Question

A $5$m long aluminium wire $(Y=7\times {10}^{10}N{m}^{-2})$ of diameter $3$mm supports a $40$kg mass. In order to have the same elongation in the copper wire $(Y=12\times {10}^{10}N{m}^{-2})$ of the same length under the same weight, the diameter of the copper wire should now be (in mm).

• A. $1.75$
• B. $1.5$
• C. $2.5$
• D. $5.0$

Answer 1

The correct option is C $2.5$

$l=\frac{FL}{\pi r^{2}Y}$
$\Rightarrow r^2\propto \frac{1}{Y}(F,L$ and $l$ are constant$)$
$\frac{r_2}{r_1}={\left[\frac{Y_1}{Y_2}\right]}^{1/2}={\left[\frac{7\times {10}^{10}}{12\times {10}^{10}}\right]}^{1/2}$
$\Rightarrow r_2=1.5\times {\left(\frac{7}{12}\right)}^{1/2}=1.145$mm
$\therefore$ diameter$=2.29$mm