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NEET2026PhysicsChapterActual

Three composite rectangular slabs A, B, and C of equal thickness and equal cross-sectional area are joined end-to-end in series. Their thermal conductivities are 3K , 2K , and K respectively. The free end of slab A is maintained at 66^ C and the free end of slab C is maintained at 0^ C . Assuming steady-state heat conduction and no heat loss from the lateral surfaces, what is the temperature difference across the mid

Options

  1. A18^ C
  2. B22^ C
  3. C36^ C
  4. D54^ C

Correct answer

A. 18^ C

Step-by-step solution

The thermal resistance of a slab is given by R = L kA . For the three slabs A, B, and C having equal thickness L and equal cross-sectional area A , the thermal resistances are: R_A = L 3KA R_B = L 2KA R_C = L KA Since the slabs are connected in series, the rate of heat flow H is the same through all of them. The temperature difference across any slab is proportional to its thermal resistance ( T = HR ). The ratio of temperature differences across the slabs is: T_A : T_B : T_C = R_A : R_B : R_C = 1 3 : 1 2 : 1 = 2 :

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