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 most heat ?
A.
$$r = 2{r_0};l = 2{l_0}$$
B.
$$r = 2{r_0};l = {l_0}$$
C.
$$r = {r_0};l = {l_0}$$
D.
$$r = {r_0};l = 2{l_0}$$
Answer :
$$r = 2{r_0};l = {l_0}$$
Solution :
As from law of heat transfer through conduction
$$\eqalign{
& H = \frac{{\Delta Q}}{{\Delta t}} = KA\left( {\frac{{{T_1} - {T_2}}}{1}} \right) \cr
& \Rightarrow H \propto \frac{{{r^2}}}{l}\,......\left( {\text{i}} \right) \cr} $$
(A) When $$r = 2{r_0};l = 2{l_0}$$
$$H \propto \frac{{{{\left( {2{r_0}} \right)}^2}}}{{2{l_0}}} \Rightarrow H \propto \frac{{2r_0^2}}{{{l_0}}}$$
(B) When $$r = 2{r_0};l = {l_0}$$
$$H \propto \frac{{{{\left( {2{r_0}} \right)}^2}}}{{{l_0}}} \Rightarrow H \propto \frac{{4r_0^2}}{{{l_0}}}$$
(C) When $$r = {r_0};l = {l_0} \Rightarrow H \propto \frac{{r_0^2}}{{{l_0}}}$$
(D) When $$r = {r_0};l = 2{l_0} \Rightarrow H \propto \frac{{r_0^2}}{{2{l_0}}}$$
It is obvious that heat conduction will be more in case (B).