Question

A composite block is made of slabs $A,B,C,D$ and $E$ of different thermal conductivities (given in terms of a constant $K$) and sizes (given in terms of length, $L$) as shown in Fig. All slabs are of same width.

• A. Heat flow through $A$ and $E$ slabs are same
• B. Heat flow through slab $E$ is maximum
• C. Temperature difference across slab $E$ is smallest
• D. Heat flow through $C=$ heat flow through $B+$ heat flow through $D$

Answer 1

Answer: (A), (C), (D)

The correct options are
A Heat flow through $A$ and $E$ slabs are same
C Temperature difference across slab $E$ is smallest
D Heat flow through $C=$ heat flow through $B+$ heat flow through $D$

a. At steady state, heat flow through $A$ and $E$ are same. So option (a) is correct and option (b) is incorrect.

c. $△T=H\times R$, where $H$ is heat current.

${}^{\prime }{H}^{\prime }$ is same for $A$ and $E$ but $R$ is smallest for $E$. So temperature difference across slab $E$ is smallest. So option (c) is correct.

d. $H_B=\frac{△T}{R_B},H_C=\frac{△T}{R_C}$ and $H_D=\frac{△T}{R_D}$,

From this option if, $H_C=H_B+H_D$

$\Rightarrow \frac{1}{R_C}=\frac{1}{R_B}+\frac{1}{R_D}$

$\Rightarrow \frac{4K\left(2Kb\right)}{4L}=\frac{2K\left(Lb\right)}{4L}+\frac{5K\left(Lb\right)}{4L}$

which is true. Here $b$ is the width of slab.

Hence option (d) is also correct.