The end $A$ of a rod $AB$ of length $1\,m$ is maintained at $80\,^oC$ and the end $B$ at $0\,^oC.$ The temperature at a distance of $60\,\,c.m.$ from the end $A$ is......... $^oC$
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$\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{AC}}=\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\mathrm{AB}}$
$\Rightarrow \frac{\mathrm{KA}}{1}[80-0]=\frac{\mathrm{KA}}{0.6}[80-0]$
$\therefore \theta=32^{\circ} \mathrm{C}$
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