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 $K_{1}: K_{2}=\frac{5}{3}$
$\therefore K_{1}\left(100^{\circ}-\theta\right)=K_{2}\left(\theta-20^{\circ}\right)$
$\Rightarrow \frac{K_{1}}{K_{2}}=\frac{\theta-20^{\circ}}{100^{\circ}-\theta}$
$\Rightarrow \frac{5}{3}=\frac{\theta-20^{\circ}}{100^{\circ}-\theta}$
$\Rightarrow 8 \theta=560^{\circ}$
$\Rightarrow \theta=70^{\circ}$