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Consider two rods of same length and different specific heats $(s_{1} a n d s_{2})$ , conductivities $K_{1}$ and $K_{2}$ and areas of cross-section ( $A_{1} a n d A_{2}$ )and both giving temperature $T_{1} a n d T_{2}$ at their ends. If the rate of heat loss due to conduction is equal, then:

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

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

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

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

answer is A.

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Detailed Solution

According to Illustration, rate of heat loss in both rods are equal

${(\frac{d Q}{d t})}_{1} = {(\frac{d Q}{d t})}_{2}$

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

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

[As $Δ θ_{1} = Δ θ_{2} = (T_{1} - T_{2})$ ]and $l_{1} = l_{2} (g i v e n)$

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