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Conduction Law

Two bars of s...

Question

Two bars of same length and same cross-sectional area but of different thermal conductivities $K_{1}$ and $K_{2}$ are joined end to end as shown in the figure. One end of the compound bar is at temperature $T_{1}$ and the opposite end at temperature $T_{2}$ (where $T_{1} > T_{2}$ ).The temperature of the junction is

\frac{K_{1} T_{1} + K_{2} T_{2}}{K_{1} + K_{2}}

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\frac{K_{1} T_{2} + K_{2} T_{1}}{K_{1} + K_{2}}

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\frac{K_{1} (T_{1} + T_{2})}{K_{2}}

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\frac{K_{2} (T_{1} + T_{2})}{K_{1}}

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Solution

The correct option is A $\frac{K_{1} T_{1} + K_{2} T_{2}}{K_{1} + K_{2}}$

Let L and A be length and area of cross-section of each bar respectively.
$∴$ Heat current through the bar 1 is
$H_{1} = \frac{K_{1} A (T_{1} - T_{0})}{L}$
Here $T_{0}$ is junction temperature.
Heat current through the bar 2 is
$H_{2} = \frac{K_{2} A (T_{0} - T_{2})}{L}$
At steady state, $H_{1} = H_{2}$
$∴ H_{1} = \frac{K_{1} A (T_{1} - T_{0})}{L} = H_{1} = \frac{K_{2} A (T_{0} - T_{2})}{L}$
$K_{1} (T_{1} - T_{0}) = K_{2} (T_{0} - T_{2})$
$K_{1} T_{1} - K_{1} T_{0} = K_{2} T_{0} - K_{2} T_{2}$
$K_{1} T_{0} + K_{2} T_{0} = K_{1} T_{1} + K_{2} T_{2}$
$T_{0} (K_{1} + K_{2}) = K_{1} T_{1} + K_{2} T_{2}$
$T_{0} = \frac{K_{1} T_{1} + k_{2} T_{2}}{(K_{1} + K_{2})} \dots (i)$

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Two bars of same length and same cross-sectional area but of different thermal conductivities K1 and K2 are joined end to end as shown in the figure. One end of the compound bar is at temperature T1 and the opposite end at temperature T2 (where T1>T2).The temperature of the junction is