Question

Three rods of same dimensions have thermal conductivity $3K,2K$ and $K$. They are arranged as show in the figure below. Then the temperature of the junction in steady state is ________.

• A. ${\frac{100}{3}}^{\circ }C$
• B. ${\frac{200}{3}}^{\circ }C$
• C. ${75}^{\circ }C$
• D. ${\frac{50}{3}}^{\circ }C$

Answer 1

Answer: (B)

${\frac{200}{3}}^{\circ }C$

Given: Three rods of same dimensions have thermal conductivity 3K,2K and K. They are arranged as show in the figure below.

To find the temperature of the junction in steady state

Solution:

The amount of heat transmitted from one end to other end is

$Q=\frac{KAT(t_1-t_2)}{d}$

where $T$ is the time, $t_1-t_2$ is the temperature difference

Hence at the junction,

$Q=Q_1+Q-2⟹\frac{3K\left(A\right)T(100-t)}{d}=\frac{2K\left(A\right)T(t-50)}{d}+\frac{KAT(t-0)}{d}$

here $t$ is the temperature of the junction

$3(100-t)=2(t-50)+(t-0)⟹300-3t=2t-100+t⟹6t=400⟹t=\frac{400}{6}={\frac{200}{3}}^{\circ }C$

is the temperature of the junction in steady state