Question

Four rods with different radii $r$ and length $l$ are used to connect two heat reservoirs at different temperature. Which one will conduct most heat ?

• A. $r = 1 \, cm, l = 1 \, m $
• B. $r = 1 \, cm , l = \frac{1}{2} m$
• C. $r = 2 \, cm, l = 2 \, m $
• D. $r = 2 \, cm , l = \frac{1}{2} m$

Answer 1

The Correct Option is D

Key Idea: Rate of Heat Flow

The rate of heat flow is directly proportional to the following:

\[\frac{dQ}{dt} \propto \frac{A \Delta T}{l}\]

Consider a rod maintained at a temperature difference of 

\[\left(T_{1} - T_{2}\right)\]

 between its ends. The rate of heat conduction is given by:

\[\frac{dQ}{dt} = \frac{K A \left(T_{1} - T_{2}\right)}{l}\]

In this case, we are given the following parameters:

\[r = 2 \, \text{cm}\]\[l = \frac{1}{2} \, \text{m}\]

The rod with the above parameters will conduct the most heat because the rate of conduction depends on the ratio of \(\frac{A}{l}\). Therefore, the rate of heat conduction increases as the cross-sectional area \(A\) increases and as the length \(l\) decreases.

Answer 2

Concept:
The rate of heat conduction through a rod is given by the formula:

\(Q = \frac{k A \Delta T}{l}\)
 
where:
 

• $Q$ is the heat conducted per unit time
• $k$ is the thermal conductivity (same for all rods)
• $A$ is the cross-sectional area of the rod
• $\Delta T$ is the temperature difference (same for all rods)
• $l$ is the length of the rod

Since $k$ and $\Delta T$ are same for all, the heat conduction is directly proportional to $\frac{A}{l}$:

\(Q \propto \frac{A}{l} = \frac{\pi r^2}{l}\)

Now let's compare $\frac{r^2}{l}$ for each option:

Option 1:
\(r = 1\, cm = 0.01\, m,\ l = 1\, m \Rightarrow \frac{r^2}{l} = \frac{(0.01)^2}{1} = 1 \times 10^{-4}\)

Option 2:
\(r = 1\, cm = 0.01\, m,\ l = \frac{1}{2}\, m \Rightarrow \frac{r^2}{l} = \frac{(0.01)^2}{0.5} = 2 \times 10^{-4}\)

Option 3:
\(r = 2\, cm = 0.02\, m,\ l = 2\, m \Rightarrow \frac{r^2}{l} = \frac{(0.02)^2}{2} = \frac{4 \times 10^{-4}}{2} = 2 \times 10^{-4}\)

Option 4:
\(r = 2\, cm = 0.02\, m,\ l = \frac{1}{2}\, m \Rightarrow \frac{r^2}{l} = \frac{(0.02)^2}{0.5} = \frac{4 \times 10^{-4}}{0.5} = 8 \times 10^{-4}\)

Conclusion:
Option 4 has the highest $\frac{r^2}{l}$ value, so it will conduct the most heat.

Final Answer:
\(\boxed{r = 2\, cm,\ l = \frac{1}{2}\, m}\)