Q. A wire extends by 1 mm when a force is applied. Double the force is applied to another wire of same material and length but half the radius of cross-section. The elongation of the wire in mm will be
A) 8
B) 4
C) 2
D) 1
Explanation
(a) $Y = \frac{Fl}{A\mathrm{\Delta}l}\ or\ \mathrm{\Delta}l\ \propto \frac{F}{r^{2}}$
Or $\frac{\mathrm{\Delta}l_{2}}{\mathrm{\Delta}l_{1}} = \frac{F_{2}}{F_{1}}\ \times \frac{r_{1}^{2}}{r_{2}^{2}}$
Or $\frac{\mathrm{\Delta}l_{2}}{\mathrm{\Delta}l_{1}} = \ 2 \times 2 \times 2 = 8\ $
Or $\ \mathrm{\Delta}l_{2} = 8\mathrm{\Delta}l_{1} = 8 \times 1\ mm = 8\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.