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NEET PHYSICSThermal Properties of MatterEasy

Question

The two ends of a rod of length LL and a uniform cross-sectional area AA are kept at two temperatures T1T_1 and T2T_2 (T1>T2T_1 > T_2). The rate of heat transfer dQdt\frac{dQ}{dt} through the rod in a steady state is given by:

A

dQdt=KL(T1T2)A\frac{dQ}{dt} = \frac{KL(T_1 - T_2)}{A}

B

dQdt=K(T1T2)LA\frac{dQ}{dt} = \frac{K(T_1 - T_2)}{LA}

C

dQdt=KLA(T1T2)\frac{dQ}{dt} = KLA(T_1 - T_2)

D

dQdt=KA(T1T2)L\frac{dQ}{dt} = \frac{KA(T_1 - T_2)}{L}

Step-by-Step Solution

In steady state, the rate of heat flow dQdt\frac{dQ}{dt} through a uniform rod is given by Fourier's law of heat conduction: dQdt=KA(T1T2)L\frac{dQ}{dt} = \frac{KA(T_1 - T_2)}{L}, where KK is the thermal conductivity of the material, AA is the cross-sectional area, T1T_1 and T2T_2 are the temperatures at the two ends (T1>T2T_1 > T_2), and LL is the length of the rod.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermal Properties of Matter. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSThermal Properties of Matterlengthuniformcrosssectionaltemperaturestransfer

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