Two different metal rods of the same length have their ends kept at the same temperatures -1 and -2 with -2 - -1- 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

Rate of heat flow is Q/t=kA(θ_2 θ_1)/l Since (θ_2 θ_1) and l are the same for the two rods, the rate of flow Q/t will be the same if the product kA is the same ...

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Standard XII

Physics

Conduction Law: Differential Form

Two different...

Question

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

The correct option is B k1k2=A2A1 Rate of heat flow is Qt=kA(θ2−θ1)l Since (θ2−θ1) and l are the same for the two rods, the rate of flow Qt will be the same if the product kA is the same for the two rods, i.e., if k1A1=k2A2⇒k1k2=A2A1 Hence, the correct choice is (b).