Q. The length of a wire is increased by 1 mm on the application of a given load. In a wire of the same material, but of length and radius twice that of the first, on the application of the same load, extension is
A) 0.25 cm
B) 0.5 cm
C) 2 mm
D) 4 mm
Explanation
(b) $Y = \frac{Fl}{A\mathrm{\Delta}l}\ or\ \ \mathrm{\Delta}l = \frac{Fl}{AY} = \frac{Fl}{\pi r^{2}Y}$
In the given problem, $\mathrm{\Delta}l = \frac{1}{r^{2}};$ when both $l\ and\ r\ $are double, $\mathrm{\Delta}l\ $ is halved.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.