Question

A steel wire is lf length 1 metre and area of cross-section 1 cm². If Young's modulus of steel is 10¹¹ Newton/meter², the force required to elongate the wire of 1 mm is
1. 10³ Newton
2. 10⁴ Newton
3. 10⁵ Newton
4. 10¹¹ Newton

Answer 1

Correct Answer - Option 2 : 10⁴ Newton

Concept:

Young’s modulus: 

• Young's modulus a modulus of elasticity, applicable to the stretching of wire, etc., equal to the ratio of the applied load per unit area of the cross-section to the increase in length per unit length.
• It is denoted as E or Y.
• The unit of Young’s modulus is N m-2.

\(\text{Y}=\frac{\text{ }\!\!σ\!\!\text{ }}{\epsilon }\)

Where σ = stress, ϵ = strain in wire.

• Young’s Modulus Formula by using other quantities:

\({\rm{Y}} = \frac{{{\rm{F}}{{\rm{L}}_0}}}{{{\rm{A\Delta L}}}}\)

Where F = force exerted under tension, A = actual cross-sectional area, L0 = actual length, ΔL = change in length.

Calculation:

Given, 

Actual length of wire L₀ = 1 m

Actual cross sectional area A = 1 cm² = 10 ^-4 m²

ΔL = change in length = 1 mm = 10 ^-3 m

Youngs modulus Y = 1011 N/m2

Force required F

Now, using these parameters in the formula of Youngs Modulus. 

\(\)\(10^{11} = \frac{F(1)}{10^{-4}10^{-3}}\)

\(\implies 10^{11} = \frac{F(1)}{10^{-7}}\)

⇒ F = 10¹¹ × 10 ^-7 = 10⁴ N

So, the correct option is 104 N