Q. 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) $0.002\ m$
Explanation
(a) $l = \frac{FL}{\pi r^{2}Y}\ \therefore l \propto \frac{L}{r^{2}}\ \lbrack Y$ and $F$ are constant]
$\frac{l_{2}}{l_{1}} = \frac{L_{2}}{L_{1}} \times \left( \frac{r_{1}}{r_{2}} \right)^{2} = (2) \times \left( \frac{1}{2} \right)^{2} = \frac{1}{2}$
$\Rightarrow l_{2} = \frac{l_{1}}{2} = \frac{0.01m}{2} = 0.005m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.