Q. Two similar wires under the same load yield elongation of $0.1\ mm$ and $0.05\ mm$ respectively. If the area of cross-section of the first wire is $4\ mm^{2}$, then the area of cross section of the second wire is
A) $6\ mm^{2}$
B) $8\ mm^{2}$
C) $10\ mm^{2}$
D) $12\ mm^{2}$
Explanation
(b) $l = \frac{FL}{AY}\ \therefore l \propto \frac{1}{A}\ \ \ \ \lbrack F,L$ and $Y$ are constant]
$\frac{A_{2}}{A_{1}} = \frac{l_{1}}{l_{2}} \Rightarrow A_{2} = A_{1}\left( \frac{0.1}{0.05} \right) = 2A_{1} = 2 \times 4 = 8mm^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.