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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?
Solution 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.
Solution 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`
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