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Two identical rods AB and CD, each of length L, cross-sectional area A and thermal conductivity k are connected as shown in the figure. Ends A, C and D are maintained at temperatures $T_{1} = 20^{o} C, T_{2} = 30^{o} C$ and $T_{3} = 40^{o} C$ , respectively. The temperature at B is

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$32^{o} C$

$33^{o} C$

$34^{o} C$

$35^{o} C$

answer is A.

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Detailed Solution

$i_{1} = i_{2} + i_{3}$

$\frac{30 - T_{B}}{l / 2 K A} = \frac{T_{B} - 40}{l / 2 K A} + \frac{T_{B} - 20}{l / K A}$

$60 - 2 T_{B} = 2 T_{B} - 80 + T_{B} - 20$

$60 - 2 T_{B} = 3 T_{B} - 100$

$60 + 100 = 5 T_{B}$

$T_{B} = \frac{160}{5} = 32^{o} C$

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