Two wires, one made of copper and other of steel are joined end to...

Young's modulus for steel

${Y}_{S} = \dfrac{FLS}{{A}_{S} ∆ {l}_{S}}$

Where,~${Y}_{S}$~is Young's modulus,~${L}_{S}$~is length,~${A}_{S}$~is area of cross section,~$∆ {l}_{S}$~is change in length,~$F$~is force applied for steel wire respectively.

~$⇒ {L}_{S} = \dfrac{{Y}_{S}{A}_{S} ∆ {l}_{S}}{F}$~.......(i)

Young's modulus for copper
${Y}_{C} = \dfrac{FLC}{{A}_{C} ∆ {l}_{C}}$

Where,~${Y}_{C}$~is Young modulus,~${L}_{C}$~is length,~${A}_{C}$~is area of cross section,~$∆ {l}_{C}$~is change in length,~$F$~is force applied for copper~wire respectively.

$⇒ {L}_{C} = \dfrac{{Y}_{C}{A}_{C} ∆ {l}_{C}}{F}$~........(ii)

Dividing equation (ii) by equation (i), we get
$\dfrac{LC}{LS} = \dfrac{{Y}_{C}{A}_{C} ∆ {l}_{C}}{F} \times \dfrac{F}{{Y}_{S}{A}_{S} ∆ {l}_{S}}$

or $\dfrac{LC}{LS} = \left(\dfrac{YC}{YS}\right) \times\left(\dfrac{AC}{AS}\right) \times \left(\dfrac{∆ {l}_{C}}{∆ {l}_{S}}\right)</p><p>Given,~${Y}_{S} = 2 $\times {10}^{11}$N / m^{2}$</p><p>${Y}_{C} = 1 . 1 $\times {10}^{11}$N / m^{2}$</p><p>${A}_{C} = 2{A}_{S}$</p><p>$∆ {l}_{C} = ∆ {l}_{S}$</p><p>$⇒ $\dfrac{LC}{LS} = \left(\dfrac{1 . 1 \times {10}^{11}}{2 \times {10}^{11}}\right) \times \left(\dfrac{2{A}_{S}}{AS}\right) \times \left(\dfrac{∆ {l}_{C}}{∆ {l}_{C}}\right)</p><p>$∴ $\dfrac{LC}{LS} =$1 . 1$<br /></p>$