A steel rod has a radius of 20 mm and a length of 2.0 m. A force of 62.8 kN stretches it along its length. Young's modulus of steel is 2.0 × 10¹¹ N/m². The longitudinal strain produced in the wire is _____________ × 10⁻⁵ Numerical answer type. Work it out, then open the answer.

Answer: 25. Strain = (stress)/(Y) = (F)/(π r² Y) Stress = 62.8 × 10³3.14 × (0.02)² = 62.8 × 10³1.256 × 10⁻³ = 5 × 10⁷ N/m² Strain = 5 × 10⁷2 × 10¹¹ = 2.5 × 10⁻⁴ = 25 × 10⁻⁵ Answer: 25

A steel rod has a radius of 20 mm and a length of 2.0 m. A force of… | JEE Main 2023 (6 Apr, Shift 1)

Q12JEE Main 2023 · 6 APR · SHIFT 1Easy

Elastic Behaviour, Stress & Strain

A steel rod has a radius of $20\text{ mm}$ and a length of $2.0\text{ m}$ . A force of $62.8\text{ kN}$ stretches it along its length. Young's modulus of steel is $2.0 \times 10^{11}\text{ N/m}^2$ . The longitudinal strain produced in the wire is _____________ $\times 10^{-5}$

Numerical answer type. Work it out, then open the answer.

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Answer: 25

Strain $= \frac{\text{stress}}{Y} = \frac{F}{\pi r^2 Y}$
Stress $= \frac{62.8 \times 10^{3}}{3.14 \times (0.02)^2} = \frac{62.8 \times 10^3}{1.256 \times 10^{-3}} = 5 \times 10^{7}\ \text{N/m}^2$
Strain $= \frac{5 \times 10^{7}}{2 \times 10^{11}} = 2.5 \times 10^{-4} = 25 \times 10^{-5}$
Answer: $25$

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