Q. A copper wire of length $4.0m$ and area of cross-section $1.2cm^{2}$ is stretched with a force of $4.8 \times 10^{3}N$. If Young’s modulus for copper is $1.2 \times 10^{11}N/m^{2}$, the increase in the length of the wire will be
A) $1.33\ mm$
B) $1.33\ cm$
C) $2.66\ mm$
D) $2.66\ cm$
Explanation
(a) $l = \frac{FL}{AY} = \frac{4.8 \times 10^{3} \times 4}{1.2 \times 10^{- 4} \times 1.2 \times 10^{11}} = 1.33\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.