Question

A hanging wire made of an alloy of iron with diameter 0.09cm is initially 2.2m long. When a 66kg mass is hung from it, the wire stretches an amount of 1.12cm. A mole of iron has a mass of 56g, and its density is 7.87 $\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$. Based on these experimental measurement, what is Young’s modulus for this alloy iron.

Short answer

The Young’s modulus of the iron alloy is $2.0\times {10}^{11}{\text{N/m}}^2$.

Step-by-step solution

Identification of given data

The given data is listed below.

• The diameter of wire is $D=0.09\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)=0.09\times {10}^{-2}\text{m}$.
• The length of the wire is $L=2.2\text{m}$.
• Mass of the hanging object is $m=66\text{kg}$.
• The elongation is wire is $\Delta L=1.12\text{cm}$.
• Molecular mass of the iron is $M_{Fe}=56\text{g}$.
• The density of iron is ${\rho }_{Fe}=7.87{\text{g/cm}}^3\left(\frac{1000{\text{kg/m}}^3}{1{\text{g/cm}}^3}\right)=7.87\times {10}^3{\text{kg/m}}^3$.

Concept/ Definition of Young’s modulus

Young's modulus is defined as the proportion of longitudinal strain to longitudinal stress.

Young's modulus or elastic modulus of a material can be understood as a measure of its stiffness that is constant throughout a wide range of stresses in most materials.

Determination of Young’s modulus of iron alloy

The force on the hanging object is given by

$F=mg$

Here m is the mass of the object and g is the acceleration due to gravity.

Substitute all the values in the above.

$$\begin{gathered} F=66\text{kg}\times 9.8{\text{m/s}}^2 \\ =646.8\text{N} \end{gathered}$$

The cross-sectional area of the wire is given by

$A=\frac{\pi D^2}{4}$

Here D is the diameter of wire.

Substitute the values in the above.

$$\begin{gathered} A=\frac{\pi {\left(0.09\times {10}^{-2}\text{m}\right)}^2}{4} \\ =6.36\times {10}^{-7}{\text{m}}^2 \end{gathered}$$

Young’s modulus of the wire is given by

$Y=\frac{FL}{A\Delta L}$

Here F is the force on the object, L is the length of the wire, A is the cross-sectional area of the wire and$∆L$ is the elongation in the wire.

Substitute all the values in the above expression.

$$\begin{gathered} Y=\frac{\left(646.8\text{N}\right)\left(2.2\text{m}\right)}{\left(6.36\times {10}^{-7}{\text{m}}^2\right)\left(1.12\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)} \\ =2.0\times {10}^{11}{\text{N/m}}^2 \end{gathered}$$

Thus, the Young’s modulus of the iron alloy is $2.0\times {10}^{11}{\text{N/m}}^2$.