Question

A uniform cylindrical rod of length ‘L’ and radius ‘r’, is made from a material whose Young's modulus of elasticity equals ‘Y’. When this rod is heated by temperature ‘T’ and simultaneously subjected to a net longitudinal compressional force ‘F’, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to:
1. \(\frac{{9{\rm{F}}}}{{\left( {{\rm{\pi }}{{\rm{r}}^2}{\rm{YT}}} \right)}}\)
2. \(\frac{{6{\rm{F}}}}{{\left( {{\rm{\pi }}{{\rm{r}}^2}{\rm{YT}}} \right)}}\)
3. \(\frac{{3{\rm{F}}}}{{\left( {{\rm{\pi }}{{\rm{r}}^2}{\rm{YT}}} \right)}}\)
4. \(\frac{{\rm{F}}}{{\left( {3{\rm{\pi }}{{\rm{r}}^2}{\rm{YT}}} \right)}}\)

Answer 1

Correct Answer - Option 3 : \(\frac{{3{\rm{F}}}}{{\left( {{\rm{\pi }}{{\rm{r}}^2}{\rm{YT}}} \right)}}\)

Concept:

Given, length of the uniform cylindrical rod = L

Radius of the uniform cylindrical rod = R

Young’s modulus of the uniform cylindrical rod = Y

Uniform rod heated to the temperature = T

Longitudinal compressive force acting on the rod = F

From question, we need to find the coefficient of volumetric expansion.

The volumetric expansion is:

α_V = 3α_L

The linear expansion is given by the formula:

\(\frac{{{\rm{\Delta L}}}}{{\rm{L}}} = {{\rm{\alpha }}_{\rm{L}}}{\rm{\Delta T}}\) 

\({{\rm{\alpha }}_{\rm{L}}} = \frac{{{\rm{\Delta L}}}}{{{\rm{L}}\left( {{\rm{\Delta T}}} \right)}}\) 

The young modulus is given by the formula:

\({\rm{Y}} = \frac{{{\rm{Stress}}}}{{{\rm{Strain}}}}\) 

Now, the stress (force per unit area applied to the material) is:

\({\rm{Stress}} = \frac{{\rm{F}}}{{\rm{A}}}\) 

Where,

A = Area of the rod = πr²

\(\therefore {\rm{Stress}} = \frac{{\rm{F}}}{{{\rm{\pi }}{{\rm{r}}^2}}}\) 

Now, the strain (ratio of change in length to original length) is:

\({\rm{Strain}} = \frac{{{\rm{\Delta L}}}}{{\rm{L}}}\) 

Calculation:

On substituting stress and strain in young’s modulus,

\( \Rightarrow {\rm{Y}} = \frac{{\left( {\frac{{\rm{F}}}{{{\rm{\pi }}{{\rm{r}}^2}}}} \right)}}{{\left( {\frac{{{\rm{\Delta L}}}}{{\rm{L}}}} \right)}}\)

\( \Rightarrow {\rm{Y}} = \frac{{\rm{F}}}{{{\rm{\pi }}{{\rm{r}}^2}}} \times \frac{{\rm{L}}}{{{\rm{\Delta L}}}}\)

\( \Rightarrow \frac{{\rm{L}}}{{{\rm{\Delta L}}}} = \frac{{{\rm{Y}}\left( {{\rm{\pi }}{{\rm{r}}^2}} \right)}}{{\rm{F}}}\)

\( \Rightarrow \frac{{{\rm{\Delta L}}}}{{\rm{L}}} = \frac{{\rm{F}}}{{{\rm{Y}}\left( {{\rm{\pi }}{{\rm{r}}^2}} \right)}}\)

Now, substituting above equation in linear expansion,

\( \Rightarrow {{\rm{\alpha }}_{\rm{L}}} = \frac{{\rm{F}}}{{\left( {{\rm{Y}}\left( {{\rm{\pi }}{{\rm{r}}^2}} \right)} \right){\rm{\Delta T}}}}\) 

\(\therefore {{\rm{\alpha }}_{\rm{L}}} = \frac{{\rm{F}}}{{{\rm{\pi }}{{\rm{r}}^2}{\rm{Y\Delta T}}}}\) 

Now, substituting linear expansion in volumetric expansion,

\( \Rightarrow {{\rm{\alpha }}_{\rm{V}}} = 3\left( {\frac{{\rm{F}}}{{{\rm{\pi }}{{\rm{r}}^2}{\rm{Y\Delta T}}}}} \right)\) 

\(\therefore {{\rm{\alpha }}_{\rm{V}}} = \frac{{3{\rm{F}}}}{{{\rm{\pi }}{{\rm{r}}^2}{\rm{Y\Delta T}}}} = \frac{{3{\rm{F}}}}{{{\rm{\pi }}{{\rm{r}}^2}{\rm{YT}}}}\)