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One end of a thermally insulated rod is kept at a temperature $T_1$ and the other at $T_2$ . The rod is composed of two sections of length $l_1$ and $l_2$ and thermal conductivities $K_1$ and $K_2$ respectively. The temperature at the interface of the two section is
$\frac{{({K_2}{l_1}{T_1} + {K_1}{l_2}{T_2})}}{{({K_2}{l_1} + {K_1}{l_2})}}$
$\frac{{({K_1}{l_2}{T_1} + {K_2}{l_1}{T_2})}}{{({K_1}{l_2} + {K_2}{l_1})}}$
$\frac{{({K_1}{l_1}{T_1} + {K_2}{l_2}{T_2})}}{{({K_1}{l_1} + {K_2}{l_2})}}$
$\frac{{({K_2}{l_2}{T_1} + {K_1}{l_2}{T_2})}}{{({K_1}{l_1} + {K_2}{l_2})}}$
Solution
$\frac{{{K_1}A\left( {{T_1} – T} \right)}}{{\ell .}} = \frac{{{K_2}A\left( {T – {T_2}} \right)}}{{{\ell _2}}},$
$\therefore T = \frac{{{K_1}{T_1}{\ell _2} + {K_2}{T_2}{\ell _1}}}{{{K_2}{\ell _1} + {K_1}{\ell _2}}} = \frac{{{K_1}{\ell _2}{T_1} + {K_2}{\ell _1}{T_2}}}{{{K_1}{\ell _2} + {K_2}{\ell _1}}}.$
