Question

A copper wire and a steel wire of the same length and same cross section are joined end to end to form a composite wire. The composite wire is hung from a rigid support and a load is suspended from the other end. If the increase in length of the composite wire is $2.4mm$, then the increase in lengths of steel and copper wires are:

$(Y_{cu}=10\times {10}^{10}N/m^2,Y_{steel}=2\times {10}^{11}N/m^2)$

• A. $1.2mm$, $1.2mm$
• B. $0.6mm$, $1.8mm$
• C. $0.8mm$, $1.6mm$
• D. $0.4mm$, $2.0mm$

Answer 1

Answer: (C)

$0.8mm$, $1.6mm$

Given : $L_{cu}=L_{steel}=L$ $A_{cu}=A_{steel}=A$

Let the increase in the lengths of individual wires be $\Delta l_{cu}$ and $\Delta l_{steel}$ .

Total increase in the length of composite wire $\Delta l_{total}=2.4mm$

$\therefore$ $\Delta l_{cu}+\Delta l_{steel}=2.4$ ...........(1)

Also $Y=\frac{FL}{A\Delta l}$ $⟹\Delta l=\frac{FL}{AY}$

$\therefore$ $\Delta l_{cu}=\frac{F{L}_{cu}}{A_{cu}{Y}_{cu}}=\frac{FL}{A{Y}_{cu}}$

Similarly $\Delta l_{steel}=\frac{F{L}_{steel}}{A_{steel}{Y}_{steel}}=\frac{FL}{A{Y}_{steel}}$

Dividing these two equations we get $\frac{\Delta l_{cu}}{\Delta l_{steel}}=\frac{Y_{steel}}{Y_{cu}}$

$\therefore$ $\frac{\Delta l_{cu}}{\Delta l_{steel}}=\frac{2\times {10}^{11}}{10\times {10}^{10}}=2$ $⟹$ $\Delta l_{cu}=2\Delta l_{steel}$ ............(2)

From (1) and (2), $2\Delta l_{steel}+\Delta l_{steel}=2.4$ $⟹$ $\Delta l_{steel}=0.8mm$

$\therefore$ $\Delta l_{cu}=2\times 0.8=1.6mm$