Q. An iron rod of length $2m$ and cross section area of $50mm^{2}$, stretched by $0.5mm$, when a mass of $250kg$ is hung from its lower end. Young’s modulus of the iron rod is
A) $19.6 \times 10^{10}N/m^{2}$
B) $19.6 \times 10^{15}N/m^{2}$
C) $19.6 \times 10^{18}N/m^{2}$
D) $19.6 \times 10^{20}N/m^{2}$
Explanation
(a) $Y = \frac{MgL}{Al} = \frac{250 \times 9.8 \times 2}{50 \times 10^{- 6} \times 0.5 \times 10^{- 3}}$
$= 19.6 \times 10^{11}N/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.