First slide

Conduction

Question

Equal temperature difference exists between the ends of two metallic rods 1 and 2 of equal length. Their thermal conductivities are K₁ and K₂ and cross-sectional areas are respectively A₁ and A₂ The condition for equal rate of heat transfer will be

Easy

A

$K_{1} A_{2} = K_{2} A_{1}$
B

$K_{1} A_{2} = {K_{2}}^{2} A_{1}$
C

$K_{1} A_{1} = K_{2} A_{2}$
D

$K_{1} A_{1}^{2} = K_{2} A_{2}^{2}$

Solution

The rate of flow of heat is given by $\frac{Q}{t} = \frac{KA (T_{1} - T_{2})}{l}$
Since [T₁ - T₂) and / are the same for two rods, the rate of flow of heat will be the same, if $K_{1} A_{1} = K_{2} A_{2}$

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