Question

Three rods of same dimensions have thermal conductivity $3k,2k$ and $k$. They are arranged as shown in figure. What will be the temperature of junction in steady state?

Answer 1

Let $T$ be the temperature of junction,
Then $Q_1=Q_2+Q_3$
$\frac{100-T}{\frac{L}{3kA}}=\frac{T-50}{\frac{L}{2}kA}+\frac{T-0}{\frac{L}{kA}}$
$3(100-T)=2(T-50)+T-0$
$300-3T=2T-100+T$
$6T=400$
$T=\frac{400}{6}{}^{\circ }C=\frac{200}{3}{}^{\circ }C$.