Q. Two wires of the same material have lengths in the ratio 1 : 2 and their radii are in the ratio 1 : $\sqrt{2\ .}\ $ If they are stretched by applying equal forces, the increase in their lengths will be in the ratio
A) $\sqrt{2}\ :2$
B) 2 : $\sqrt{2}$
C) 1 : 1
D) 1 : 2
Explanation
(c) $Y = \frac{Fl}{\pi r^{2}\mathrm{\Delta}l}\ \ or\ \ \mathrm{\Delta}l = \frac{F}{\pi r^{2}Y}$
$\mathrm{\Delta}l\ \ \propto \frac{1}{r^{2}},\ \mathrm{\Delta}l^{'} \propto \ \frac{2l}{\left( \sqrt{2r} \right)^{2}}\ \ or\ \mathrm{\Delta}l^{'} \propto \frac{1}{r^{2}}$
$\therefore\frac{\mathrm{\Delta}l}{\mathrm{\Delta}l^{'}} = 1\ \ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.