NEET Physics Question - Thermal Physics

The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_1$ and $T_2$ ($T_1 > T_2$). The rate of heat transfer, $frac{dQ}{dt}$ through the rod in steady state is given by: (2009)

$frac{dQ}{dt} = frac{k (T_1 - T_2)}{LA}$

$frac{dQ}{dt} = k L A (T_1 - T_2)$

$frac{dQ}{dt} = frac{k A (T_1 - T_2)}{L}$

$frac{dQ}{dt} = frac{k L (T_1 - T_2)}{A}$

Solution:

Rate of heat transfer $frac{dQ}{dt} = frac{k A (T_1 - T_2)}{L}$

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