Three rods of the same dimensions have thermal conductivities $3k, 2k$ and $k$. They are arranged as shown, with their ends at $100\,^oC, 50\,^oC$ and $0\,^oC$. The temperature of their junction is
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$\frac{3 \mathrm{K}(100-\theta)}{\mathrm{L}}=\frac{2 \mathrm{K}(\theta-50)}{\mathrm{L}}+\frac{\mathrm{K}(\theta-0)}{\mathrm{L}}$
$300-3 \theta=2 \theta-100+\theta$
$400=6 \theta$
$\theta=\frac{200}{3}^{\circ} \mathrm{C}$
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