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10-2.Transmission of Heat
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Three rods made of the same material and having same cross-sectional area but different lengths $10\, cm, 20\, cm$ and $30\, cm$ are joined as shown. The temperature of the junction is......... $^oC$
A
$10.8$
B
$14.6$
C
$16.4$
D
$18.2$
Solution

Let $\theta$ be the temperature of junction, $\mathrm{H}_{1}$ $\mathrm{H}_{2}$ and $\mathrm{H}_{3}$ the heat currents. Then
$\mathrm{H}_{1}=\mathrm{H}_{2}+\mathrm{H}_{3}$
or $\frac{30-\theta}{\left(\frac{30}{\mathrm{KA}}\right)}=\frac{\theta-20}{\left(\frac{20}{\mathrm{KA}}\right)}+\frac{\theta-10}{\left(\frac{10}{\mathrm{KA}}\right)}$
or $2\left(30^{\circ}-\theta\right)=3(\theta-20)+6(\theta-10)$
or $\theta=16.36^{\circ} \mathrm{C} \approx 16.4^{\circ} \mathrm{C}$
Standard 11
Physics
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