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Step-by-Step Solution
Step 1: Recall the formula for heat conduction
The rate of heat conduction through a cylindrical rod of length $l$ and cross-sectional area $A$ (with temperature difference $T_1 - T_2$ at its ends) is given by
$Q = \frac{K \, A \, (T_1 - T_2) \, t}{l} = \frac{K \, \pi \, r^2 \, (T_1 - T_2) \, t}{l},$
where:
$K$ is the thermal conductivity of the material.
$r$ is the radius of the rod.
$l$ is the length of the rod.
$t$ is the time duration for which heat flows.
Step 2: Identify the key factor determining maximum heat flow
The ratio that primarily decides the amount of heat conducted is $A/l$, since $A = \pi r^2$. Thus, the rod with the largest value of $\frac{\pi r^2}{l}$ will conduct the most heat.
Step 3: Compare options based on $r$ and $l$
Option 1: $r = r_0,\; l = l_0 \quad \rightarrow \quad \frac{\pi {r_0}^2}{l_0}.$
Option 2: $r = 2r_0,\; l = l_0 \quad \rightarrow \quad \frac{\pi \,(2r_0)^2}{l_0} = \frac{4 \,\pi \,{r_0}^2}{l_0}.$
Option 3: $r = r_0,\; l = 2l_0 \quad \rightarrow \quad \frac{\pi \,{r_0}^2}{2l_0}.$
Option 4: $r = 2r_0,\; l = 2l_0 \quad \rightarrow \quad \frac{\pi \,(2r_0)^2}{2l_0} = \frac{4\, \pi \,{r_0}^2}{2l_0} = \frac{2\, \pi \,{r_0}^2}{l_0}.$
Clearly, Option 2 has the largest cross-sectional area relative to its length ($4 \,\pi {r_0}^2 / l_0$), making it the best conductor among the given choices.
Step 4: State the final answer
The rod with $r = 2r_0$ and $l = l_0$ (Option 2) conducts the most heat.
