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

we know, Young's modulus , Y = FL/A∆L
where F is tension , A is cross sectional area , L is the length of wire and ∆L is change in length.

case 1 :- $L_1=4.7m$
$A_1=3\times10^{-5}m^2$
Let load is F and extension in wire is l
then, $Y_1=\frac{FL_1}{A_1l}$
$Y_1=\frac{4.7F}{3\times10^{-5}l}$....(1)

case 2 :- $L_2=3.5m$
$A_2=4\times10^{-5}m^2$
load is same e.g., F and extension in wire is also same e.g., l
then, $Y_2=\frac{FL_2}{A_2l}$
$Y_2=\frac{3.5F}{4\times10^{-5}l}$....(2)

from equations (1) and (2),

$\frac{Y_1}{Y_2}=\frac{4.7}{3.5}\frac{4}{3}\\=\frac{188}{105}$

Answer 2

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 = (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.