The Strain Formula gives the ratio of the change in dimension of a body to its original dimension when an external force is applied. Expressed as ε = ΔL / L₀, this formula is a cornerstone of Class 11 Physics (NCERT Chapter 9 — Mechanical Properties of Solids). It is equally critical for JEE Main, JEE Advanced, and NEET aspirants who encounter stress-strain problems every year. This article covers the complete strain formula, its types, derivation, a full formula sheet, three solved examples at progressive difficulty levels, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Strain Formulas at a Glance
Quick reference for the most important strain-related formulas.
- Longitudinal Strain: \( \varepsilon = \dfrac{\Delta L}{L_0} \)
- Volumetric Strain: \( \varepsilon_v = \dfrac{\Delta V}{V_0} \)
- Shear Strain: \( \gamma = \tan\theta \approx \theta \) (for small angles)
- Young’s Modulus: \( Y = \dfrac{\text{Stress}}{\text{Longitudinal Strain}} = \dfrac{F/A}{\Delta L/L_0} \)
- Bulk Modulus: \( B = -\dfrac{P}{\Delta V / V_0} \)
- Modulus of Rigidity: \( G = \dfrac{\text{Shear Stress}}{\text{Shear Strain}} \)
- Poisson’s Ratio: \( \sigma = -\dfrac{\varepsilon_{\text{lateral}}}{\varepsilon_{\text{longitudinal}}} \)
What is the Strain Formula?
The Strain Formula defines strain as the fractional change in the size or shape of a body due to an applied external force. Strain is a dimensionless quantity because it is the ratio of two lengths (or two volumes). It has no SI unit.
In NCERT Class 11 Physics, Chapter 9 (Mechanical Properties of Solids), strain is introduced alongside stress to describe how materials respond to external forces. The concept is fundamental to understanding elasticity, Hooke’s Law, and the elastic moduli of materials.
There are three main types of strain. Longitudinal strain occurs when a body is stretched or compressed along its length. Volumetric strain occurs when a body experiences a change in volume under uniform pressure. Shear strain occurs when a body is deformed by a tangential force, causing angular displacement.
Understanding the strain formula helps students predict whether a material will return to its original shape (elastic behaviour) or undergo permanent deformation (plastic behaviour). This distinction is crucial in both CBSE board exams and competitive entrance tests.
Strain Formula — Expression and Variables
The general longitudinal strain formula is:
\[ \varepsilon = \frac{\Delta L}{L_0} \]
where \( \Delta L = L_f – L_0 \) is the change in length and \( L_0 \) is the original length.
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( \varepsilon \) | Longitudinal Strain | Dimensionless (no unit) |
| \( \Delta L \) | Change in length (\( L_f – L_0 \)) | Metre (m) |
| \( L_0 \) | Original length | Metre (m) |
| \( \varepsilon_v \) | Volumetric Strain | Dimensionless (no unit) |
| \( \Delta V \) | Change in volume | m³ |
| \( V_0 \) | Original volume | m³ |
| \( \gamma \) | Shear Strain | Dimensionless (radians) |
| \( \theta \) | Angle of shear | Radians |
Types of Strain and Their Expressions
1. Longitudinal (Tensile/Compressive) Strain:
\[ \varepsilon = \frac{\Delta L}{L_0} = \frac{L_f – L_0}{L_0} \]
2. Volumetric Strain:
\[ \varepsilon_v = \frac{\Delta V}{V_0} \]
3. Shear Strain:
\[ \gamma = \tan\theta \approx \theta \quad (\text{for small } \theta) \]
Derivation of the Strain Formula
Consider a wire of original length \( L_0 \) fixed at one end. An external tensile force F is applied at the free end. The wire elongates by \( \Delta L \), reaching a final length \( L_f \).
By definition, strain measures how much the body deforms relative to its original size. We write:
\[ \varepsilon = \frac{\text{Change in length}}{\text{Original length}} = \frac{L_f – L_0}{L_0} = \frac{\Delta L}{L_0} \]
Since both \( \Delta L \) and \( L_0 \) are measured in metres, the metres cancel. Strain is therefore a pure number with no units. This derivation holds for any linear dimension. For three-dimensional deformation, the same logic extends to volumetric strain using volume instead of length.
Complete Mechanical Properties of Solids Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Longitudinal Strain | \( \varepsilon = \Delta L / L_0 \) | ΔL = change in length, L&sub0; = original length | Dimensionless | Class 11, Ch 9 |
| Volumetric Strain | \( \varepsilon_v = \Delta V / V_0 \) | ΔV = change in volume, V&sub0; = original volume | Dimensionless | Class 11, Ch 9 |
| Shear Strain | \( \gamma = \tan\theta \approx \theta \) | θ = angle of shear deformation | Dimensionless (rad) | Class 11, Ch 9 |
| Stress | \( \sigma = F / A \) | F = applied force, A = cross-sectional area | Pa (N/m²) | Class 11, Ch 9 |
| Young’s Modulus | \( Y = \dfrac{F L_0}{A \Delta L} \) | F = force, L&sub0; = original length, A = area, ΔL = elongation | Pa (N/m²) | Class 11, Ch 9 |
| Bulk Modulus | \( B = -\dfrac{P V_0}{\Delta V} \) | P = pressure, V&sub0; = original volume, ΔV = change in volume | Pa (N/m²) | Class 11, Ch 9 |
| Modulus of Rigidity | \( G = \dfrac{F/A}{\theta} \) | F/A = shear stress, θ = shear strain | Pa (N/m²) | Class 11, Ch 9 |
| Hooke’s Law | \( \sigma = Y \varepsilon \) | σ = stress, Y = Young’s modulus, ϵ = strain | Pa | Class 11, Ch 9 |
| Poisson’s Ratio | \( \sigma_p = -\dfrac{\varepsilon_{\text{lat}}}{\varepsilon_{\text{long}}} \) | ϵ_lat = lateral strain, ϵ_long = longitudinal strain | Dimensionless | Class 11, Ch 9 |
| Elastic Potential Energy Density | \( u = \dfrac{1}{2} \times \text{Stress} \times \text{Strain} \) | u = energy per unit volume | J/m³ | Class 11, Ch 9 |
| Elongation of Wire | \( \Delta L = \dfrac{F L_0}{A Y} \) | F = force, L&sub0; = length, A = area, Y = Young’s modulus | m | Class 11, Ch 9 |
Strain Formula — Solved Examples
Example 1 (Class 9-10 Level): Basic Longitudinal Strain
Problem: A rubber band has an original length of 10 cm. When stretched, its length becomes 14 cm. Calculate the longitudinal strain produced in the rubber band.
Given:
- Original length, \( L_0 = 10 \) cm = 0.10 m
- Final length, \( L_f = 14 \) cm = 0.14 m
Step 1: Find the change in length.
\( \Delta L = L_f – L_0 = 14 – 10 = 4 \) cm = 0.04 m
Step 2: Apply the strain formula.
\( \varepsilon = \dfrac{\Delta L}{L_0} = \dfrac{4}{10} = 0.4 \)
Step 3: Check units. Both lengths are in cm; they cancel. Strain is dimensionless.
Answer
Longitudinal Strain = 0.4 (dimensionless)
Example 2 (Class 11-12 Level): Young’s Modulus and Strain
Problem: A steel wire of length 2.0 m and cross-sectional area 2.0 × 10²³¹&sup0; m² is subjected to a tensile force of 200 N. The Young’s modulus of steel is 2.0 × 10¹¹ Pa. Find (a) the longitudinal strain and (b) the elongation of the wire.
Given:
- \( L_0 = 2.0 \) m
- \( A = 2.0 \times 10^{-6} \) m²
- \( F = 200 \) N
- \( Y = 2.0 \times 10^{11} \) Pa
Step 1: Calculate stress.
\( \sigma = \dfrac{F}{A} = \dfrac{200}{2.0 \times 10^{-6}} = 1.0 \times 10^{8} \) Pa
Step 2: Use Young’s modulus to find strain.
\( Y = \dfrac{\sigma}{\varepsilon} \Rightarrow \varepsilon = \dfrac{\sigma}{Y} = \dfrac{1.0 \times 10^{8}}{2.0 \times 10^{11}} = 5.0 \times 10^{-4} \)
Step 3: Find elongation using the strain formula.
\( \varepsilon = \dfrac{\Delta L}{L_0} \Rightarrow \Delta L = \varepsilon \times L_0 = 5.0 \times 10^{-4} \times 2.0 = 1.0 \times 10^{-3} \) m
Answer
(a) Longitudinal Strain = 5.0 × 10²³&sup4; (dimensionless)
(b) Elongation = 1.0 mm (1.0 × 10²³³ m)
Example 3 (JEE/NEET Level): Volumetric Strain and Bulk Modulus
Problem: A solid copper sphere of volume 0.50 m³ is placed in a hydraulic press where the pressure is increased by 2.0 × 10&sup7; Pa. The bulk modulus of copper is 1.4 × 10¹¹ Pa. Calculate (a) the volumetric strain and (b) the decrease in volume of the sphere.
Given:
- \( V_0 = 0.50 \) m³
- \( \Delta P = 2.0 \times 10^{7} \) Pa
- \( B = 1.4 \times 10^{11} \) Pa
Step 1: Write the bulk modulus formula.
\( B = -\dfrac{\Delta P}{\Delta V / V_0} \)
Step 2: Solve for volumetric strain \( (\Delta V / V_0) \).
\( \dfrac{\Delta V}{V_0} = -\dfrac{\Delta P}{B} = -\dfrac{2.0 \times 10^{7}}{1.4 \times 10^{11}} = -1.43 \times 10^{-4} \)
The negative sign indicates a decrease in volume (compression).
Step 3: Calculate the actual change in volume.
\( |\Delta V| = 1.43 \times 10^{-4} \times 0.50 = 7.14 \times 10^{-5} \) m³
Step 4: State the volumetric strain magnitude.
Volumetric Strain \( = 1.43 \times 10^{-4} \) (dimensionless)
Answer
(a) Volumetric Strain = 1.43 × 10²³&sup4; (compression)
(b) Decrease in volume = 7.14 × 10²³&sup5; m³
CBSE Exam Tips 2025-26
- Remember that strain is dimensionless. A very common board exam question asks for the SI unit of strain. The correct answer is “no unit” or “dimensionless.” Never write Pa or N/m² for strain.
- Distinguish between stress and strain clearly. Stress has units (Pa), but strain does not. Confusing the two costs marks in 2-mark definition questions.
- Learn all three types of strain. CBSE 2025-26 board papers frequently ask students to define or compare longitudinal, volumetric, and shear strain. Keep a short note for each.
- Practice the Young’s Modulus link. Most 3-mark and 5-mark numerical problems combine the strain formula with Young’s modulus. We recommend practising at least 10 such problems before the board exam.
- Use consistent units. Convert all lengths to metres and all forces to Newtons before substituting. Unit errors are the most common reason for losing marks in numericals.
- Revise Hooke’s Law. The NCERT Class 11 Chapter 9 text explicitly connects strain to Hooke’s Law. Expect a 2-mark question linking the two in 2025-26 board exams.
Common Mistakes to Avoid
- Mistake 1 — Assigning units to strain. Students often write the unit of strain as “m” or “m/m.” Strain is a pure ratio and is always unitless. Write “no unit” or “dimensionless.”
- Mistake 2 — Using final length instead of change in length. The formula uses \( \Delta L = L_f – L_0 \), not \( L_f \) alone. Always subtract the original length first.
- Mistake 3 — Confusing compressive and tensile strain signs. Tensile strain is positive (elongation). Compressive strain is negative (shortening). In volumetric strain, a decrease in volume gives a negative value. Always check the sign.
- Mistake 4 — Applying the longitudinal strain formula to volumetric problems. When pressure is applied uniformly in all directions, use the volumetric strain formula \( \varepsilon_v = \Delta V / V_0 \) and the Bulk Modulus, not Young’s Modulus.
- Mistake 5 — Ignoring Poisson’s effect. When a wire is stretched longitudinally, its lateral (transverse) dimensions decrease. JEE problems sometimes ask for lateral strain. Do not assume only one type of strain occurs.
JEE/NEET Application of the Strain Formula
In our experience, JEE aspirants encounter the strain formula in at least two to three questions per paper, either directly or as part of a larger elasticity problem. NEET typically includes one concept-based question on the type of strain or elastic moduli. Here are the key application patterns to master.
Pattern 1: Strain from Young’s Modulus
JEE Main frequently provides the Young’s modulus, applied force, and wire dimensions, then asks for elongation or strain. The approach is:
- Calculate stress: \( \sigma = F/A \)
- Use \( Y = \sigma / \varepsilon \) to find strain: \( \varepsilon = \sigma / Y \)
- Find elongation: \( \Delta L = \varepsilon \times L_0 \)
This three-step chain is the most tested pattern in JEE Main 2023, 2024, and expected in 2025.
Pattern 2: Comparing Strains in Composite Wires
JEE Advanced problems often involve two wires of different materials connected in series under the same load. Since both wires carry the same force, the wire with the smaller Young’s modulus will have the larger strain. The formula \( \varepsilon = F / (AY) \) makes this comparison straightforward.
Pattern 3: Volumetric Strain and Compressibility
NEET and JEE Main both test the relationship between bulk modulus, pressure, and volumetric strain. Compressibility is defined as \( k = 1/B \). A higher bulk modulus means smaller volumetric strain for the same applied pressure. This concept appears as both a numerical and a conceptual MCQ.
Our experts suggest memorising the values of Young’s modulus for steel (\( 2 \times 10^{11} \) Pa), copper (\( 1.2 \times 10^{11} \) Pa), and aluminium (\( 7 \times 10^{10} \) Pa). These appear directly in JEE and NEET numericals without being given in the question.
FAQs on Strain Formula
Related Physics Formulas
Strengthen your understanding of elasticity and fluid mechanics by exploring these related formula articles on ncertbooks.net:
- Visit our Complete Physics Formulas Hub for a full list of Class 11 and Class 12 formula articles.
- Learn how pressure and depth relate in fluids with the Buoyancy Formula, which pairs naturally with volumetric strain concepts.
- Understand fluid flow and pressure changes with Bernoulli’s Equation Formula, another key Class 11 topic.
- Explore light and optics with the Brewster’s Law Formula for complete CBSE Class 12 Physics coverage.
For the official NCERT syllabus and chapter list, refer to the NCERT official website.