Calculating Longitudinal Strain in a Stretched Cylindrical Rod

Q: Why is the length L = 2.0m not needed in the calculation?

Strain is defined as (Δ L)/L = F/AY. Since the question asks directly for strain, the original length L cancels out and is extra information.

A steel rod has a radius of 20mm and a length of 2.0m. A force of 62.8kN stretches it along its length. Young's modulus of steel is 2.0 × 10^11N/m^2. The longitudinal strain produced in the wire is _______ × 10^(-5)

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Question type: Numerical
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

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Why is the length

L = 2.0 m

not needed in the calculation?

Strain is defined as $\frac{Δ L}{L} = \frac{F}{A Y}$ . Since the question asks directly for strain, the original length $L$ cancels out and is extra information.

Answer

Accepted answer: 25

The longitudinal strain produced in the steel rod is 25 × 10⁻⁵.

Why each option works or fails

Step-by-step solution

Given: Radius $r = 20 mm = 20 \times 10^{- 3} m = 2 \times 10^{- 2} m$ , length $L = 2.0 m$ , applied force $F = 62.8 kN = 62.8 \times 10^{3} N \approx 20 π \times 10^{3} N$ , Young's modulus $Y = 2.0 \times 10^{11} {N/m}^{2}$ .
Find: Longitudinal strain $ε = \frac{Δ L}{L}$ expressed as a factor of $10^{- 5}$ .
Visualise: A cylindrical steel rod of cross-sectional area $A = π r^{2}$ is subjected to an axial tensile force $F$ , causing longitudinal elongation.
Strategise: From Hooke's law: Stress $= Y \times Strain ⟹ Strain = \frac{F}{A \cdot Y} = \frac{F}{π r^{2} Y}$ . Note that $62.8 \approx 20 π$ , which simplifies cancellation with $π$ .
Execute: Calculate the cross-sectional area: $A = π (2 \times 10^{- 2})^{2} = 4 π \times 10^{- 4} m^{2} = 4 (3.14) \times 10^{- 4} = 1.256 \times 10^{- 3} m^{2}$ . Now, compute strain: $ε = \frac{62.8 \times 10^{3}}{(4 \times 3.14 \times 10^{- 4}) (2.0 \times 10^{11})} = \frac{62.8 \times 10^{3}}{2.512 \times 10^{8}} = 25 \times 10^{- 5}$ .
Verify: Strain is dimensionless. $25 \times 10^{- 5} = 2.5 \times 10^{- 4}$ (or $0.025 %$ ), which is well within the elastic limit for steel rod under moderate stress.

Calculating Longitudinal Strain in a Stretched Cylindrical Rod

What feels right?

The longitudinal strain produced in the steel rod is 25 × 10⁻⁵.

Why each option works or fails

Solution

01Given

02Find

03Visualise

04Strategise

05Execute

✓Verify

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Calculating Longitudinal Strain in a Stretched Cylindrical Rod

The longitudinal strain produced in the steel rod is 25 × 10⁻⁵.

01Given

02Find

03Visualise

04Strategise

05Execute

✓Verify

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