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:

Let the linear coefficient of thermal expansion of the material be denoted by $$\alpha$$ and its coefficient of volume expansion by $$\beta$$. For any isotropic solid we have the standard relation $$\beta = 3\alpha$$ because uniform heating produces the same fractional change along all three mutually perpendicular directions.

When the temperature of the rod is raised by $$T$$, the natural (unconstrained) increase in its length would be given by the well-known thermal expansion formula

$$\Delta L_{\text{thermal}} = \alpha L T.$$

Simultaneously, the rod is subjected to a uniform compressive longitudinal force $$F$$. The mechanical (elastic) shortening produced by this force is obtained from the definition of Young’s modulus. First, the longitudinal stress is

$$\text{Stress} = \frac{F}{A},$$

where $$A$$ is the cross-sectional area. For a cylinder of radius $$r$$ we have $$A = \pi r^{2}.$$

Young’s modulus $$Y$$ is defined by the relation

$$Y = \frac{\text{Stress}}{\text{Strain}} \;\;\; \Longrightarrow \;\;\; \text{Strain} = \frac{\text{Stress}}{Y}.$$

The longitudinal strain is the fractional change in length, so

$$\text{Strain} = \frac{\Delta L_{\text{elastic}}}{L} = \frac{F}{A Y}.$$

Since the force is compressive, this change in length is a decrease, hence

$$\Delta L_{\text{elastic}} = -\,\frac{F L}{A Y} = -\,\frac{F L}{\pi r^{2} Y}.$$

The problem states that the rod’s overall length does not change at all, which means the algebraic sum of the thermal increase and the elastic decrease is zero:

$$\Delta L_{\text{thermal}} + \Delta L_{\text{elastic}} = 0.$$

Substituting the two expressions we have just obtained,

$$\alpha L T \;+\;\left(-\,\frac{F L}{\pi r^{2} Y}\right) = 0.$$

The common factor $$L$$ can be cancelled on both sides, giving

$$\alpha T - \frac{F}{\pi r^{2} Y} = 0.$$

Solving for $$\alpha$$ we obtain

$$\alpha = \frac{F}{\pi r^{2} Y T}.$$

Finally, using the earlier relation $$\beta = 3\alpha$$ for an isotropic solid, we substitute this value of $$\alpha$$ to get

$$\beta \;=\; 3 \times \frac{F}{\pi r^{2} Y T} \;=\; \frac{3F}{\pi r^{2} Y T}.$$

Hence, the correct answer is Option D.