In the figure shown below, two rods are maintained at the same temperature difference between the ends. As a result, heat is flowing through two rods made of the same material. If the diameters of the rods are in the ratio $1 : 4$ and their lengths are in the ratio $3 : 2$ , then the ratio of the rates of flow of heat through them is $1 : n$ . Find $n$

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Solution

Given that,
$(Δ T)_{1} = (Δ T)_{2}$ and $k_{1} = k_{2}$
Ratio of diameters of rods $1$ and $2$ , $(\frac{d_{1}}{d_{2}}) = \frac{1}{4}$
Ratio of lengths of rods $1$ and $2$ , $(\frac{l_{1}}{l_{2}}) = \frac{3}{2}$
We know that,
Rate of flow of heat in rod $1$ and $2$ ,
$q_{1} = \frac{k_{1} A_{1} (T_{1} - T_{2})}{l_{1}}$ ;
$q_{2} = \frac{k_{2} A_{2} (T_{1} - T_{2})}{l_{2}}$
$∴ \frac{q_{1}}{q_{2}} = (\frac{A_{1}}{A_{2}}) \times (\frac{l_{2}}{l_{1}})$
$\Rightarrow \frac{q_{1}}{q_{2}} = {(\frac{d_{1}}{d_{2}})}^{2} \times (\frac{l_{2}}{l_{1}}) = \frac{1}{16} \times \frac{2}{3} = \frac{1}{24}$

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