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 :

Properties of Matter PYQ — JEE Main 2019 · 12 Apr · Shift 2 · Q51 with Solution

View written solutionFree

Correct answer: A

The rod is heated and compressed simultaneously, yet its length remains unchanged.

If the coefficient of linear expansion is $\alpha$ , then the increase in length due to heating is

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

Longitudinal strain produced by compressive force $F$ is

$\text{strain} = \frac{F}{AY}$

So decrease in length is

$\Delta L_{\text{compress}} = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$

Since the rod's length remains unchanged,

$\Delta L_{\text{thermal}} = \Delta L_{\text{compress}}$

So,

$\alpha L T = \frac{FL}{\pi r^2 Y}$

Cancel $L$ :

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

Hence,

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

For an isotropic solid,

$\gamma = 3\alpha$

Therefore,

$\gamma = 3\left(\frac{F}{\pi r^2 Y T}\right)$

$\boxed{\gamma = \frac{3F}{\pi r^2 Y T}}$

This corresponds to Option A:

$\boxed{\frac{3F}{\pi r^2 Y T}}$

Stored correct answer: A

Our derived answer: A

So they agree.