Q. A wire of area of cross-section $10^{- 6}m^{2}$ is increased in length by 0.1%. The tension produced is $1000\ N$. The Young’s modulus of wire is
A) $10^{12}N/m^{2}$
B) $10^{11}N/m^{2}$
C) $10^{10}N/m^{2}$
D) $10^{9}N/m^{2}$
Explanation
(a) $Y = \frac{FL}{Al} = \frac{1000 \times 100}{10^{- 6} \times 0.1} = 10^{12}N/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.