Q. Two pieces of wire $A\ and\ B$ of the same material have their lengths in the ratio 1 : 2, and their diameters are in the ratio 2 : 1. If they are stretched by the same force, their elongations will be in the ratio
A) 2 : 1
B) 1 : 4
C) 1 : 8
D) 8 : 1
Explanation
(c) $Y = \frac{F}{\pi r^{2}}\ \times \frac{L}{\mathrm{\Delta}L} = \ \frac{F\ \times 2L}{x(r/2)^{2}\ \mathrm{\Delta}L}\ \ \ or\ \frac{\mathrm{\Delta}L}{\mathrm{\Delta}L^{'}} = \frac{1}{8}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.