Question

A wire extends by $1mm$ when a force is applied. Double the force is applied to another wire of same material and length but of half the radius of cross section.The elongation of wire in mm will be

• A. $8$
• B. $4$
• C. $2$
• D. $1$

Answer 1

The correct option is A $8$

$Y=\frac{Fl}{A\Delta l}$

$\Delta l=\frac{Fl}{\pi r^{2}Y}$

$\frac{\Delta l_1}{\Delta l_2}=\frac{Fl}{\pi r^{2}Y}\times \frac{Y\pi {\left(\frac{r}{2}\right)}^2}{2Fl}$

$\frac{\Delta l_1}{\Delta l_2}=\frac{1}{8}$

$8\Delta l_1=\Delta l_2$

So, $\Delta l_2=8\left(1\right)$

$=8mm$