Question

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

Answer 1

Answer: 0.55

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
$l_1+l_2=1.2cm$ .... (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),
$l_1+l_2=\frac{F\times 100}{a\times 12\times 10^{11}}+\frac{F\times 200}{a\times 20\times 10^{11}}$
$=\frac{F}{a}\left[8.33\times 10^{-11}+10\times 10^{-11}\right]$
$\therefore l_1+l_2=\frac{F}{a}\times 18.33\times 10^{-11}\dots$..(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 $\frac{F}{a}$,
$l_1=\frac{1.2\times 100}{18.33\times 10^{-11}\times 12\times 10^{11}}$
$=0.5456cm$
$\therefore l_1=0.55cm$
....(Rounding off to $2$ decimal places)