pascal’s law.

Answer:

Pascal’s law states that pressure applied at any point of an enclosed liquid is transmitted undiminished to every part of the liquid and the walls of the container.

Applications include hydraulic lifts and hydraulic brakes.

6.5. What do you mean by the elasticity of a solid?

Answer:

Elasticity is the property of a solid material by virtue of which it tends to regain its original shape and size after the removal of deforming forces.

6.6. What is Hooke’s law? Does an object remain elastic beyond the elastic limit? Give a reason.

Answer:

Hooke’s law states that, within the elastic limit, stress is directly proportional to strain.

No. Beyond the elastic limit, the material undergoes permanent plastic deformation and does not return to its original shape.

6.7. Distinguish between force and pressure.

Answer:

Force: A push or pull acting on an object. It is measured in newtons (N).
Pressure: The force acting normally per unit surface area. It is measured in pascals (Pa).

\[ P=\frac{F}{A} \]

6.8. What is the relationship between liquid pressure and the depth of the liquid?

Answer:

Liquid pressure is directly proportional to the depth \(h\) of the liquid.

\[ P=\rho gh \]

Therefore, pressure increases linearly with depth.

6.9. What is the basic principle used to measure atmospheric pressure with a simple mercury barometer?

Answer:

The atmospheric pressure supporting the mercury column equals the hydrostatic pressure exerted by the vertical mercury column.

\[ P=\rho gh \]

6.10. State the basic principle used in the hydraulic brake system of automobiles.

Answer:

Hydraulic brakes work on Pascal’s law. Pressure applied to the master-cylinder piston is transmitted equally and undiminished through the brake oil to operate the wheel-brake cylinders.

C. Constructed Response Questions

6.1. A spring having spring constant \(k\) hangs vertically. For each arrangement, complete the table.

Answer:

(i) Series arrangement: Two identical springs are connected end-to-end.

Total extension: \(2x\)
Equivalent spring constant: \(\displaystyle \frac{k}{2}\)

(ii) Parallel arrangement: Two identical springs are connected side-by-side.

Total extension: \(\displaystyle \frac{x}{2}\)
Equivalent spring constant: \(2k\)

6.2. Springs are made of steel instead of iron. Why?

Answer:

Steel has a higher elastic limit and greater stiffness than iron. It can undergo repeated deformation within its elastic limit without permanent distortion.

6.3. Which of the following materials is more elastic? (a) Iron or rubber (b) Air or water

Answer:

(a) Iron or rubber: Iron is more elastic than rubber in the physics sense because it has a higher Young’s modulus and requires a much larger deforming stress to produce unit strain.

(b) Air or water: Water is more resistant to compression than air. Therefore, water has a much higher bulk modulus and is considered more elastic in terms of resistance to compression.

6.4. How does the water pressure one metre below the surface of a swimming pool compare with the water pressure one metre below the surface of a very large and deep lake?

Answer:

The pressure is the same because pressure depends only on depth, density of water and gravitational acceleration.

\[ P=\rho gh \]

At a depth of 1 m, the pressure is the same in both bodies of water, regardless of their total volume.

6.5. What will happen to the pressure in all parts of a confined liquid if pressure is increased in one part? Give an example.

Answer:

According to Pascal’s law, the pressure increases by the same amount in all parts of the liquid.

Example: Pressing the brake pedal in a car increases pressure uniformly throughout the brake fluid.

6.6. If some air remains trapped within the top of the mercury column of a barometer, how would it affect the height?

Answer:

The trapped air exerts pressure on the mercury column, causing the mercury height to be lower than the true barometric height.

6.7. How is a long neck not a problem for a giraffe when it raises its neck suddenly?

Answer:

Giraffes have specialized blood vessels with elastic walls, valves and a powerful heart that help regulate blood pressure and prevent excessive pressure changes when they raise or lower their long necks.

6.8. The end of the glass tube used in a simple barometer is not properly sealed, and a leak is present. What will be its effect?

Answer:

Air will enter the tube, destroying the vacuum at the top and causing the mercury column to fall. This will make the barometer inaccurate or unusable.

6.9. Comment on the statement: “Density is a property of a material, not the property of an object made of that material.”

Answer:

Density is an intensive property defined as mass per unit volume.

\[ \rho=\frac{m}{V} \]

Although mass and volume vary between objects, their ratio—the density—remains constant for a given pure material under specified conditions.

6.10. How is the load of a large structure estimated by an engineer?

Answer:

Engineers estimate the load by calculating the dead load (the self-weight of materials) and live loads, including occupants, furniture, wind and seismic forces. They then analyse how the loads are distributed across structural columns and foundations.

D. Comprehensive Questions

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

Answer:

Hooke’s law states that stress is directly proportional to strain within the elastic limit.

Applications include:

Spring balances
Mechanical watches and clocks
Seismic vibration dampeners

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

Answer:

Mercury barometer: A barometer measures atmospheric pressure by balancing the height of a mercury column against atmospheric pressure. The pressure is calculated using:

\[ P=\rho gh \]

Manometer: A manometer measures gas pressure or pressure differences using a liquid column in a U-shaped tube. The pressure difference is determined from the difference in liquid levels.

6.3. Describe Pascal’s law. State its applications with examples.

Answer:

Pascal’s law states that pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid.

Applications include:

Hydraulic lifts: Used to raise vehicles.
Hydraulic presses: Used to compress materials or apply large forces.
Hydraulic brakes: Used to transmit force from the brake pedal to the wheel brakes.

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

Answer:

Liquid pressure depends on:

Depth of the liquid \((h)\)
Density of the liquid \((\rho)\)
Gravitational acceleration \((g)\)

It is determined using:

\[ \boxed{P=\rho gh} \]

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

Answer:

Air has weight and exerts atmospheric pressure on surfaces. Atmospheric pressure is approximately \(101.3\,\text{kPa}\) at sea level.

Applications include:

Drinking through a straw: Atmospheric pressure pushes the liquid up the straw when the pressure inside it is reduced.
Working of a syringe: A pressure difference allows liquid to enter or leave the syringe.
Suction cups: Reduced pressure beneath the cup allows atmospheric pressure to hold it against a surface.

E. Numerical Problems

6.1. A spring is stretched 20 mm by a load of 40 N. Calculate the value of the spring constant. If an object causes an extension of 16 mm, what will be its weight?

Answer:

Given:

\(x_1=20\,\text{mm}=0.02\,\text{m}\)
\(F_1=40\,\text{N}\)

(a) Spring constant

Using Hooke’s law:

\[ k=\frac{F_1}{x_1} \]

\[ k=\frac{40}{0.02} \]

\[ \boxed{k=2000\,\text{N/m}=2\,\text{kN/m}} \]

(b) Weight of the object

Given:

\[ x_2=16\,\text{mm}=0.016\,\text{m} \]

Using \(F_2=kx_2\):

\[ F_2=2000\times0.016 \]

\[ \boxed{F_2=32\,\text{N}} \]

6.2. The mass of 5 litres of milk is 4.5 kg. Find its density in SI units.

Answer:

Given:

\(m=4.5\,\text{kg}\)
\(V=5\,\text{litres}=5\times10^{-3}\,\text{m}^3\)

Density:

\[ \rho=\frac{m}{V} \]

\[ \rho=\frac{4.5}{5\times10^{-3}} \]

\[ \boxed{\rho=900\,\text{kg/m}^3} \]

6.3. When a solid of mass 60 g is lowered into a measuring cylinder, the water level rises from 40 cm³ to 44 cm³. Calculate the density of the solid.

Answer:

Given:

\(m=60\,\text{g}=0.06\,\text{kg}\)
Initial volume \(=40\,\text{cm}^3\)
Final volume \(=44\,\text{cm}^3\)

Volume of the solid:

\[ V=44-40=4\,\text{cm}^3 \]

In SI units:

\[ V=4\times10^{-6}\,\text{m}^3 \]

Density:

\[ \rho=\frac{m}{V} \]

\[ \rho=\frac{0.06}{4\times10^{-6}} \]

\[ \boxed{\rho=1.5\times10^4\,\text{kg/m}^3} \]

Equivalently, the density is \(15\,\text{g/cm}^3\).

6.4. A block of density \(8\times10^2\,\text{kg/m}^3\) has a volume of 60 cm³. Find its mass.

Answer:

Given:

\(\rho=800\,\text{kg/m}^3\)
\(V=60\,\text{cm}^3=60\times10^{-6}\,\text{m}^3\)

Using:

\[ m=\rho V \]

\[ m=800\times(60\times10^{-6}) \]

\[ \boxed{m=0.048\,\text{kg}=48\,\text{g}} \]

6.5. A brick measures 5 cm × 10 cm × 20 cm. If its mass is 5 kg, calculate the maximum and minimum pressure which the brick can exert on a horizontal surface.

Answer:

Given:

\(m=5\,\text{kg}\)
\(g=10\,\text{m/s}^2\)

Weight:

\[ F=mg=5\times10=50\,\text{N} \]

Possible contact areas:

\[ A_1=0.05\times0.10=0.005\,\text{m}^2 \]

\[ A_2=0.10\times0.20=0.02\,\text{m}^2 \]

Maximum pressure: This occurs over the smallest contact area.

\[ P_{\max}=\frac{50}{0.005} \]

\[ \boxed{P_{\max}=1\times10^4\,\text{Pa}} \]

Minimum pressure: This occurs over the largest contact area.

\[ P_{\min}=\frac{50}{0.02} \]

\[ \boxed{P_{\min}=2500\,\text{Pa}=2.5\times10^3\,\text{Pa}} \]

6.6. What will be the height of the column in a barometer at sea level if mercury is replaced by water of density \(1000\,\text{kg/m}^3\)? The density of mercury is \(13.6\times10^3\,\text{kg/m}^3\).

Answer:

For the same atmospheric pressure:

\[ \rho_{\text{Hg}}gh_{\text{Hg}} = \rho_{\text{water}}gh_{\text{water}} \]

Cancelling \(g\):

\[ \rho_{\text{Hg}}h_{\text{Hg}} = \rho_{\text{water}}h_{\text{water}} \]

Using \(h_{\text{Hg}}=0.76\,\text{m}\):

\[ 13600\times0.76=1000\times h_{\text{water}} \]

\[ h_{\text{water}}=\frac{13600\times0.76}{1000} \]

\[ \boxed{h_{\text{water}}=10.3\,\text{m}} \]

6.7. In the hydraulic brake system of a car, the force exerted normally on a 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?

Answer:

Given:

\(F_1=500\,\text{N}\)
\(A_1=5\,\text{cm}^2=5\times10^{-4}\,\text{m}^2\)
\(A_2=20\,\text{cm}^2=20\times10^{-4}\,\text{m}^2\)

(a) Pressure transferred to the brake oil

\[ P=\frac{F_1}{A_1} \]

\[ P=\frac{500}{5\times10^{-4}} \]

\[ \boxed{P=1.0\times10^6\,\text{Pa}} \]

(b) Force on the second piston

\[ F_2=PA_2 \]

\[ F_2=(1.0\times10^6)(20\times10^{-4}) \]

\[ \boxed{F_2=2000\,\text{N}} \]

6.8. Find the water pressure on a deep-sea diver at a depth of 10 m, where the density of seawater is 1030 kg/m³.

Answer:

Given:

\(\rho=1030\,\text{kg/m}^3\)
\(g=10\,\text{m/s}^2\)
\(h=10\,\text{m}\)

Using:

\[ P=\rho gh \]

\[ P=1030\times10\times10 \]

\[ \boxed{P=103000\,\text{Pa}=1.03\times10^5\,\text{N/m}^2} \]

This is the water’s gauge pressure. Absolute pressure would also include atmospheric pressure.

6.9. The cross-sectional areas of the small and large pistons of a hydraulic press are 10 cm² and 100 cm², respectively. What force should be exerted on the small piston to lift a car of weight 4000 N?

Answer:

Given:

\(A_1=10\,\text{cm}^2\)
\(A_2=100\,\text{cm}^2\)
\(F_2=4000\,\text{N}\)

Using Pascal’s law:

\[ \frac{F_1}{A_1}=\frac{F_2}{A_2} \]

\[ \frac{F_1}{10}=\frac{4000}{100} \]

\[ F_1=\frac{4000\times10}{100} \]

\[ \boxed{F_1=400\,\text{N}} \]

6.10. In a hot-air balloon, the following data was recorded: (a) What would the air pressure have been at sea level? (b) At what height would the air pressure be 90 kPa?

Answer:

(a) Air pressure at sea level

Extrapolating to sea level, at \(0\,\text{m}\):

\[ \boxed{P\approx1.01\times10^5\,\text{Pa}\approx101\,\text{kPa}} \]

(b) Height at which the air pressure is 90 kPa

\[ \boxed{h\approx1.02\,\text{km}\approx1020\,\text{m}} \]

Note: The original source refers to recorded data for the hot-air balloon but does not include the data table or graph. These numerical answers cannot be independently verified without that data.

6.11. If the pressure in a hydraulic press is increased by an additional 10 N/cm², how much extra load will the output platform support if its cross-sectional area is 50 cm²?

Answer:

Given:

Additional pressure, \(\Delta P=10\,\text{N/cm}^2\)
Area, \(A=50\,\text{cm}^2\)

Additional load:

\[ F=\Delta P\times A \]

\[ F=10\times50 \]

\[ \boxed{F=500\,\text{N}} \]

6.12. The force exerted normally on the hydraulic brake system of a car, with its piston of cross-sectional area 5 cm², is 500 N. What will be the: (a) pressure transferred to the brake oil? (b) force on the brake piston of cross-sectional area 20 cm²?

Answer:

Given:

\(F_1=500\,\text{N}\)
\(A_1=5\,\text{cm}^2\)
\(A_2=20\,\text{cm}^2\)

(a) Pressure transferred to the brake oil

\[ P=\frac{F_1}{A_1}=\frac{500}{5} \]

\[ P=100\,\text{N/cm}^2 \]

Converting to SI units:

\[ \boxed{P=1.0\times10^6\,\text{N/m}^2} \]

(b) Force on the second piston

\[ F_2=PA_2 \]

\[ F_2=100\times20 \]

\[ \boxed{F_2=2000\,\text{N}} \]

Notes by Sir Dilawar

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