Practice question
Question
A copper rod of length 2.0 m and cross-sectional area 2.5 × 10⁻⁶m² is subjected to a tensile force of 500 N . If the Young's modulus of copper is 1.1 × 10¹¹N/m², what is the strain produced?
Explanation
Given:
A copper rod of length 2.0 m and cross-sectional area 2.5 × 10⁻⁶m² is subjected to a tensile force of 500 N . If the Young's modulus of copper is 1.1 × 10¹¹N/m², what is the strain produced?
These values define the system as per NCERT data.
Formula:
Stress: Stress = F/A = frac5002.5 × 10⁻⁶= 2 × 10⁸N/m².
This is standard NCERT relation.
Substitution & Calculation:
Young's modulus: Y = fracStressStrain . Strain: Strain = fracStressY = frac2 × 10⁸¹.1 × 10¹¹approx 1.82 × 10⁻³.
Result:
The computed value matches expected outcome and confirms correct choice as per NCERT.
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