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 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 = 3.0×10–5 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 :

$Y1 =\frac{ F_1}{ A_1} =\frac{ L_1 }{ ΔL}$

$=\frac{ F × 4.7 }{ 3.0 × 10 ^{- 5} × ΔL} ...(i)$

Young’s modulus of the copper wire :

$Y_2 = \frac{F_2}{ A_2} × \frac{l_2}{ΔL_2}$

$=\frac{F\times3.5}{4.0\times^{-5}\times ΔL} ......(ii)$

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

$\frac{Y_1 }{ Y_2} = \frac{4.7 × 4.0 × 10 ^{- 5} }{ 3.0 × 10^{ - 5} × 3.5} = 1.79:1$
The ratio of Young’s modulus of steel to that of copper is 1.79 : 1.