Question

Three rods of identical cross - section and lengths are made of three different materials of thermal conductivity $K_1,K_2\text{and}{K}_3,$ respectively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at ${100}^{\circ }\text{C}$ and the other at $0^{\circ }\text{C}$ (see figure). If the joints of the rod are at ${70}^{\circ }\text{C}$ and ${20}^{\circ }\text{C}$ in steady state and there is no loss of energy from the surface of the rod, the correct relationship between $K_1,K_2\text{and}{K}_3$ is :

• A. $K_1:K_3=2:3,K_2:K_3=2:5$
• B. $K_1<K_2<K_3$
• C. $K_1:K_2=5:2,K_1:K_3=3:5$
• D. $K_1>K_2>K_3$

Answer 1

The correct option is A $K_1:K_3=2:3,K_2:K_3=2:5$

As the rods are identical, so they have same length $\left(l\right)$ and area of cross-section $\left(A\right)$. They are connected in series. So, heat current will be same for all rods.

Heat current $={\left(\frac{\Delta Q}{\Delta t}\right)}_{AB}={\left(\frac{\Delta Q}{\Delta t}\right)}_{BC}={\left(\frac{\Delta Q}{\Delta t}\right)}_{CD}$

$\Rightarrow \frac{(100-70)K_{1}A}{l}=\frac{(70-20)K_{2}A}{l}=\frac{(20-0)K_{3}A}{l}$

$\Rightarrow K_1(100-70)=K_2(70-20)=K_3(20-0)$

$\Rightarrow K_1\left(30\right)=K_2\left(50\right)=K_3\left(20\right)$

$\Rightarrow \frac{K_1}{10}=\frac{K_2}{6}=\frac{K_3}{15}$

$\Rightarrow K_1:K_2:K_3=10:6:15$

$\therefore K_1:K_3=2:3,K_2:K_3=2:5$

Hence, option $\left(\text{A}\right)$ is correct.