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 joint is ....... $^oC$
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Let $\theta$ be the temperature of junction and $H_{1}, H_{2}$ and $H_{3}$ the heat currents.
Then
$H_{1}=H_2+H_{3}$
$\frac{30-\theta}{\left(\frac{30}{K A}\right)}=\frac{\theta-20}{\left(\frac{20}{K A}\right)}+\frac{\theta-10}{\left(\frac{30}{K A}\right)}$
or $2(30-\theta)=3(\theta-20)+3(2 \theta-20)$
or $\theta=16.36^{\circ} \mathrm{C} \approx 16.4^{\circ} \mathrm{C}$
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