Search any question & find its solution
Question:
Answered & Verified by Expert
One rod of length 2 m and thermal conductivity 50 unit is attached to another rod of length 1 m and thermal conductivity 100 unit. Temperature of free ends are $70^{\circ} \mathrm{C}$ and $50^{\circ} \mathrm{C}$ respectively. Then temperature of junction point will be
Options:
Solution:
1423 Upvotes
Verified Answer
The correct answer is:
$54^{\circ} \mathrm{C}$
Let $\theta$ be the junction temperature, then given situation is shown in the following figure.
The rate of heat current due to conduction in both rod will be same, i.e, $\mathrm{H}_1=\mathrm{H}_2$
$\begin{aligned}
\Rightarrow \quad \mathrm{K}_1 \frac{A \Delta \theta_1}{\mathrm{l}_1} & =\mathrm{K}_2 \mathrm{~A} \cdot \frac{\Delta \theta_2}{\mathrm{l}_2} \\
\mathrm{~K}_1 \frac{\Delta \theta_1}{\mathrm{l}_1} & =\mathrm{K}_2 \frac{\Delta \theta_2}{\mathrm{l}_2} \\
50 \frac{(70-\theta)}{2} & =100 \frac{(\theta-50)}{1} \\
5 \theta & =270 \Rightarrow \theta=54^{\circ} \mathrm{C}
\end{aligned}$
The rate of heat current due to conduction in both rod will be same, i.e, $\mathrm{H}_1=\mathrm{H}_2$
$\begin{aligned}
\Rightarrow \quad \mathrm{K}_1 \frac{A \Delta \theta_1}{\mathrm{l}_1} & =\mathrm{K}_2 \mathrm{~A} \cdot \frac{\Delta \theta_2}{\mathrm{l}_2} \\
\mathrm{~K}_1 \frac{\Delta \theta_1}{\mathrm{l}_1} & =\mathrm{K}_2 \frac{\Delta \theta_2}{\mathrm{l}_2} \\
50 \frac{(70-\theta)}{2} & =100 \frac{(\theta-50)}{1} \\
5 \theta & =270 \Rightarrow \theta=54^{\circ} \mathrm{C}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.