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Elasticity is one of the shortest chapters in Class 11 physics, and one of the easiest places to lose marks for silly reasons. The ideas are few: stress, strain, and the modulus that links them. The errors are many: radius and diameter mixed up, millimetres left as millimetres, a wire's own weight forgotten. This post gives you the definitions, four worked problems and the checks that catch slips.
What do stress, strain and Young's modulus mean?
The NCERT Class 11 chapter Mechanical Properties of Solids defines them this way. Stress is the restoring force per unit area, F/A, measured in pascals. Strain is the fractional change in size, such as ΔL/L for a stretched wire, so it has no units. Within the elastic limit, stress is proportional to strain (Hooke's law), and for stretching or compression the ratio is Young's modulus:
Y = (F/A) / (ΔL/L) = FL / (AΔL)
NCERT uses Y = 2.0 × 10¹¹ Pa for structural steel and 1.1 × 10¹¹ Pa for copper in its examples. Y is a property of the material. It does not depend on the length or thickness of the particular wire.
Elasticity, Hooke's law and the three moduli are in the JEE (Main) 2026 syllabus, and "Young's modulus of elasticity of the material of a metallic wire" is on its experiment list. The CBSE 2026-27 curriculum has the same experiment for Class 11. Check the current year's documents for your exam.
Why is steel "more elastic" than rubber?
This one surprises students. In physics, more elastic means harder to deform for the same stress, so a larger modulus. NCERT notes that stretching a thin steel wire of cross-section 0.1 cm² by 0.1% takes 2000 N, while aluminium, brass and copper wires of the same size need 690 N, 900 N and 1100 N, and concludes that steel is more elastic than all three. Rubber stretches easily, so its Young's modulus is small. If a question asks which is more elastic, compare moduli, not how far each one stretches.
Worked example: stretching a steel wire
A steel wire 2.0 m long and 1.0 mm in diameter holds a 10 kg load. Take Y = 2.0 × 10¹¹ Pa and g = 9.8 m/s². Find the stress, strain and extension.
Force: F = mg = 10 × 9.8 = 98 N. Use weight, not mass.
Area: radius = 0.50 mm = 5.0 × 10⁻⁴ m, so A = π(5.0 × 10⁻⁴)² ≈ 7.85 × 10⁻⁷ m².
Stress: 98 / 7.85 × 10⁻⁷ ≈ 1.25 × 10⁸ Pa.
Strain: 1.25 × 10⁸ / 2.0 × 10¹¹ ≈ 6.2 × 10⁻⁴.
Extension: strain × L ≈ 6.2 × 10⁻⁴ × 2.0 ≈ 1.25 × 10⁻³ m, or about 1.25 mm.
Is the wire still in its elastic range? NCERT gives the yield strength of mild steel as about 300 × 10⁶ Pa. Our stress, 1.25 × 10⁸ Pa, is well below that, so Hooke's law applies.
Energy stored. For a stretched wire, U = ½ × F × ΔL = 0.5 × 98 × 1.25 × 10⁻³ ≈ 0.061 J. Per unit volume, this is ½ × stress × strain.
Two ratio problems JEE likes
Same load, different wire. Two wires of the same material carry the same load. Wire B is twice as long as wire A and has half its diameter. Since ΔL = FL/(AY) and A goes as d², ΔL goes as L/d². For wire B that is 2 / (½)² = 8, so B stretches eight times as much.
A wire under its own weight. A uniform wire hangs from the ceiling. The tension is largest at the top (the whole weight) and zero at the bottom, so on average each part feels half the weight. The extension is half of what you would get by hanging the full weight from the end of a weightless wire: ΔL = ρgL² / (2Y), where ρ is density. Students often forget the half.
How do errors add up in the Young's modulus practical?
In the experiment, Y = 4FL / (πd²ΔL), where d is the wire's diameter measured with a screw gauge. Because d is squared, its relative error counts twice:
ΔY/Y = ΔF/F + ΔL/L + 2(Δd/d) + Δ(ΔL)/ΔL
If d = 0.50 mm is read with an uncertainty of 0.01 mm, that is a 2% error in d and a 4% contribution to Y. That is why the method asks you to measure the diameter at several points and average. Our guide to how measurement errors combine explains the rule, and the post on preparing for the practical exam and viva covers the questions examiners ask.
Which other mistakes should you watch for?
Using diameter as radius. This makes the area four times too large.
Mixing up stress and pressure. They share the unit, but stress is an internal restoring force per area in a solid.
Applying Hooke's law past the elastic limit. If the stress is above the yield strength, the linear formula does not hold.
Forgetting that strain has no units. If yours has units, something is wrong.
If the idea of stress inside a solid still feels abstract, the Fermi Concept Library covers every concept in physics, each with its own lesson, so you can revise stress and strain before going back to problems.
A practice routine
Day 1: Five direct problems like the worked example, converting every unit on a separate line.
Day 2: Four ratio problems with different lengths, diameters and materials.
Day 3: Two problems on a wire under its own weight and two on elastic energy.
Day 4: Write out the error formula for the practical and work out the percentage error from a sample set of readings.
Tonight, redo the worked example for a copper wire (Y = 1.1 × 10¹¹ Pa) and check that the extension is about 1.8 times larger.
