Q. There are two wires of same material and same length while the diameter of second wire is 2 times the diameter of first wire, then ratio of extension produced in the wires by applying same load will be
A) 1 : 1
B) 2 : 1
C) 1 : 2
D) 4 : 1
Explanation
(d) $l = \frac{FL}{AY}\therefore l \propto \frac{1}{r^{2}}\lbrack F,L\ $and$\ Y$ are constant]
$\frac{l_{1}}{l_{2}} = \left( \frac{r_{2}}{r_{1}} \right)^{2} = (2)^{2} = 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.