First slide

Conduction

Question

Two metal rods 1 and 2 of same lengths have same temperature difference between their ends. Their thermal conductivities are $K_{1} a n d K_{2}$ and cross sectional areas $A_{1} a n d A_{2}$ , respectively. If the rate of heat conduction in 1 is four times that in 2 , then

Moderate

A

$K_{1} A_{1} = 4 K_{2} A_{2}$
B

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

$4 K_{1} A_{1} = K_{2} A_{2}$
D

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

Solution

Let L be length of each rod. Rate of heat flow in rod 1 for the temperature difference $Δ T$ is
$H_{1} = \frac{K_{1} A_{1} Δ T}{L}$
Rate of heat flow in rod 2 for the same difference } $Δ T$ is
$H_{2} = \frac{K_{2} A_{2} Δ T}{L}$
As per question
$H_{1} = 4 H_{2} \frac{K_{1} A_{1} Δ T}{L} = 4 \frac{K_{2} A_{2} Δ T}{L} K_{1} A_{1} = 4 K_{2} A_{2}$

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