9th Class Physics Chapter 6 Notes PDF 2026 – Mechanical Properties of Matter

Stress $= \frac{F}{A}$ (force per unit area)

Strain $= \frac{change in length}{original length}$

Young's modulus $E = \frac{stress}{strain}$

Rule

Hooke's law: within the elastic limit, extension is directly proportional to the applied force: $F = k x$ ( $k$ = spring constant).

Example 1 — spring constant

A spring stretches $2$ cm under a $40$ N load. Find the spring constant.

$k = \frac{F}{x} = \frac{40}{0.02}$ .

Answer: $2000$ N/m

$P = \frac{F}{A}$ , SI unit pascal (Pa) $=$ N/m $^{2}$

Example 2 — pressure

A force of $10$ N acts on an area of $0.01$ m $^{2}$ . Find the pressure.

$P = \frac{F}{A} = \frac{10}{0.01}$ .

Answer: $1000$ Pa

Remember

Spreading the same force over a larger area lowers the pressure — why a camel's wide feet don't sink into sand, and why elephants' large feet support their weight.

Definition

Atmospheric pressure is the force per unit area exerted by the weight of the air above a point; it decreases with altitude (less air above at higher points). A mercury barometer measures it directly from the height of a mercury column it can support.

$ρ = \frac{m}{V}$

Example 3 — density

Find the density of a substance of mass $2$ kg and volume $0.002$ m $^{3}$ .

$ρ = \frac{m}{V} = \frac{2}{0.002}$ .

Answer: $1000$ kg/m $^{3}$

Quick Recap

●Stress $= F / A$ , strain $= Δ L / L$ , Young's modulus $=$ stress/strain
●Hooke's law: $F = k x$ , within the elastic limit
●Pressure $P = F / A$ ; spreading force over a larger area lowers pressure
●Atmospheric pressure decreases with altitude; measured by a barometer
●Density $ρ = m / V$

9th Class Physics Chapter 6 Notes

Elasticity, Stress and Strain

Pressure

Atmospheric Pressure

Density

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