A copper wire and a steel wire of the same cross – sectional area and of length 1 m and 2 m respectively are connected end to end. A force is applied, which stretches their combined length by 1 cm. Find how much each wire is elongated. (Young’s modulus of copper and steel are [latex]1.2\times 10^{11}N/m^2[/latex] & [latex]2\times 10^{11}N/m^2[/latex] respectively).

Solution
Combined elongation (e) = [latex]e_c + e_s[/latex] = 1 cm = 0.01 m

For Copper wire:
Length [latex](l_c)[/latex] = 1 m

[latex]Y_c = \frac{Stress}{Strain}[/latex]

Or, [latex]Stress = Y_c\times Strain[/latex]
Or, [latex]Stress = Y_c\times\frac{e_c}{l_c}[/latex] ………………. (i)

Similarly, for Steel wire:
Length [latex](l_s)[/latex] = 2 m
[latex]Stress = Y_s\times\frac{e_s}{l_s}[/latex] ………………….. (ii)

Since, Area is same for both wires. The stress developed is same.
So, from eqn. (i) & (ii),

[latex]Y_c\times\frac{e_c}{l_c} = Y_s\times\frac{e_s}{l_s}[/latex]