Question

A metal bar of 10 mm diameter when subjected to a pull of 23.5 kN gave an elongation of 0.3 mm on a gauge length of 200 mm. The Young's modulus of elasticity of the metal will nearly be
1. 200 kN/mm²
2. 300 kN/mm²
3. 360 kN/mm²
4. 400 kN/mm²

Answer 1

Correct Answer - Option 1 : 200 kN/mm²

Concept:

The Young’s modulus of elasticity is the ratio of stress to strain i.e.

\(E = \frac{\sigma }{\epsilon}\)

Wherein,

E = Young’s modulus of elasticity

σ = Stress (Force per unit area)

ϵ = Strain (Ratio of Change in length to original length)

Note that, here we should take actual area of cross-section because on stretching the bar its diameter reduces owing to Poison’s effect. Since Poison’s ratio is not given, we assume actual area is equal to the original given area.

Calculation:

Given,

Force, F = 23.5 kN; diameter, d = 10 mm; 

Gauge length (original length), l  = 200 mm

Increase in length, Δl = 0.3 mm

\({\rm{Stress}},{\rm{\;\sigma }} = {\rm{\;}}\frac{{23.5\; \times \;1000}}{{\frac{{\rm{\pi }}}{4}\; \times \;{{10}^2}}} = 299.36{\rm{\;MPa}}\)

\({\rm{Strain}},\epsilon {\rm{}} = {\rm{\;}}\frac{{{\rm{\Delta L}}}}{{\rm{L}}} = \frac{{0.3}}{{200}} = 1.5 \times {\rm{\;}}{10^{ - 3}}\)

∴ \(E = \frac{\sigma }{\epsilon} = \frac{{299.36}}{{1.5\; \times {\rm{\;}}{{10}^{ - 3}}}}\) = 199.6 × 10³ MPa

⇒ E ≈ 200 kN/mm²