Question

A copper wire and steel wire of the same diameter and of lengths $100\, cm$ and $200\, cm$ respectively are connected end to end and a force is applied, which stretches their combined length by $1.2 \,cm$. Find by what amount (in $cm$ ) copper wire is elongated, if Young's modulus for copper and steel are $12 \times 10^{11}$ dyne $cm ^{-2}$ and $20 \times 10^{11}$ dyne $cm ^{-2}$ respectively.

Answer 1

Let $L_{1}, L_{2}$ be the lengths and $Y_{1}, Y_{2}$ be the values of Young's modulus of the copper and steel wires respectively.
If $l_{1}$ and $l_{2}$ are respective increase in the lengths of the two wires, then
$l_{1}+l_{2}=1.2 \,cm$ .... (i)
Let $F$ be the force applied to the wires. If a is their area of cross-section, then
$ Y_{1}=\frac{F \times L_{1}}{a \times l_{1}} $
$\therefore l_{1}=\frac{F \times L_{1}}{a \times Y_{1}}=\frac{F \times 100}{a \times 12 \times 10^{11}}$
Also, $ Y_{2}=\frac{F \times L_{2}}{a \times l_{2}}$ .... (ii)
$\therefore l_{2}=\frac{F \times L_{2}}{a \times Y_{2}}=\frac{F \times 200}{a \times 20 \times 10^{11}}$ .... (iii)
Adding equations (ii) and (iii),
$l_{1}+l_{2} =\frac{ F \times 100}{ a \times 12 \times 10^{11}}+\frac{ F \times 200}{ a \times 20 \times 10^{11}} $
$=\frac{ F }{ a }\left[8.33 \times 10^{-11}+10 \times 10^{-11}\right]$
$\therefore l_{1}+l_{2}=\frac{ F }{ a } \times 18.33 \times 10^{-11} \ldots$..(iv)
From equations (i) and (iv),
$\frac{ F }{ a } \times 18.33 \times 10^{-11}=1.2$
$\therefore \frac{ F }{ a }=\frac{1.2}{18.33 \times 10^{-11}}$
In the equation (ii), substituting for $\frac{F}{a}$,
$l_{1} =\frac{1.2 \times 100}{18.33 \times 10^{-11} \times 12 \times 10^{11}}$
$=0.5456\, cm$
$\therefore l_{1} =0.55 \,cm$
....(Rounding off to $2$ decimal places)