Young's Modulus — Solving Strain Problems

Young’s Modulus — Solving Strain Problems

Young’s modulus is the single most tested elastic constant in JEE Main, NEET, and CBSE Class 11. Once we understand what it represents physically, every numerical reduces to plugging into one formula. Let’s build the intuition first, then walk through the standard problem types you’ll see in any paper.

When we pull a wire, it stretches. Some materials stretch a lot (rubber), some barely budge (steel). Young’s modulus $Y$ is the number that quantifies this difference — and it’s defined cleanly so the same number describes a hair-thin steel wire and a steel girder.

Key Terms & Definitions

Stress ( $\sigma$ ) — Force per unit cross-sectional area. SI unit: N/m² or pascal (Pa). Tensile stress is what we deal with in elongation problems.

$\sigma = \frac{F}{A}$

Strain ( $\varepsilon$ ) — Fractional change in length. Dimensionless.

$\varepsilon = \frac{\Delta L}{L_0}$

Young’s modulus ( $Y$ ) — Ratio of stress to strain in the linear (elastic) regime. SI unit: N/m². Typical values: steel $\approx 2 \times 10^{11}$ Pa, copper $\approx 1.1 \times 10^{11}$ Pa, rubber $\approx 10^7$ Pa.

$Y = \frac{\sigma}{\varepsilon} = \frac{F/A}{\Delta L/L_0} = \frac{FL_0}{A\Delta L}$

Rearranging for elongation:

$\Delta L = \frac{FL_0}{AY}$

Elastic limit — The stress beyond which the material no longer obeys $\sigma = Y\varepsilon$ . Below this, the wire returns to its original length when load is removed; above, permanent deformation sets in.

The Three Standard Problem Types

Type 1 — Direct Substitution

Find one of $\{F, A, L_0, \Delta L, Y\}$ given the other four. Just plug in.

Example. A copper wire of length $1$ m, area $0.5$ mm², is stretched by $200$ N. Find $\Delta L$ . ( $Y = 1.1 \times 10^{11}$ Pa.)

$\Delta L = \frac{200 \times 1}{0.5 \times 10^{-6} \times 1.1 \times 10^{11}} = \frac{200}{5.5 \times 10^4} \approx 3.64 \times 10^{-3} \text{ m} = 3.64 \text{ mm}$

Type 2 — Wire Under Its Own Weight

A wire hangs from the ceiling. Each element is stretched by the weight of wire below it — so strain isn’t uniform.

For a wire of length $L$ , density $\rho$ , hanging vertically:

$\Delta L = \frac{\rho g L^2}{2Y}$

The factor of 2 comes from integrating: the average tension is half the total weight.

Type 3 — Composite Wires (Series)

Two wires of different materials joined end-to-end. Same tension acts on both, so:

$\Delta L_{\text{total}} = \Delta L_1 + \Delta L_2 = \frac{F L_1}{A_1 Y_1} + \frac{F L_2}{A_2 Y_2}$

Think of each wire as a “spring” with stiffness $k = AY/L$ . Springs in series add reciprocally.

Solved Examples — Easy to Hard

Easy (CBSE Level)

A steel wire $2$ m long, area $1$ mm², is loaded with $100$ N. Find elongation. ( $Y = 2 \times 10^{11}$ Pa.)

$\Delta L = \frac{100 \times 2}{10^{-6} \times 2 \times 10^{11}} = \frac{200}{2 \times 10^5} = 10^{-3} \text{ m} = 1 \text{ mm}$

Medium (JEE Main)

A steel wire of length $L$ and a copper wire of equal length and area are connected end-to-end and stretched by force $F$ . Find the ratio of their elongations.

Same $F$ , same $A$ , same $L$ , different $Y$ :

$\frac{\Delta L_{\text{steel}}}{\Delta L_{\text{copper}}} = \frac{Y_{\text{copper}}}{Y_{\text{steel}}} = \frac{1.1}{2.0} = 0.55$

Steel stretches 55% as much as copper under identical conditions.

Hard (JEE Advanced)

A uniform steel wire of length $L$ , mass $m$ , hangs vertically from a ceiling. A weight $W$ is attached to the lower end. Find the total elongation.

Two contributions: (i) elongation due to $W$ , (ii) elongation due to the wire’s self-weight.

$\Delta L = \frac{WL}{AY} + \frac{mgL}{2AY} = \frac{L}{AY}\left(W + \frac{mg}{2}\right)$

The half-weight term comes from integrating tension along the wire — this is the JEE Advanced trick.

Exam-Specific Tips

JEE Main: Direct substitution with unit conversion is the staple. Always convert mm² to m² (factor of $10^{-6}$ ) and mm to m (factor of $10^{-3}$ ). Roughly 1 out of every 3 papers has a Young’s modulus question worth 4 marks.

NEET: Conceptual MCQs about which material stretches more, or about elastic vs plastic regime. Less computation, more theory. Memorise typical $Y$ values in the rough order: rubber ≪ wood ≪ aluminium ≪ copper ≪ steel.

CBSE Boards: Definitions, units, dimensional analysis, and one numerical (3 marks) appear almost every year. Stress-strain curve labelling (proportional limit, elastic limit, yield point, breaking point) is asked in 2-mark theory.

Common Mistakes to Avoid

Mistake 1 — Unit blunders. $1$ mm² $= 10^{-6}$ m², not $10^{-3}$ . Wrong conversion gives an answer 1000 times off.

Mistake 2 — Using diameter as radius. If diameter is $0.5$ mm, radius is $0.25$ mm. Then $A = \pi r^2$ , not $\pi d^2$ .

Mistake 3 — Forgetting the factor of 2 in self-weight problems. The full weight acts only at the top. Average tension is $mg/2$ , so elongation has the factor of 1/2.

Mistake 4 — Mixing up series and parallel composites. End-to-end wires have the same force (series). Side-by-side wires have the same elongation (parallel). The formulas differ.

Mistake 5 — Using the elastic formula beyond the elastic limit. $\sigma = Y\varepsilon$ only holds below the proportional limit. For high stresses, the material yields and the linear formula fails.

Practice Questions

A wire of length $4$ m and area $2$ mm² stretches by $1$ mm under a $50$ N load. Find Young’s modulus.
A steel rod $1$ m long, area $1$ cm², is heated from $0°C$ to $100°C$ but its ends are clamped. Find the thermal stress. ( $Y = 2 \times 10^{11}$ Pa, $\alpha = 12 \times 10^{-6}$ /°C.)
A composite rod consists of $1$ m of steel and $1$ m of copper, both with area $1$ mm², joined end-to-end. A tension of $100$ N is applied. Find the total elongation.
A wire of length $L$ and Young’s modulus $Y$ supports a mass $m$ . If the wire is replaced by another of the same material but with double length and half the area, find the new elongation in terms of the original.
A steel wire $2$ m long stretches $1$ mm under a load. If a copper wire of identical dimensions replaces it, find the new elongation. ( $Y_{\text{steel}}/Y_{\text{copper}} = 1.82$ .)
A wire of cross-section $A$ and Young’s modulus $Y$ has its temperature lowered by $\Delta T$ while clamped at both ends. Find the tensile force induced.
In a Searle’s apparatus, a wire of length $1$ m and diameter $0.5$ mm shows an elongation of $0.2$ mm under a $20$ N load. Compute $Y$ .
Two wires $A$ and $B$ of the same material, length, and cross-section are stretched by forces $F$ and $2F$ . Compare strain and stored elastic energy.

Q1: $Y = FL/(A\Delta L) = (50 \times 4)/(2 \times 10^{-6} \times 10^{-3}) = 1 \times 10^{11}$ Pa.

Q2: Thermal strain $= \alpha\Delta T = 12 \times 10^{-6} \times 100 = 1.2 \times 10^{-3}$ . Thermal stress $= Y\varepsilon = 2 \times 10^{11} \times 1.2 \times 10^{-3} = 2.4 \times 10^8$ Pa.

Q3: $\Delta L = \dfrac{FL}{A}\left(\dfrac{1}{Y_s} + \dfrac{1}{Y_c}\right) = \dfrac{100 \times 1}{10^{-6}}\left(\dfrac{1}{2 \times 10^{11}} + \dfrac{1}{1.1 \times 10^{11}}\right) \approx 1.4$ mm.

FAQs

Q: Why doesn’t Young’s modulus appear in the formula for sound speed in a rod?

It does — for longitudinal waves in a solid rod, $v = \sqrt{Y/\rho}$ . The bulk modulus shows up for sound in fluids; Young’s modulus governs longitudinal vibrations in thin solids.

Q: Is Young’s modulus the same in tension and compression?

For most metals in the elastic regime, yes — the linear stress-strain curve is symmetric. For materials like concrete, the compressive modulus differs slightly from the tensile one, but JEE problems treat them as equal.

Q: Why is steel preferred for construction over copper, despite copper being more conductive?

Construction needs a high modulus (low strain under load) and high yield strength. Steel wins on both. Copper is great for wiring (electrical conduction) but yields at much lower stress.

Q: How is Young’s modulus measured in a lab?

Searle’s apparatus is the standard CBSE setup: two identical wires, one as reference, one loaded with weights. A vernier measures the elongation of the loaded wire to 0.01 mm precision.

Q: What is the elastic potential energy stored in a stretched wire?

$U = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume} = \frac{1}{2}\sigma\varepsilon V$ . Equivalently, $U = \frac{1}{2}F\Delta L$ — same form as a spring with $k = AY/L$ .