Q. A weight of $200\ kg$ is suspended by vertical wire of length $600.5cm$. The area of cross-section of wire is $1mm^{2}$. When the load is removed, the wire contracts by $0.5\ cm$. The Young’s modulus of the material of wire will be
A) $2.35 \times 10^{12}N/m^{2}$
B) $1.35 \times 10^{10}N/m^{2}$
C) $13.5 \times 10^{11}N/m^{2}$
D) $23.5 \times 10^{9}N/m^{2}$
Explanation
(a) $F = 2000N,L = 6m,\ l = 0.5cm,\ A = 10^{- 6}m^{2}$
$Y = \frac{FL}{Al} = \frac{2000 \times 6}{10^{- 6} \times 0.5 \times 10^{- 2}} = 2.35 \times 10^{12}N/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.