Question

A copper wire of length 1.0 m and a steel wire of length 0.5 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 mm. If the Young's moduli of copper and steel are $1.0\times {10}^{11}N{m}^{-2}$ and $2.0\times {10}^{11}N{m}^{-2}$, the total extension of the composite wire is

• A. 1.25 mm
• B. 1.50 mm
• C. 1.75 mm
• D. 2.00 mm

Answer 1

The correct option is A

1.25 mm

When a wire of 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
$l=\frac{FL}{AY}$
Since F and A are the same for the two wires, we have
For copper wire $l_c=\frac{F{L}_c}{A{Y}_c}$
For steel wire $l_s=\frac{F{L}_s}{A{Y}_s}$
$\therefore l_s=l_c\left(\frac{L_s}{L_c}\right)\left(\frac{Y_c}{Y_s}\right)$
$=1mm\times \left(\frac{0.5m}{1.0m}\right)\times \left(\frac{1.0\times {10}^{11}N{m}^{-2}}{2.0\times {10}^{11}N{m}^{-2}}\right)$
$=0.25mm$
$\therefore$ Total extension = 1 + 0.25 = 1.25 mm.
Hence, the correct choice is (a).