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Stress, strain and Young's modulus: Class 11 practice

Stress = force ÷ area, strain = extension ÷ original length (no unit). Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Scientific notation
  • Area of a circle

The idea behind stress, strain and young's modulus

Stress = force ÷ area, strain = extension ÷ original length (no unit). Young's modulus Y = stress ÷ strain = FL/(AΔL).

Step-by-step method

  1. Convert area to m² and lengths to m.
  2. Stress = F/A; strain = ΔL/L.
  3. Y = stress/strain, or rearrange for ΔL.

Worked example 1 (Basic)

A 300 N load hangs on a wire of area 1 mm². Find the stress.

  1. Stress = 300/1e-6
  2. = 3.00e+8 N/m²
  3. Answer: 3.00e+8 N/m²

Worked example 2 (Standard)

A 3 m wire of area 3 mm² (Y = 7.0e+10 N/m²) carries 700 N. Find its extension.

  1. ΔL = FL/(AY)
  2. = 10.000 mm
  3. Answer: 10.000 mm

Worked example 3 (Challenge)

A 5 m wire of area 5 mm² (Y = 1.2e+11 N/m²) carries 400 N. Find its extension.

  1. ΔL = FL/(AY)
  2. = 3.333 mm
  3. Answer: 3.333 mm

Common mistakes and misconceptions

  • Leaving area in mm²
  • Giving strain a unit

Practice questions

  1. (Basic) A 1900 N load hangs on a wire of area 5 mm². Find the stress.
  2. (Basic) A 1200 N load hangs on a wire of area 1 mm². Find the stress.
  3. (Standard) A 1 m wire of area 1 mm² (Y = 1.2e+11 N/m²) carries 1900 N. Find its extension.
  4. (Standard) A 3 m wire of area 4 mm² (Y = 2.0e+11 N/m²) carries 500 N. Find its extension.
  5. (Challenge) A 3 m wire of area 4 mm² (Y = 7.0e+10 N/m²) carries 1300 N. Find its extension.
  6. (Challenge) A 1 m wire of area 5 mm² (Y = 1.2e+11 N/m²) carries 1200 N. Find its extension.

Answer key with full working

  • Q1: Stress = 1900/5e-6 → = 3.80e+8 N/m² → Answer: 3.80e+8 N/m²
  • Q2: Stress = 1200/1e-6 → = 1.20e+9 N/m² → Answer: 1.20e+9 N/m²
  • Q3: ΔL = FL/(AY) → = 15.833 mm → Answer: 15.833 mm
  • Q4: ΔL = FL/(AY) → = 1.875 mm → Answer: 1.875 mm
  • Q5: ΔL = FL/(AY) → = 13.929 mm → Answer: 13.929 mm
  • Q6: ΔL = FL/(AY) → = 2.000 mm → Answer: 2.000 mm

Syllabus note

This skill is listed under "Mechanical Properties of Solids" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is stress, strain and young's modulus in?

"Mechanical Properties of Solids", Class 11 Physics (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

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