A copper wire of length 1.0 ~m and a steel wire of length 0.5 ~m...

A copper wire of length

1.0 \mathrm{~m}

and a steel wire of length

0.5 \mathrm{~m}

having equal cross-sectional areas are joined end to end. The composite wire is stretched by a certain load which stretches the copper wire by

1 \mathrm{~mm}

. If the Young's modulii of copper and steel are respectively

1.0 \times 10^{11}

\mathrm{Nm}^{-2}

and

2.0 \times 10^{11} \mathrm{Nm}^{-2}

, the total extension of the composite wire is :

Options

A $1.75 \mathrm{~mm}$
B $2.0 \mathrm{~mm}$
C $1.50 \mathrm{~mm}$
D $1.25 \mathrm{~mm}$

Correct answer

D. $1.25 \mathrm{~mm}$

Step-by-step solution

\mathrm{Y}_{\mathrm{c}} \times

\left(\Delta \mathrm{L}_{\mathrm{c}} / \mathrm{L}_{\mathrm{c}}\right)=\mathrm{Y}_{\mathrm{s}} \times\left(\Delta \mathrm{L}_{\mathrm{s}} / \mathrm{L}_{\mathrm{s}}\right)

\Rightarrow 1 \times 10^{11} \times\left(\frac{1 \times 10^{-3}}{1}\right)=2 \times 10^{11} \times\left(\frac{\Delta \mathrm{L}_{\mathrm{s}}}{0.5}\right) \therefore \Delta \mathrm{L}_{\mathrm{s}}=\frac{0.5 \times 10^{-3}}{2}=0.25 \mathrm{~mm}

Therefore, total extension of the composite

\begin{aligned} & \text{ wire }=\Delta \mathrm{L}_{\mathrm{c}}+\Delta \mathrm{L}_{\mathrm{s}} \\ & =1 \mathrm{~mm}+0.25 \mathrm{~m}=1.25 \mathrm{~m} \end{aligned}

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