NEET Physics Question - Thermal Physics

Consider two rods of same length and different specific heats ($S_1$, $S_2$), conductivities ($K_1$, $K_2$) and area of cross-sections ($A_1$, $A_2$) and both having temperature $T_1$ and $T_2$ at their ends. If rate of loss of heat due to conduction is equal, then (2002)

$K_1 A_1 = K_2 A_2$

$frac{K_1 A_1}{S_1} = frac{K_2 A_2}{S_2}$

$K_2 A_1 = K_1 A_2$

$frac{K_2 A_1}{S_2} = frac{K_1 A_2}{S_1}$

Solution:

Rate of heat loss $frac{dQ}{dt} = frac{K A (T_1 - T_2)}{l}$
Given $(frac{dQ}{dt})_1 = (frac{dQ}{dt})_2 Rightarrow frac{K_1 A_1 (T_1 - T_2)}{l} = frac{K_2 A_2 (T_1 - T_2)}{l}$
$Rightarrow K_1 A_1 = K_2 A_2$

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Solution:

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