Question

A wire is stretched by 0.01 m by a certain force F. Another wire of same material whose diameter and length are double to the original wire is stretched by the same force. Then its elongation will be

• A. 0.005 m
• B. 0.01 m
• C. 0.02 m
• D. 0.002 m

Answer 1

The correct option is A 0.005 m

$l=\frac{FL}{\pi r^{2}Y}\therefore l\propto \frac{L}{r^2}$ [As F and Y are constants]
$\frac{l_2}{l_1}=\left(\frac{L_2}{L_1}\right){\left(\frac{r_1}{r_2}\right)}^2=\left(2\right)\times {\left(\frac{1}{2}\right)}^2=\frac{1}{2}\Rightarrow l_2=\frac{l_1}{2}=\frac{0.01}{2}=0.005m$.