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Conduction Law

Three rods ...

Question

Three rods $A, B$ and $C$ have the same dimensions. Their thermal conductivities $k_{A}, k_{B}$ and $k_{C}$ respectively. $A$ and $B$ are placed end to end, with the free ends kept at a certain temperature difference. $C$ is placed separately with its ends kept at same temperature difference. The two arrangements conduct heat at the same rate $k_{C}$ equal to

A

k_{A} + k_{B}

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B

\frac{k_{A} k_{B}}{k_{A} + k_{B}}

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C

\frac{1}{2} (k_{A} + k_{B})

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D

2 \frac{k_{A} k_{B}}{k_{A} + k_{B}}

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Solution

The correct option is A $\frac{k_{A} k_{B}}{k_{A} + k_{B}}$
Given that three rods A, B and C have same dimensions. Their thermal conductivities $k_{A}$ , $k_{B}$ and $k_{C}$ respectively. A and B are placed end to end,
with the free ends kept at a certain temperature difference. C is placed separately with its ends kept at same temperature difference.
Heat flow through A and B rods at thermal equilibrium is equal
$H_{A} = H_{B} = \frac{k_{A} a (T - T_{1})}{L} = \frac{k_{B} a (T_{2} - T)}{L} (k_{A} + k_{B}) T = k_{A} T_{1} + k_{B} T_{2} T = \frac{k_{A} T_{1} + k_{B} T_{2}}{k_{A} + k_{B}}$
where T is the temperature of the common end Of A and B, $T_{1}$ and $T_{2}$ are temperatures at the other ends of A and B, $a$ is area and $L$ is the length of the each rod
Heat flow through C
$H_{C} = \frac{k_{C} a (T_{2} - T_{1})}{L}$
If the rate of heat flow for the above two arrangements are equal
$H_{A} = H_{B} = H_{C}$
$\frac{k_{A} a (\frac{k_{A} T_{1} + k_{B} T_{2}}{k_{A} + k_{B}} - T_{1})}{L} = \frac{k_{C} a (T_{2} - T_{1})}{L} \frac{k_{A} (k_{A} T_{1} + k_{B} T_{2} - k_{A} T_{1} - k_{B} T_{1})}{k_{A} + k_{B}} = k_{C} (T_{2} - T_{1}) \frac{k_{A} k_{B}}{k_{A} + k_{B}} = k_{C}$
option (B) is the correct answer.

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