Q. The adjacent graph shows the extension ($l$) of a wire of length 1m suspended from the top of a roof at one end and with a load W connected to the other end. If the cross-sectional area of the wire is $10^{- 6\ }m^{2\ }$, calculate the Young’s modulus of the material of the wire.
A) $2 \times 10^{11\ }{Nm}^{- 2}$
B) $2 \times 10^{- 11\ }{Nm}^{- 2}$
C) $3 \times 10^{12\ }{Nm}^{- 2}$
D) $2 \times 10^{13\ }{Nm}^{- 2}$
Explanation
(a) Extensions $\mathrm{\Delta}l = \left( \frac{L}{YA} \right) \bullet W$
$ie,$ graph is a straight line passing through origin (as shown in question also), the slope of which is$\frac{L}{YA}$
Slope$= \left( \frac{L}{YA} \right)$
Y =$\left( \frac{L}{A} \right)\left( \frac{1}{slope} \right)$
$= \left( \frac{1.0}{10^{- 6}} \right)\ \frac{(80 - 20)}{(4 - 1) \times 10^{- 4}}$
$= 2.0 \times 10^{11}{Nm}^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.