Question

Consider a compound slab consisting of two different materials having equal lengths, thickness and thermal conductivities \( K \) and \( 2K \) respectively. The equivalent thermal conductivity of the slab is

• A. \( \sqrt{2}K \)
• B. \( 3K \)
• C. \( \frac{4}{3}K \)
• D. \( \frac{2}{3}K \)

Hint:

When materials are in series, the equivalent thermal conductivity is the inverse sum of the individual conductivities.

Answer 1

The Correct Option is C

The thermal conductivity of the compound slab is determined by the formula for two materials in series: \[ \frac{1}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2} \] Since \( K_1 = K \) and \( K_2 = 2K \), we get: \[ \frac{1}{K_{eq}} = \frac{1}{K} + \frac{1}{2K} = \frac{3}{2K} \] Thus, \[ K_{eq} = \frac{2}{3}K \]