Question

A piece of copper wire has twice the radius of steel wire. One end of the copper wire is joined to one end of steel wire so that both of them can be subjected to the same longitudinal force. $Y$ for steel is twice that of copper. When the length of copper wire is increased by $1$%, the steel wire will be stretched by

• A. $2$% of its original length
• B. $1$% of its original length
• C. $4$% of its original length
• D. $0.5$% of its original length

Answer 1

The correct option is A $2$% of its original length

$Y=\frac{F/A}{\Delta l/l}$

$\frac{\Delta l}{l}=\frac{F}{YA}$

$\frac{\Delta l_c}{l_c}=\frac{F}{Y_c{A}_c}=\frac{F}{Y_c\pi (r_c{)}^2}$------(1)

$\frac{\Delta l_s}{l_s}=\frac{F}{Y_s{A}_s}=\frac{F}{Y_s\pi (r_s{)}^2}$------(2)

Dividing (1) by (2)

$\frac{\Delta l_c/l_c}{\Delta l_s/l_s}=\frac{Y_s(r_s{)}^2}{Y_c(r_c{)}^2}=\frac{Y_s}{Y_c}\times {\left(\frac{r_s}{r_c}\right)}^2=\left(2\right)\times (\frac{1}{2}{)}^2$

$\Rightarrow \frac{\Delta l_c/l_c}{\Delta l_s/l_s}=\frac{1}{2}$

$\therefore \Delta l_s/l_s=2\Delta l_c/l_c$

$=2$%