NEET 2024 Physics Question 11 Solution
Question
The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young's modulus, respectively, are 8×10⁸ N m⁻² and 2×10¹¹ N m⁻², is:
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The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young's modulus, respectively, are 8×10⁸ N m⁻² and 2×10¹¹ N m⁻², is:
ΔL_max = (σ_max / Y) × L = (8 × 10⁸ / 2 × 10¹¹) × 1 = 4 × 10⁻³ m = 4 mm!
Up to the elastic limit, deformation is completely reversible, and Hooke’s law holds rigorously. Young’s modulus Y is defined as the ratio of tensile stress σ to tensile strain ε:
To find the maximum elongation ΔL_max without exceeding the elastic limit, we use the maximum allowable stress σ_max:
Substituting σ_max = 8 × 10⁸ N m⁻², L = 1 m, and Y = 2 × 10¹¹ N m⁻²:
Since 1 m = 1000 mm:
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Off by a factor of 10.
Fails to divide by the factor of 2 in Young’s modulus.
Accurate evaluation: ΔL = 4 × 10⁻³ m = 4 mm.
Off by a factor of 10.
Students frequently mix up 10⁻³ m as 0.4 mm or 40 mm when rushing.
"Tensile strain is the ratio of tensile stress to Young’s modulus (ΔL/L = σ/Y)."
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