Question

The area of cross section of a rod is given by $A={{A}_{0}}\left( 1+\alpha x \right)$, where ${{A}_{0}}$ and $\alpha $ are constants and $x$ is the distance from one end. The thermal conductivity of the material is $K$ . What is the thermal resistance of the rod if its length is ${{l}_{0}}$?
$A)\text{ }K{{A}_{0}}\alpha \ln \left( 1+\alpha {{l}_{0}} \right)$
$B)\text{ }\dfrac{1}{K{{A}_{0}}\alpha }\ln \left( 1+\alpha {{l}_{0}} \right)$
$C)\text{ }\dfrac{\alpha }{K{{A}_{0}}}\ln \left( 1+\alpha {{l}_{0}} \right)$
$D)\text{ }\dfrac{K{{A}_{0}}}{\alpha }\ln \left( 1+\alpha {{l}_{0}} \right)$

Answer 1

Hint: We will use the direct formula for the thermal resistance of an infinitesimal length of the rod in terms of its thermal conductivity, length and cross sectional area and then integrate it for the whole length of the rod to get the thermal resistance of the whole rod.

Formula used:
$R=\dfrac{x}{KA}$

Complete step by step answer:
We will use the direct formula for the thermal resistance of a body in terms of its length, cross sectional area and thermal conductivity.
The thermal resistance $R$ of a body of length $x$, cross sectional area $A$ and thermal conductivity $K$ is given by
$R=\dfrac{x}{KA}$ --(1)
Now, let us analyze the question.
The area of cross section of the rod is given by $A={{A}_{0}}\left( 1+\alpha x \right)$, where ${{A}_{0}}$ and $\alpha $ are constants and $x$ is the distance from one end.
The full length of the rod is ${{l}_{0}}$.
The thermal conductivity of the material of the rod is $K$.
Let the thermal resistance of the full rod be $R$.
Now, we will consider an infinitesimal strip of length $dx$ at a position $x$ from one end.
The cross sectional area of this strip will be $A={{A}_{0}}\left( 1+\alpha x \right)$.
Let the thermal resistance of this small strip be $dR$.
Therefore, using (1), we get,
$dR=\dfrac{dx}{KA}=\dfrac{dx}{K{{A}_{0}}\left( 1+\alpha x \right)}$ --(2)
Upon integrating (2) for the whole length of the rod, we will get the thermal resistance for the rod.
Therefore, integrating (2) with the respective variables, we get
$\int\limits_{0}^{R}{dR}=\int\limits_{0}^{{{l}_{0}}}{\dfrac{1}{K{{A}_{0}}\left( 1+\alpha x \right)}dx}=\dfrac{1}{K{{A}_{0}}}\int\limits_{0}^{{{l}_{0}}}{\dfrac{1}{\left( 1+\alpha x \right)}dx}$ --(3)
Now, we put $1+\alpha x=m$
Differentiating it on both sides we get
$d\left( 1+\alpha x \right)=0+\alpha dx=dm$
$\therefore \alpha dx=dm$
$\therefore dx=\dfrac{dm}{\alpha }$
When $x=0,m=1;x={{l}_{0}},m=1+\alpha {{l}_{0}}$
Putting these values in (3), we get
\[\int\limits_{0}^{R}{dR}=\dfrac{1}{K{{A}_{0}}}\int\limits_{1}^{1+\alpha {{l}_{0}}}{\dfrac{1}{m}\dfrac{dm}{\alpha }}\]
$\Rightarrow \left[ R \right]_{0}^{R}=\dfrac{1}{K{{A}_{0}}\alpha }\left[ \ln m \right]_{1}^{1+\alpha {{l}_{0}}}$ $\left( \because \int{\dfrac{1}{m}dm}=\ln m \right)$
$\Rightarrow \left[ R-0 \right]=\dfrac{1}{K{{A}_{0}}\alpha }\left[ \ln \left( 1+\alpha {{l}_{0}} \right)-\ln \left( 1 \right) \right]$
$\Rightarrow R=\dfrac{1}{K{{A}_{0}}\alpha }\left[ \ln \left( 1+\alpha {{l}_{0}} \right)-0 \right]$ $\left( \because \ln \left( 1 \right)=0 \right)$
 $\Rightarrow R=\dfrac{1}{K{{A}_{0}}\alpha }\left[ \ln \left( 1+\alpha {{l}_{0}} \right) \right]=\dfrac{1}{K{{A}_{0}}\alpha }\ln \left( 1+\alpha {{l}_{0}} \right)$
$\Rightarrow R=\dfrac{1}{K{{A}_{0}}\alpha }\ln \left( 1+\alpha {{l}_{0}} \right)$
Therefore, we have got the expression for the thermal resistance of the rod.

So, the correct answer is “Option B”.

Note:
Whenever the area of cross section of a rod is varying, it is best to find the thermal resistance of the rod by first considering an infinitesimal strip of the rod and then by using any relation between two parameters of the rod (for example in this question, the cross sectional area and the distance from one end), integrate for the whole rod.
The thermal resistance of the rod is a measure of the resistance offered by the rod to the conduction of heat through the length of the rod. It can be seen from the formula (1) that analogous to electrical resistance offered by a conductor which is directly proportional to the length and inversely proportional to the cross sectional area, the thermal resistance also depends upon these parameters in the same way.