Gujarati
Hindi
10-2.Transmission of Heat
hard

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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