Question

A steel wire of length 4.7 m and cross-sectional area 3.0 × 10^–5 m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10^–5 m² under a given load. What is the ratio of 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^–5 m²

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:

`Y_1 = F_1/A_1 xx L_1/(triangleL)`

`= (F xx 4.7)/(3.0 xx 10^(-5) xx triangleL)`...(i)

Young’s modulus of the copper wire:

`Y_2 = F_2/A_2 xx L_2/triangle_2`

`= (Fxx3.5)/(4.0 xx 10^(-5) xx triangleL)` ...(ii)

Dividing (i) by (ii), we get:

`Y_1/Y_2 = (4.7xx 4.0xx 10^(-5))/(3.0xx10^(-5)xx 3.5) = 1.79:1`

The ratio of Young’s modulus of steel to that of copper is 1.79 : 1.

Answer 2

For steel `I_1 = 4.7m`, `A_1 = 3.0 xx 10^(-5) m^2` 

if F newton is the stretching force and `trianglel` metre the extension in each case, then

`Y_1  = (Fl_1)/(A_1trianglel)`

`=>Y_1 = (Fxx 4.7)/(3.0xx10^(-5)xxtrianglel)`  ...(i)

For copper `l_2 = 3.5m`, `A_2 = 4.0 xx 10^(-5) m^2`

Now `Y_2 = (Fxx3.5)/(4.0xx10^(-5)xx trianglel)` ...(ii)

Dividing (i) by (ii) we get

`Y_1/Y_2 = 4.7/(3.0xx10^(-5)) xx (4.0xx10^(-5))/3.5 = (4.7xx4.0)/(3.0xx3.5) =1.79`