Q107 / 150 · MHT CET 2026 (17 April Shift 2) · Physics

In a composite slab there are two materials having coefficients of thermal conductivity K and 2K, thickness x and 4x repectively. The temperature of the two outer surfaces of a composite slab are $T_2$ and $T_1$ $(T_2 > T_1)$. $T_2$ is on side K and $T_1$ is on side 2K. The rate of heat transfer through the slab in a steady state is $\left[\dfrac{A(T_2 - T_1)K}{x}\right] \cdot f$, where f is equal to

  • A
    $1$
  • B
    $\dfrac{1}{2}$
  • C
    $\dfrac{2}{3}$
  • Correct answer
    D
    $\dfrac{1}{3}$

Solution

The thermal resistance of a slab is given by $R = \dfrac{L}{KA}$, where $L$ is the thickness, $K$ is the thermal conductivity, and $A$ is the area of cross-section.

For the first material, the thermal resistance is $R_1 = \dfrac{x}{KA}$.

For the second material, the thermal resistance is $R_2 = \dfrac{4x}{2KA} = \dfrac{2x}{KA}$.

Since the two materials form a composite slab in series, the equivalent thermal resistance is:

$R_{eq} = R_1 + R_2 = \dfrac{x}{KA} + \dfrac{2x}{KA} = \dfrac{3x}{KA}$

The rate of heat transfer through the slab in a steady state is given by:

$H = \dfrac{T_2 - T_1}{R_{eq}} = \dfrac{T_2 - T_1}{\dfrac{3x}{KA}} = \dfrac{KA(T_2 - T_1)}{3x}$

This can be rewritten as:

$H = \left[\dfrac{A(T_2 - T_1)K}{x}\right] \cdot \dfrac{1}{3}$

Comparing this with the given expression $H = \left[\dfrac{A(T_2 - T_1)K}{x}\right] \cdot f$, we get:

$f = \dfrac{1}{3}$

Answer: $\dfrac{1}{3}$

Hence, the correct answer is option: 'D'.

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