Two different metal rods of the same length have their ends kept at the same temperatures $θ_{1}$ and $θ_{2}$ with $θ_{2}$ > $θ_{1}$ . If $A_{1}$ and $A_{2}$ are their cross-sectional areas and $k_{1}$ and $k_{2}$ their thermal conductivities, the rate of heat flow in the two rods will be same if

$\frac{k_{1}}{k_{2}} = \frac{A_{1}}{A_{2}}$

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$\frac{k_{1}}{k_{2}} = \frac{A_{2}}{A_{1}}$

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$\frac{k_{1}}{k_{2}} = \frac{A_{1} θ_{2}}{A_{2} θ_{1}}$

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$\frac{k_{1}}{k_{2}} = \frac{A_{1} θ_{1}}{A_{2} θ_{2}}$

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Solution

The correct option is B
$\frac{k_{1}}{k_{2}} = \frac{A_{2}}{A_{1}}$

Rate of heat flow is $\frac{Q}{t} = \frac{k A (θ_{2} - θ_{1})}{l}$
Since $(θ_{2} - θ_{1})$ and l are the same for the two rods, the rate of flow $\frac{Q}{t}$ will be the same if the product kA is the same for the two rods, i.e., if
$k_{1} A_{1} = k_{2} A_{2} \Rightarrow \frac{k_{1}}{k_{2}} = \frac{A_{2}}{A_{1}}$
Hence, the correct choice is (b).

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Two different metal rods of the same length have their ends kept at the same temperatures θ1 and θ2 with θ2 > θ1. If A1 and A2 are their cross-sectional areas and k1 and k2 their thermal conductivities, the rate of heat flow in the two rods will be same if

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