MHT CET · Physics · Thermal Properties of Matter
A composite slab consists of two materials having coefficients of thermal conductivity K and 2 K , thickness x and 4 x respectively. The temperatures of two outer surfaces of a composite slab are \(T_2\) and \(T_1\) respectively ( \(\mathrm{T}_2>\mathrm{T}_1\) ). The rate of heat transfer through the slab in a steady state is \(\left[\frac{A\left(T_2-T_1\right) K}{x}\right] f\), where \(f\) is equal to
- A \(1\)
- B \(\frac{2}{3}\)
- C \(\frac{1}{2}\)
- D \(\frac{1}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\(R_{total} = R_1 + R_2 = \frac{x}{KA} + \frac{4x}{2KA}\) \(R_{total} = \frac{x}{KA} + \frac{2x}{KA} = \frac{3x}{KA}\)
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