A composite block is made of slabs A, B, C, D | vedclass

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(C) The thermal resistance of a slab is given by $R = \frac{L}{kA}$,where $L$ is the length,$k$ is thermal conductivity,and $A$ is the cross-sectional

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$(A, C, D)$

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(C) The thermal resistance of a slab is given by $R = \frac{L}{kA}$,where $L$ is the length,$k$ is thermal conductivity,and $A$ is the cross-sectional area. Let $b$ be the width of the slabs.$R_A = \frac{L}{2K(4Lb)} = \frac{1}{8} \frac{L}{K b} = \frac{R_0}{8}$$R_B = \frac{4L}{3K(Lb)} = \frac{4}{3} \frac{L}{K b} = \frac{4R_0}{3}$$R_C = \frac{4L}{4K(2Lb)} = \frac{1}{2} \frac{L}{K b} = \frac{R_0}{2}$$R_D = \frac{4L}{5K(Lb)} = \frac{4}{5} \frac{L}{K b} = \frac{4R_0}{5}$$R_E = \frac{L}{6K(4Lb)} = \frac{1}{24} \frac{L}{K b} = \frac{R_0}{24}$$(i)$ Since slabs $A$ and $E$ are in series with the parallel combination of $B, C, D$,the total heat current $Q$ passes through both $A$ and $E$. Thus,heat flow through $A$ and $E$ is the same. Statement $(A)$ is correct.$(ii)$ The temperature difference across a slab is $\Delta T = Q \cdot R$. Since $R_E$ is the smallest resistance,the temperature difference across $E$ is the smallest. Statement $(C)$ is correct.$(iii)$ For the parallel section,the heat current $Q$ splits into $i_B, i_C, i_D$. Since they are in parallel,the potential difference $\Delta T_{BCD}$ is the same for all. Thus $i = \frac{\Delta T}{R}$.$i_C = \frac{\Delta T}{R_C} = \frac{2\Delta T}{R_0}$,$i_B = \frac{\Delta T}{R_B} = \frac{3\Delta T}{4R_0}$,$i_D = \frac{\Delta T}{R_D} = \frac{5\Delta T}{4R_0}$.Summing $i_B + i_D = \frac{3\Delta T}{4R_0} + \frac{5\Delta T}{4R_0} = \frac{8\Delta T}{4R_0} = \frac{2\Delta T}{R_0} = i_C$. Statement $(D)$ is correct.Therefore,$(A, C, D)$ are correct.

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