Question

Which of the following circular rods (given radius $r$ and length $l$ ), each made of the same material and whose ends are maintained at the same temperature will conduct the most heat?
A. $r = {r_0};l = {l_0}$
B. $r = 2{r_0};l = {l_0}$
C. $r = {r_0};l = 2{l_0}$
D. $r = 2{r_0};l = 2{l_0}$

Answer 1

Hint: In this type of question, we must know the concept about thermal conductance i.e., whose thermal conductance is more, more the heat that material will conduct. Here we will match all the situations given above with the formula and will get the required solution.

Formula used:
Heat conduction through a rod is rate of change of heat is $\left( {\dfrac{{\Delta Q}}{{\Delta t}}} \right)$
$
  \therefore H = \left( {\dfrac{{\Delta Q}}{{\Delta t}}} \right) \\
   = KA\left( {\dfrac{{{T_1} - {T_2}}}{l}} \right) \\
 $
Where,
$H$ is the heat,
$K$ is the thermal conductivity of material,
$l$ is the length,
$A$ is the area of cross section and
${T_1} - {T_2}$ temperature difference.

Complete answer:
Now, let’s match each situation,
$H \propto \dfrac{{{r^2}}}{l}$
A. When,
$r = {r_0};l = {l_0}$
i.e.,$H \propto \dfrac{{r_0^2}}{{{l_0}}}$
B. $r = 2{r_0};l = {l_0}$
i.e., $H \propto \dfrac{{4r_0^2}}{{{l_0}}}$
C. $r = {r_0};l = 2{l_0}$
i.e., $H \propto \dfrac{{{r_0}}}{{2{l_0}}}$
D. $r = 2{r_0};l = 2{l_0}$
i.e., $H \propto \dfrac{{2r_0^2}}{{{l_0}}}$
So, from the above situation we can say that the heat conduction will be more in case B.
Hence, the correct option is B.

Note:
The thermal conductivity of a material is a measure of its ability to conduct heat. It is commonly denoted by $k, \lambda$, or $\kappa$. Heat transfer occurs at a lower rate in materials of low thermal conductivity than in materials of high thermal conductivity.