Solution:
Solution:
$$H = \frac{KAd\theta}{L} \propto \frac{A}{L } = \frac{\pi r^{2}}{L} $$
$$ \left(\frac{A}{L}\right)_{1} = \frac{\pi\left(1\right)^{2}}{1} = 2\pi\,\, units \hspace10mm \left(\frac{A}{L}\right)_{2} = \frac{\pi\left(2\right)^{2}}{\frac{1}{2}} = 8\pi \,\, units.$$
$$ \left(\frac{A}{L}\right)_{3} = \frac{\pi\left(1\right)^{2}}{1} = \pi \,\, units \hspace10mm \left(\frac{A}{L}\right)_{_4 }= \frac{\pi\left(2\right)^{2}}{2} = 2\pi\,\, units$$
So rod (2) will conduct most heat per second.|