Q. A stress of 1 kg $mm^{2}$is applied on a wire. If the modulus of elasticity of the wire is $10^{10}\ dyne\ cm^{- 2},$ then the percentage increase in the length of the wire will be
A) 0.0098%
B) 0.98%
C) 9.8%
D) 98%
Explanation
(b) $\frac{\mathrm{\Delta}l}{l} = \frac{stress}{Y} = \frac{1000 \times 980)/\left( 10^{- 1} \right)^{2}}{10^{10}} = 0.0098$
% increase in length of wire
= $\frac{\mathrm{\Delta}l}{l} \times 100 = 0.0098\ \times 100 = 0.98\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.