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 1.0×1011 Nm−2 and 2.0×1011 Nm−2, the total extension of the composite wire is
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=FLAY
Since F and A are the same for the two wires, we have
For copper wire lc=FLcAYc
For steel wire ls=FLsAYs
∴ls=lc(LsLc)(YcYs)
=1 mm×(0.5 m1.0 m)×(1.0×1011 Nm−22.0×1011 Nm−2)
=0.25 mm
∴ Total extension = 1 + 0.25 = 1.25 mm.
Hence, the correct choice is (a).