Two different metal rods of the same length their ends kept at the same temperatures -1 and -2 with -2 - -1 If A1 and A2 are their cross-sectional area and K1 and K2 their thermal conductivities- the rate of flow of heat in the two rods will be the same if

Rate of flow is Q/T=KA(θ_1 θ_2)/l Since (θ_1 θ_2) 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 f ...

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Standard XIII

Physics

Conduction Law

Two different...

Question

Two different metal rods of the same length 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 area and $K_{1}$ and $K_{2}$ their thermal conductivities, the rate of flow of heat in the two rods will be the same if

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

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

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$\frac{A_{1}}{A_{2}}$ $= \frac{K_{1} θ 1}{K_{2} θ_{2}}$

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

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Solution

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

Rate of flow is $\frac{Q}{T} = \frac{K A (θ_{1} - θ_{2})}{l}$ Since $(θ_{1} - θ_{2})$ 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 $k_{1} A_{1} = K_{2} A_{2}$ or $\frac{A_{1}}{A_{2}} = \frac{K_{2}}{K_{1}}$ Hence the correct choice is (b)

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