Question

A metal rod that is 4.00 m long and $\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{2}}$in cross-sectional area is found to stretch 0.20 cm under a tension of 5000 N. What is Young’s modulus for this metal?

Short answer

$2\times {10}^{11}Pa$

Step-by-step solution

Given information

Length: $l_0=4.00m$,

Area: $A=0.50c{m}^2$,

Elongation: $∆l=0.20cm$,

Tension Force: F = 4000 N .

Concept/Formula used

$\mathit{Y}\mathbf{=}\frac{{\mathbf{l}}_{\mathbf{0}}\mathbf{F}}{\mathbf{∆}\mathbf{l}}$

Where, Y is Young’s modulus, l₀ is length of muscle, F is muscle force, A is cross-sectional area and $\mathbf{∆}\mathbf{l}$ is elongation.

Young’s modulus Calculation

$$\begin{gathered} Y=\frac{(4.00m)(5000N)}{(0.50\times {10}^{-4}{m}^2)(0.20\times {10}^{-2}m)} \\ =2\times {10}^{11}Pa \end{gathered}$$