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 $O . K$.

$i_C=\frac{\Delta \theta}{R_C}=\frac{\Delta \theta}{R / 2}=\frac{2 \Delta \theta}{R}$

$i_B=\frac{\Delta \theta}{R_B}=\frac{\Delta \theta}{4 R / 3}=\frac{3 \Delta \theta}{4 R}$

$i_D=\frac{\Delta \theta}{R_D}=\frac{\Delta \theta}{4 R / 5}=\frac{5 \Delta \theta}{4 R}$

$i_C=i_B+i_D, \text { (iv) is } O . K$