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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 :-
A
$K_1A_1 = K_2A_2$
B
$\dfrac{K_1A_1}{S_1} = \dfrac{K_2A_2}{S_2}$
C
$K_2A_1 = K_1A_2$
D
$\dfrac{K_2A_1}{S_2} = \dfrac{K_1A_2}{S_1}$
Detailed Solution
Rate of heat flow by conduction in the steady state: $\dfrac{dQ}{dt} = \dfrac{KA}{L}(T_1 - T_2)$
It depends on conductivity, area, length and temperature difference, but not on specific heat (specific heat matters only while the rod is heating up).
Both rods have the same length L and the same end temperatures $T_1$ and $T_2$.
Equal rates: $\dfrac{K_1A_1}{L}(T_1 - T_2) = \dfrac{K_2A_2}{L}(T_1 - T_2)$
$\Rightarrow K_1A_1 = K_2A_2$
It depends on conductivity, area, length and temperature difference, but not on specific heat (specific heat matters only while the rod is heating up).
Both rods have the same length L and the same end temperatures $T_1$ and $T_2$.
Equal rates: $\dfrac{K_1A_1}{L}(T_1 - T_2) = \dfrac{K_2A_2}{L}(T_1 - T_2)$
$\Rightarrow K_1A_1 = K_2A_2$