Question

Two rods having thermal conductivity in the ratio of $5:3$ and having equal length and equal cross-sectional area. are joined face to face. If the temperature of free end of first rod is $100^{\circ }C$ and temperature of free end of second rod is $20^{\circ }C$ , temperature of junction will be:

• A. $90^{\circ }C$
• B. $85^{\circ }C$
• C. $70^{\circ }C$
• D. $50^{\circ }C$

Answer 1

Answer: (C)

In the steady state, the rate of flow of heat in both the conductors will be the same. Taking the two conductors with same cross-section area. $A$, joined in series, with same length $d$ Let $H$ be heat flow through this combination and $\theta$ is temperature of common surface. In steady state rate of flow of heat is same hence,

$H=\frac{Q}{t}=\frac{K_{1}A\left({\theta }_1-\theta \right)}{d}$
$=\frac{K_{2}A\left(\theta -{\theta }_2\right)}{d}$
Given ${\theta }_1=100^{\circ }C$,
${\theta }_2=20^{\circ }C$
and $K_1:K_2=\frac{5}{3}$
$\therefore K_1\left(100^{\circ }-\theta \right)=K_2\left(\theta -20^{\circ }\right)$
$\Rightarrow \frac{K_1}{K_2}=\frac{\theta -20^{\circ }}{100^{\circ }-\theta }$
$\Rightarrow \frac{5}{3}=\frac{\theta -20^{\circ }}{100^{\circ }-\theta }$
$\Rightarrow 8\theta =560^{\circ }$
$\Rightarrow \theta =70^{\circ }$