Q. The ratio of diameters of two wires of same materials is $n$: 1. The length of each wire is 4 m. On applying the same load, the increase in length of thin wire will be $(n > 1)$
A) $n^{2}\ times\ $
B) $n\ times\ $
C) $2n\ times\ $
D) $(2n + 1)times\ $
Explanation
(a) $Y = \ \frac{\frac{F}{A}}{\frac{\mathrm{\Delta}l}{l}} = \frac{Fl}{A\mathrm{\Delta}l}\ $
Or $\ Y = \ \frac{Fl\ \times 4}{\pi D^{2} \times \mathrm{\Delta}l}\ or\mathrm{\Delta}l\ \propto \frac{1}{D^{2}}\ or\frac{\mathrm{\Delta}L_{2}}{\mathrm{\Delta}L_{1}} = \frac{D_{1}^{2}}{D_{2}^{2}} = \frac{n^{2}}{1}\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.