Question

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

• A. 8
• B. 4
• C. 2
• D. 1

Answer 1

The correct option is A 8

$a)I=\frac{FL}{\pi r^{2}r}l\propto \frac{F}{r^2}$ (Y and L are constant)
$\frac{l_2}{l_1}=\frac{F_2}{F_1}\times (\frac{r_1}{r_2}{)}^2=2\times (2{)}^2=8$
$l_2=8l_1=8\times 1=8$