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 the figure. All slabs are of same width. Heat $'Q'$ flows only from left to right through the blocks. Then in steady state $Image$
$(A)$ 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 $D$.
IIT 2011, Advanced
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$(3)$ $A$ and $E$ are in series, hence heat current is same, $(i)$ is $O . K$.
Thermal resistances
Let $b$ is width of slabs
$R_A=\frac{L}{2 K(4 L b)}=\frac{R}{8}, R_B=\frac{4 L}{3 K(L b)}=\frac{4 R}{3}$
$R_C=\frac{4 L}{4 K(2 L b)}=\frac{R}{2}, R_D=\frac{4 L}{5 K(L b)}=\frac{4 R}{5}$
$R_E=\frac{L}{6 K(4 L b)}=\frac{R}{24}$
Temperature difference $=$ Heat current $\times$ resistance
Since resistance of $E$ is samllest, hence, temperature difference across it is minimum, $(iii)$, is $O . K$.
$i_C=\frac{\Delta \theta}{R_C}=\frac{\Delta \theta}{R / 2}=\frac{2 \Delta \theta}{R}$
$i_B=\frac{\Delta \theta}{R_B}=\frac{\Delta \theta}{4 R / 3}=\frac{3 \Delta \theta}{4 R}$
$i_D=\frac{\Delta \theta}{R_D}=\frac{\Delta \theta}{4 R / 5}=\frac{5 \Delta \theta}{4 R}$
$i_C=i_B+i_D, \text { (iv) is } O . K$
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