Question

Three rods of same dimensional have thermal conductivity 3 K, 2 K and K. They are arranged as shown in the figure below.
Then, the temperature of the junction in steady state is:

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

Answer 1

The correct option is A $\frac{200}{3}{}^{\circ }C$

According to the figure
$\Rightarrow \frac{3KA100-T}{l}=\frac{2KAT-50}{l}+\frac{KAT-0}{l}$
$300-3T=2T-100+T$
$\Rightarrow 6T=400$
Or $T=\frac{200}{3}{}^{\circ }C$