Question

Two rods A and B of different materials are welded together as shown in the figure. If their thermal conductivities are $k_1$ and $k_2$, the thermal conductivity of the composite rod will be

• A. $2(k_1+k_2)$
• B. $\frac{3}{2}(k_1+k_2)$
• C. $k_1+k_2$
• D. $\frac{1}{2}(k_1+k_2)$

Answer 1

The correct option is D

$\frac{1}{2}(k_1+k_2)$

If the ends of the rods are maintained at temperatures ${\theta }_1$ and ${\theta }_2$, the rate of flow of heat through each rod is given by
$Q_1=\frac{k_{1}A({\theta }_1-{\theta }_2)}{d}$
and $Q_2=\frac{k_{2}A({\theta }_1-{\theta }_2)}{d}$
where A is the cross-sectional area of each rod and d is the length of each rod.
Now the cross-sectional area of the composite rod is 2A. If k is the thermal conductivity of the composite rod, the rate of heat flow is
$Q=\frac{k\left(2A\right)({\theta }_1-{\theta }_2)}{d}$
But, $Q=Q_1+Q_2$
$\Rightarrow \frac{k\left(2A\right)({\theta }_1-{\theta }_2)}{d}=\frac{k_{1}A({\theta }_1-{\theta }_2)}{d}+\frac{k_{2}A({\theta }_1-{\theta }_2)}{d}\Rightarrow k=\frac{1}{2}(k_1+k_2)$
Hence, the correct choice is (d).