A steel wire of cross-sectional area 2x10⁻⁵ m² is stretched through 2 mm by a force of 4000 N. Find Young's modulus of the wire. The length of the wire is 2 m. Ans. (2 ｘ 10¹¹ Nm⁻²)


Given:

Cross-Sectional Area of the Steel Wire = A = 2 ｘ10⁻⁵ m²
Change in length after stretching by force = ΔL = 2mm = 2 ｘ10⁻³ m
Force exerted  = F = 4000 N
Original Length of steel wire  = Lₒ = 2 m

To Find:

Young's modulus of the wire = Y = ?

Solution:

As we have the formula for Young's modulus

Y = `\frac {F ｘ L}{A ｘ ΔL}` 

by putting values

Y = `\frac {4000 N ｘ 2 m}{2 ｘ10⁻⁵ m² ｘ 2 ｘ10⁻³ m}` 

Y = `\frac {2000 N}{10⁻⁵⁻³ m²}` 

Y = `\frac {2 ｘ10³ N}{10⁻⁸ m²}` 

Y = 2 ｘ10³⁺⁸ N m⁻² 

Y = 2 ｘ10¹¹ N m⁻² 

Thus Young's modulus of the wire is 2 ｘ10¹¹ N m⁻² 

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