Q. A wire of length $2\ m$ is made from $10\ cm^{3}$ of copper. A force $F$ is applied so that its length increases by $2\ mm$. Another wire of length $8\ m$ is made from the same volume of copper. If the force $F$ is applied to it, its length will increase by
A) $0.8\ cm$
B) $1.6\ cm$
C) $2.4\ cm$
D) $3.2\ cm$
Explanation
(d) $l = \frac{FL}{AY} = \frac{FL^{2}}{(AL)Y} = \frac{FL^{2}}{VY}$
$\therefore l \propto L^{2}$ if volume of the wire remains constant
$\frac{l_{2}}{l_{1}} = \left( \frac{L_{2}}{L_{1}} \right)^{2} = \left( \frac{8}{2} \right)^{2} = 16$
$\therefore l_{2} = 16 \times l_{1} = 16 \times 2 = 32mm = 3.2cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.