Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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 a steady state is given by:

$\frac{dQ}{dt} = \frac{KL(T_1 - T_2)}{A}$

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

$\frac{dQ}{dt} = KLA(T_1 - T_2)$

$\frac{dQ}{dt} = \frac{KA(T_1 - T_2)}{L}$

Step-by-Step Solution

In steady state, the rate of heat flow $\frac{dQ}{dt}$ through a uniform rod is given by Fourier's law of heat conduction: $\frac{dQ}{dt} = \frac{KA(T_1 - T_2)}{L}$ , where $K$ is the thermal conductivity of the material, $A$ is the cross-sectional area, $T_1$ and $T_2$ are the temperatures at the two ends ( $T_1 > T_2$ ), and $L$ is the length of the rod.

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