Question

The diameter of a rod is given by $d=d_0(1+ax)$ where $a$ is a constant and $x$ is the distance from one end. If the thermal conductivity of the material is $K,$ what is the thermal resistance of the rod if length is $l$ .

• A. $\frac{1}{K\pi d_0^2}$
• B. $\frac{4l}{K\pi d_0^2(al+1)}$
• C. $\frac{2l}{K\pi d_0^2(al+1{)}^2}$
• D. $\frac{2l}{K\pi d_0^2(al+1)}$

Answer 1

Answer: (B)

Thermal resistance of a rod is given by,
$R_{th}=\frac{1}{K}\frac{l}{A}$
Where,
$K\to$ Thermal conductivity of the material of the rod,
$l\to$ Length of the rod and
$A\to$ Cross-sectional area of the rod.
Given,
Diameter of the rod at a distance $x$ from one end of the rod is,
$d=d_0(1+ax)$ .
That is, the thickness of the rod increases from one end to the other end.
Let's consider an elementary disc of thickness $dx$ at a distance $x$ from one end as shown below:

Thermal resistance of this elementary disc is given by,
$dR=\frac{1}{K}\frac{dx}{A}=\frac{1}{K}\frac{dx}{\pi {\left(\frac{d}{2}\right)}^2}$
$\Rightarrow dR=\frac{4}{K}\frac{dx}{\pi {\left(d_0(1+ax)\right)}^2}$
$\Rightarrow dR=\frac{4}{\pi d_0^{2}K}\frac{dx}{{\left(1+ax\right)}^2}$
The thermal resistance of the whole rod can be determined by,
$\underset{0}{\overset{R}{\int }}dR=\frac{4}{\pi d_0^{2}K}\underset{0}{\overset{l}{\int }}\frac{dx}{{\left(1+ax\right)}^2}$
$\Rightarrow R=\frac{4}{\pi d_0^{2}K}\frac{1}{a}{\left[\frac{-1}{1+ax}\right]}_0^l$
$\Rightarrow R=\frac{4}{\pi a{d}_0^{2}K}\left[\frac{-1}{1+a\left(l\right)}-\frac{-1}{1+a\left(0\right)}\right]$
$\Rightarrow R=\frac{4}{\pi a{d}_0^{2}K}\left[1-\frac{1}{1+al}\right]$
$\Rightarrow R=\frac{4}{\pi a{d}_0^{2}K}\left[\frac{al}{1+al}\right]$
$\Rightarrow R=\frac{4l}{K\pi d_0^2\left(1+al\right)}$