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(D) In a steady state, the rate of heat flow $(H)$ through both slabs connected in series is the same.Let the temperature at the interface be $T$.The

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$1/3$

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(D) In a steady state, the rate of heat flow $(H)$ through both slabs connected in series is the same.Let the temperature at the interface be $T$.The rate of heat flow through the first slab is $H = \frac{KA(T_2 - T)}{x}$.The rate of heat flow through the second slab is $H = \frac{(2K)A(T - T_1)}{4x} = \frac{KA(T - T_1)}{2x}$.Equating the two expressions: $\frac{KA(T_2 - T)}{x} = \frac{KA(T - T_1)}{2x}$.This simplifies to $T_2 - T = \frac{T - T_1}{2}$, which gives $2T_2 - 2T = T - T_1$, or $3T = 2T_2 + T_1$, so $T = \frac{2T_2 + T_1}{3}$.Substituting $T$ back into the first equation: $H = \frac{KA}{x} \left( T_2 - \frac{2T_2 + T_1}{3} \right) = \frac{KA}{x} \left( \frac{3T_2 - 2T_2 - T_1}{3} \right) = \frac{KA(T_2 - T_1)}{3x}$.Comparing this with the given expression $\left[ \frac{A(T_2 - T_1)K}{x} \right] \cdot f$, we find $f = 1/3$.

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