A rod is $2.40 \mathrm{~m}$ long and has a diameter of $2.50 \mathrm{~mm}$. A force of $2000 \mathrm{~N}$ is applied to the end, stretching the rod by $1.40 \mathrm{~mm}$. What is the elastic modulus for this rod?
A) $6.98 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$
B) $1.75 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$
C) $1.43 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}$
D) $5.83 \times 10^{-4} \mathrm{~N} / \mathrm{m}^{2}$
E) $4.07 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}$

Question

Quizplus · Accepted Answer

The elastic modulus (E) is calculated using the formula $E = \frac{F \cdot L_0}{A \cdot \Delta L}$, where $F$ is the force applied, $L_0$ is the original length, $A$ is the cross-sectional area, and $\Delta L$ is the change in length. The area $A$ of the rod's cross-section is $\pi r^2 = \pi (1.25 \times 10^{-3} \mathrm{m})^2$. Substituting the given values, the elastic modulus is found to be $6.98 \times 10^{11} \mathrm{N/m}^2$.

A Rod Is 2.40 m2.40 \mathrm{~m}2.40 m Long and Has a Diameter Of 2.50 mm2.50 \mathrm{~mm}2.50 mm

A Rod Is $2.40 \mathrm{~m}$ Long and Has a Diameter Of $2.50 \mathrm{~mm}$