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A cylindrical steel rod of length $0.10 \,m$ and thermal conductivity $50 \,Wm ^{-1} K ^{-1}$ is welded end to end to copper rod of thermal conductivity $400 \,Wm ^{-1} K ^{-1}$ and of the same area of cross-section but $0.20 \,m$ long. The free end of the steel rod is maintained at $100^{\circ} C$ and that of the copper rod at $0^{\circ} C$. Assuming that the rods are perfectly insulated from the surrounding, the temperature at the junction of the two rods is ................... $^{\circ} C$
$20$
$30$
$40$
$50$
Solution
(a)
In steady state, rate of heat flow through both rods is equal.
$\operatorname{So},\left(\frac{k A\left(T_1-T\right)}{l}\right)_{\text {steel }}=\left(\frac{k A\left(T-T_2\right)}{l}\right)_{\text {copper }}$
Substituting given values, we have
$\frac{50 \times A (100- T ) t }{0.1}=\frac{400 \times( T -0) t }{0.2}$
$\Rightarrow 100(100-T) =400(T)$
$\Rightarrow T =20^{\circ} C$
