Question

A rod $CD$ of thermal resistance $5.0{\text{KW}}^{-1}$ is joined at the middle of an identical rod $AB$ as shown in the figure. The ends $A,B$ and $D$ are maintained at ${100}^{\circ }\text{C}$, $0^{\circ }\text{C}$ and ${25}^{\circ }\text{C}$ respectively. Find the heat current in $CD$.

• A. $2\text{W}$
• B. $3\text{W}$
• C. $4\text{W}$
• D. $5\text{W}$

Answer 1

The correct option is C $4\text{W}$


The thermal resistance of $CD$ is equal to $5{\text{KW}}^{-1}$.
As, $AB$ and $CD$ are identical rods, the thermal resistance of $AC$ is equal to that of $CB$ and each is equal to $2.5{\text{KW}}^{-1}$.

Suppose, the temperature at $C$ is $\theta$.

The heat currents through $AC,CB$ and $CD$ are given by,

${\left(\frac{dq}{dt}\right)}_{AC}=\frac{(100-\theta )}{2.5}$

${\left(\frac{dq}{dt}\right)}_{CB}=\frac{\theta -0}{2.5}$

${\left(\frac{dq}{dt}\right)}_{CD}=\frac{\theta -25}{5}$

Now,
${\left(\frac{dq}{dt}\right)}_{AC}={\left(\frac{dq}{dt}\right)}_{CB}+{\left(\frac{dq}{dt}\right)}_{CD}$

$\Rightarrow \frac{100-\theta }{2.5}=\frac{\theta -0}{2.5}+\frac{\theta -25}{5}$

$200-2\theta =2\theta +\theta -25$

$5\theta =225$

$\therefore \theta =45{}^{\circ }\text{C}$
$\therefore {\left(\frac{dq}{dt}\right)}_{CD}=\frac{45-25}{5}=4\text{W}$

Hence, $\left(C\right)$ is the correct answer.