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

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$75^{0} C$

$\frac{100}{3}^{0} C$

$\frac{50}{3}^{0} C$

$\frac{200}{3}^{0} C$

answer is B.

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Detailed Solution

According to Kirchoff 1^st law,

$i = i_{1} + i_{2}$

$(100 - T) 3 K = (T - 50) 2 K + (T - 0) K$

$T = {\frac{200}{3}}^{0} C$

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