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

• A. $9 F /\left(\pi r^{2} Y T\right)$
• B. $6 F /\left(\pi r^{2} Y T\right)$
• C. $3 F /\left(\pi r^{2} Y T\right)$
• D. $F /\left(3 \pi r^{2} Y T\right)$

Answer 1

Answer: (C)

As length of rod remains unchanged,

Strain caused by compressive forces is equal and opposite to the thermal strain.
Now, compressive strain is obtained by using formula for Young's modulus,
$Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}$
Compressive strain,
$\frac{\Delta l}{l}=\frac{F}{A Y}=\frac{F}{\pi Y r^{2}}\,\,\,$...(i)
Also, thermal strain in rod is obtained by using formula for expansion in rod,
$\Delta l=l \alpha \Delta T$
$\Rightarrow$ Thermal strain, $\frac{\Delta l}{l}=\alpha \Delta T\,\,\,$...(ii)
From Eqs. (i) and (ii), we get
$\frac{F}{\pi r^{2} Y}=\alpha T\,\,\,[\because \Delta T=T]$
$\Rightarrow \alpha=\frac{F}{\pi r^{2} Y T}$
Hence, coefficient of volumetric expansion of rod is
$\gamma=3 \alpha=\frac{3 F}{\pi r^{2} Y T}$