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Question

Let $	heta$ 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-	heta}{\left(

Vedclass · Accepted Answer

$16.4$

Vedclass · Answer

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

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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