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 g/cm. Based on these experimental measurement, what is Young’s modulus for this alloy iron

Short answer

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

Step-by-step solution

Identification of given data

The given data can be listed below,

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

Concept/Significance of young modulus

The elastic modulus of a material is defined of its stiffness that is constant throughout a wide range of stresses in most materials.

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

Determination of young 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 of gravity

Substitute all the values in the above,

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

Cross-sectional area of the wire is given by,

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

Here, D is the diameter of wire.

Substitute the value in the above,

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

Young modulus of the wire is given by,

$Y=\frac{FL}{A∆L}$

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

Substitute all the values in the above expression.

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

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