Question

A steel wire of length 4.7 m and cross-sectional area 3.0 x 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 x 10⁻⁵ m² under a given load. What is the ratio of the Young's modulus of steel to that of copper?

Answer 1

Length of the steel wire, L1 = 4.7 m
Area of cross-section of the steel wire, A1 = 3.0 × 10–5m2
Length of the copper wire, L2 = 3.5 m
Area of cross-section of the copper wire, A2 = 4.0 × 10–5 m2
Change in length = ΔL1 = ΔL2 = ΔL
Force applied in both the cases = F
Young’s modulus of the steel wire:
Y1 = (F1 / A1) (L1 / ΔL1)
= (F / 3 X 10-5) (4.7 / ΔL)     ….(i)
Young’s modulus of the copper wire:
Y2 = (F2 / A2) (L2 / ΔL2)
= (F / 4 × 10-5) (3.5 / ΔL)     ….(ii)
Dividing (i) by (ii), we get:
Y1 / Y2  =  (4.7 × 4 × 10-5) / (3 × 10-5 × 3.5)
= 1.79 : 1
The ratio of Young’s modulus of steel to that of copper is 1.79 : 1.

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Answer 2

Answer:

Y₁ : Y₂ = 9 : 5

Explanation:

Given :

For steel wire :

A₁ = 3 × 10⁻⁵ m³  and l₁ = 4.7 m

For copper wire :

A₂ =  4 × 10⁻⁵ m³  and l₂ = 3.5 m

It is said as :

Δl₁ = Δl₂ = Δl  and  F₁ = F₂ = F

We know :

Y₁ =  F₁ l₁ / A₁ Δl₁

= > F /  3 × 10⁻⁵ × 4.7 /Δl

Also Y₂ = F₂ l₂ / A₂ Δl₂

Y₂ = F /  4 × 10⁻⁵ m³ × 3.5 / Δl

We have find ratio of Y₁ / Y₂

Y₁ : Y₂ = ( F /  3 × 10⁻⁵ × 4.7 /Δl ) / ( F /  4 × 10⁻⁵ m³ × 3.5 / Δl )

Y₁ : Y₂ = 4 × 10⁻⁵ × 4.7 / 3 × 10⁻⁵ × 3.5 )

Y₁ : Y₂ = 18.5 / 10.5 ≈ 1.8

Y₁ : Y₂ = 18 / 10 .

Y₁ : Y₂ = 9 : 5

Hence the ratio of the Young's modulus of steel to that of copper 9 : 5