10-2.Transmission of Heat
medium

Three rods of the same dimension have thermal conductivities $3K$ , $2K$ and $K$ . They are arranged as shown in fig. Given below, with their ends at $100^oC, 50^oC $and $20^oC$. The temperature of their junction is ......... $^oC$

A

$60$

B

$70$

C

$50$

D

$35$

Solution

(b) Let the temperature of junction be $\theta$ then according to following figure.
$H = H_1 + H_2$

==> $\frac{{3K \times A \times (100 – \theta )}}{l} = \frac{{2KA(\theta – 50)}}{l} + \frac{{KA(\theta – 20)}}{l}$

==> $300 -3\theta = 3\theta -120 $

==> $\theta$ $= 70°C$

Standard 11
Physics

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