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Exam
NEET
Subject
Physics
Chapter
Thermal Properties of Matter

NEET · Physics · Thermal Properties of Matter

Two slabs of different materials are connected in parallel between...

Two slabs of different materials are connected in parallel between two temperature reservoirs. Both slabs have the same uniform thickness. The first slab has a cross-sectional area AA and thermal conductivity KK. The second slab has a cross-sectional area 2A2A and thermal conductivity 3K3K. What is the equivalent thermal conductivity of the composite parallel combination?

Options

  1. A2K2K
  2. B4K4K
  3. C3K4\dfrac{3K}{4}
  4. D7K3\dfrac{7K}{3}

Correct answer

D. 7K3\dfrac{7K}{3}

Step-by-step solution

For a parallel combination of thermal conductors, the total rate of heat flow is the sum of the individual rates of heat flow:
Heq=H1+H2H_{eq} = H_1 + H_2

Let ΔT\Delta T be the temperature difference and LL be the thickness of the slabs.

The equivalent thermal conductivity KeqK_{eq} is defined for the total area Aeq=A1+A2=A+2A=3AA_{eq} = A_1 + A_2 = A + 2A = 3A.

Keq(3A)ΔTL=K1A1ΔTL+K2A2ΔTL\dfrac{K_{eq} (3A) \Delta T}{L} = \dfrac{K_1 A_1 \Delta T}{L} + \dfrac{K_2 A_2 \Delta T}{L}

Substituting the given values:
Keq(3A)=K(A)+(3K)(2A)K_{eq} (3A) = K(A) + (3K)(2A)

3KeqA=KA+6KA3 K_{eq} A = KA + 6KA

3KeqA=7KA3 K_{eq} A = 7KA

Keq=7K3K_{eq} = \dfrac{7K}{3}

Answer: 7K3\dfrac{7K}{3}
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