Q. When a wire of length 10m is subjected to a force of 100 N along its length, the lateral strain produced is $0.01 \times 10^{- 3}$m. The Poisson’s ratio was found to be 0.4. If the area of cross-section of wire is $0.025\ m^{2}$, its Young’s modulus is
A) $1.6 \times 10^{8\ }{Nm}^{- 2}$
B) $2.5 \times 10^{10\ }{Nm}^{- 2}$
C) $1.25 \times 10^{11}{Nm}^{- 2}$
D) $16 \times 10^{9}{Nm}^{- 2}$
Explanation
(a) Poisson’s ratio $= \frac{Lateral\ strain}{Longitudinal\ strain}$
$ie,\ \ \ 0.4 = \frac{0.01 \times 10^{- 3}}{\frac{l}{L}}$
or $\frac{L}{l} = \frac{0.4}{0.01 \times 10^{- 3}} = 4{\times 10}^{4}$
Young’s modulus
$Y = \frac{FL}{Al}$
$= \frac{100}{0.025} \times 4 \times 10^{4} = 1.6 \times 10^{8}{Nm}^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.