Heat and Thermodynamics 3 Question 16
16. Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at $0^{\circ} C$ and $90^{\circ} C$ respectively. The temperature of junction of the three rods will be
$(2001,2 M)$
(a) $45^{\circ} C$
(b) $60^{\circ} C$
(c) $30^{\circ} C$
(d) $20^{\circ} C$
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Solution:
- Let $\theta$ be the temperature of the junction (say $B$ ). Thermal resistance of all the three rods is equal. Rate of heat flow through $A B+$ Rate of heat flow through $C B=$ Rate of heat flow through $B D$
$\therefore \quad \frac{90^{\circ}-\theta}{R}+\frac{90^{\circ}-\theta}{R}=\frac{\theta-0}{R}$
Here, $R=$ Thermal resistance
$$ \therefore \quad 3 \theta=180^{\circ} \text { or } \theta=60^{\circ} C $$
NOTE
Rate of heat flow
$$ \begin{aligned} (H) & =\frac{\text { Temperature difference }(TD)}{\text { Thermal resistance }(R)} \\ \text { where, } \quad R & =\frac{1}{KA} \\ K & =\text { Thermal conductivity of the rod. } \end{aligned} $$
This is similar to the current flow through a resistance $(R)$ where current (i) $=$ Rate of flow of charge
$$ =\frac{\text { Potential difference }(PD)}{\text { Electrical resistance }(R)} $$
Here, $R=\frac{l}{\sigma A}$ where $\sigma=$ Electrical conductivity.
