Question

Five rods of same dimensions are arranged as shown in figure. They have thermal conductivitites $K_1,$ $K_2,$ $K_3,$ $K_4$ and $K_5.$ When points $\text{A}$ and $\text{B}$ are maintained at same temperatures, no heat flows through the central rod if

• A. $K_1$=$K_4$ and $K_2$ = $K_3$
• B. $K_1$$K_4$=$K_2$$K_3$
• C. $K_1$$K_2$ = $K_3$$K_4$
• D. None of these

Answer 1

The correct option is B $K_1$$K_4$=$K_2$$K_3$

For no heat flow between $\text{A}$ and $\text{B}$, heat flow throght $\text{DA}$ will be equal to the heat flow through $\text{AC}$,

${\left(\frac{Q}{t}\right)}_{DA}={\left(\frac{Q}{t}\right)}_{AC}$

Substituting the values from figure

$\Rightarrow \frac{K_{3}A({\theta }_D-{\theta }_A)}{l}=\frac{K_{1}A({\theta }_A-{\theta }_C)}{l}$

where, ${\theta }_A,{\theta }_B,{\theta }_C,$ and ${\theta }_D$ are the tempratute at the respective points, $l$ is the length and $A$ is the cross-section area for each rod.

$\Rightarrow \frac{({\theta }_D-{\theta }_A)}{({\theta }_A-{\theta }_C)}=\frac{K_1}{K_3}...\left(1\right)$

Similarly,
${\left(\frac{Q}{t}\right)}_{DB}={\left(\frac{Q}{t}\right)}_{BC}$

$\Rightarrow \frac{K_{4}A({\theta }_D-{\theta }_B)}{l}=\frac{K_{2}A({\theta }_B-{\theta }_C)}{l}$

$\Rightarrow \frac{({\theta }_D-{\theta }_B)}{({\theta }_B-{\theta }_C)}=\frac{K_2}{K_4}...\left(2\right)$

It is given that ${\theta }_A={\theta }_B$, hence from equation $\left(1\right)$ and $\left(2\right)$, we get

$\frac{K_1}{K_3}$ = $\frac{K_2}{K_4}$

$\therefore K_1$$K_4$ = $K_2$$K_3$

Hence, option (b) is correct answer.