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=$$r_0$: $l=2$$l_0$
• D. $r=$$r_0$: $l=$$l_0$

Answer 1

Answer: (B)

$r=2$$r_0$: $l=$$l_0$

We know that $Q=\frac{T_H-T_L}{R}$
Also Thermal resistance, $R=\frac{l}{KA}=\frac{l}{K\pi r^2}$
Heat flow will be maximum when thermal resistance is minimum. From given option
(i) r = 2$r_0$: l = 2$l_0\therefore R=\frac{2l_0}{K\pi (2r_0{)}^2}=\frac{l_0}{2K\pi r_0^2}$
(ii) r = 2$r_0$: l = $l_0\therefore R=\frac{l_0}{K\pi (2r_0{)}^2}=\frac{l_0}{4K\pi r_0^2}$
(iii) r = $r_0$: l = 2$l_0\therefore R=\frac{2l_0}{K\pi (2r_0{)}^2}=\frac{2l_0}{K\pi r_0^2}$
(iv) r = $r_0$: l = $l_0\therefore R=\frac{l_0}{K\pi (r_0{)}^2}=\frac{l_0}{K\pi r_0^2}$
It is clear that for option (b) resistance is minimum, hence heat flow will be maximum.

$Power=\frac{Q}{t}=\frac{kA\Delta T}{l}$

$Power$ $\alpha$ $\frac{A}{l}$

So the rod with more radius and less length conducts more heat

So option $B$ will be the answer