A steel rod (Young's modulus = 2 × 1011 Nm–2) has an area of cross-section 3 × 10–4 m2 and length 1 m. A force of 6 × 104 N stretched it axially. The elongation of the rod is

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Stress Strain and Hookes Law New

Question

A steel rod (Young's modulus = 2 × 10¹¹ Nm^–2) has an area of cross-section 3 × 10^–4 m² and length 1 m. A force of 6 × 10⁴ N stretched it axially. The elongation of the rod is

5 × 10^–2 m

5 × 10^–3 m

10^–4 m

10^–3 m

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To find the elongation (ΔL) of the rod, we use the formula for longitudinal stress and strain from Hooke's Law: Stress = Young's Modulus × Strain.

Stress is Force (F) / Area (A), and Strain is ΔL / Original Length (L). Therefore, the formula is:

$\frac{F}{A} = Y \times \frac{Δ L}{L}$

Rearranging for ΔL: $Δ L = \frac{F \times L}{Y \times A}$

Substitute the given values: F = 6×10⁴ N, L = 1 m, Y = 2×10¹¹ Nm⁻², A = 3×10⁻⁴ m².

$Δ L = \frac{6 \times 10^{4} \times 1}{2 \times 10^{11} \times 3 \times 10^{- 4}} = \frac{6 \times 10^{4}}{6 \times 10^{7}} = 10^{- 3} m$

The elongation is 10⁻³ m.