Q143 / 200 · Manipal (MET) 2010 · Physics
Three identical thermal conductors are connected as shown in figure. Considering no heat loss due to radiation, temperature at the junction will be
- A$40^{\circ} \mathrm{C}$
- B$60^{\circ} \mathrm{C}$
- Correct answerC$50^{\circ} \mathrm{C}$
- D$35^{\circ} \mathrm{C}$
Solution
Let the temperature of junction be $\theta$. In equilibrium, rate of flow of heat through rod 1 is equal to sum of rate of flow of heat through rods 2 and 3 , ie,
$\left(\frac{d Q}{d t}\right)_1=\left(\frac{d Q}{d t}\right)_2+\left(\frac{d Q}{d t}\right)_3$
$\therefore \frac{K A(\theta-20)}{l}=\frac{K A(60-\theta)}{l}+\frac{K A(70-\theta)}{l}$
$\Rightarrow \theta-20=130-2 \theta$
$\begin{aligned}
&\text { or }\\
&3 \theta=150
\end{aligned}$
$\therefore \theta=50^{\circ} \mathrm{C}$
Hence, the correct answer is option: 'C'.