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A copper wire of length \mathrm{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 \mathrm{1.0 \times 10^{11} \mathrm{Nm}^{-2}} and \mathrm{2.0 \times 10^{11} \mathrm{Nm}^{-2}, } the total extension of the composite wire is

Option: 1

1.25 mm


Option: 2

1.50 mm


Option: 3

1.75 mm


Option: 4

2.0 mm


Answers (1)

best_answer

When a wire or length L, cross-sectional area A and
Young's modulus Y is stretched with a force F, the extension $l$ in the wire is given by
                                      \mathrm{ l=\frac{F L}{A Y} }


Since F and A are the same for the two wires, we have

For copper wire \mathrm{l_c=\frac{F L_c}{A Y_c}}

For steel wire, \mathrm{\quad l_s=\frac{F L_s}{A Y_s}}

\mathrm{ \begin{aligned} \therefore \quad l_s= & l_c \frac{L_s}{L_c} \cdot \frac{Y_c}{Y_s} \\ = & 1 \mathrm{~mm} \times\left(\frac{0.5 \mathrm{~m}}{1.0 \mathrm{~m}}\right) \\ & \times\left(\frac{1.0 \times 10^{11} \mathrm{Nm}^{-2}}{2.0 \times 10^{11} \mathrm{Nm}^{-2}}\right) \\ & =0.25 \mathrm{~mm} \end{aligned} }

\mathrm{\therefore Total \, \, extension =1+0.25=1.25 \mathrm{~mm}.}

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