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 to temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion is (nearly) equal to:
$
  {\text{A}}{\text{. }}\dfrac{F}{{3\pi {r^2}YT}} \\
  {\text{B}}{\text{. }}\dfrac{{3F}}{{\pi {r^2}YT}} \\
  {\text{C}}{\text{. }}\dfrac{{6F}}{{\pi {r^2}YT}} \\
  {\text{D}}{\text{. }}\dfrac{{9F}}{{\pi {r^2}YT}} \\
 $

Answer 1

Hint: According to given condition, in order to keep the length of the rod constant, the thermal stress due to expansion will be equal to the compressional stress acting on the rod. By equating these quantities, we can obtain the linear coefficient of expansion. The volume coefficient is three times the linear coefficient of expansion.

Complete step by step answer:
We are given a uniform cylindrical rod which has length L and radius r. The Young’s modulus of its material is given as Y. The rod is heated to a temperature T due to which it undergoes thermal expansion along its length. It is simultaneously subjected to a net longitudinal compressional force F along its length due to which its length remains unchanged. This means that stress due to thermal expansion is equal to the stress due to compression.
The thermal stress can be written in terms of the Young’s modulus and the strain since the young’s modulus is equal to the ratio of the thermal stress and the thermal strain on the given cylindrical rod. It is given as
Thermal stress $ = Y\alpha T$
Here $\alpha $ is known as the thermal coefficient of linear expansion of the material of the cylindrical rod.
The compression stress is simply equal to force applied on the rod per unit cross-sectional area of the rod and is given as
Compressional stress $ = \dfrac{F}{{\pi {r^2}}}$
Now on equating the two stresses, we get
$
  Y\alpha T = \dfrac{F}{{\pi {r^2}}} \\
   \therefore \alpha = \dfrac{F}{{\pi {r^2}YT}} \\
 $
Now we need to remember that the volume coefficient of expansion $\gamma $ is related to the linear coefficient of expansion in the following way.
$\gamma = 3\alpha $
Using the above expression here, we obtain
$\gamma = \dfrac{{3F}}{{\pi {r^2}YT}}$
This is the required value of the coefficient of volume expansion of the given cylindrical rod.

So, the correct answer is “Option B”.

Note:
It should be noted that when the rod is heated, it will expand along its length. The compressional stress is applied on the rod along the length of the rod and its direction is opposite to the direction in which the rod is undergoing thermal expansion.