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10-2.Transmission of Heat
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Four rods of same material and having the same cross section and length have been joined, as shown. The temperature of junction of four rods will be........ $^oC$

A
$20$
B
$30$
C
$45$
D
$60$
Solution
$\frac{{{\rm{KA}}(90 – \theta )}}{l} = \frac{{{\rm{KA}}(\theta – 30)}}{l} + \frac{{{\rm{KA}}(\theta – {\rm{0}})}}{l} + \frac{{{\rm{KA}}(\theta – 60)}}{l}$
$90-\theta=3 \theta-90^{\circ}$
$\theta=45^{\circ}$ $C$
Standard 11
Physics
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