Verified Solution • 2025–26 Session moderate

QAIPMT-2009-Ph-2178: NEET 2009 Physics - Class 11 AIPMT-2009-Ph-2178 Solution: The two ends of a rod of length L and a uniform cross-sectional …

The two ends of a rod of length L and a uniform cross-sectional area A 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:

Multiple Choice Options

  • A dQdt=k(T1−T2)LA\frac{dQ}{dt}=\frac{k(T_1-T_2)}{LA}
  • B dQdt=kLA(T1−T2)\frac{dQ}{dt}=kLA(T_1-T_2)
  • C dQdt=kA(T1−T2)L\frac{dQ}{dt}=\frac{kA(T_1-T_2)}{L}
  • D dQdt=kL(T1−T2)A\frac{dQ}{dt}=\frac{kL(T_1-T_2)}{A}

Step-by-Step Verified Solution

Similar to I = V/R d Q d t = k A L ( T 1 _ T 2 ) k = conductivity of the rod.

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