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 

 

Thermal resistance  $R=\frac{1}{KA}$
$\therefore R_A=\frac{L}{\left(2K\right)\left(4Lw\right)}$ (Here w=width)
$=\frac{1}{8Kw}$

$$\begin{gathered} R_B=\frac{4L}{3K\left(Lw\right)}=\frac{4}{3Kw} \\ {R}_C=\frac{4L}{\left(4K\right)\left(2Lw\right)}=\frac{1}{2Kw} \\ {R}_D=\frac{4L}{\left(5K\right)\left(Lw\right)}=\frac{4}{5Kw} \\ {R}_E=\frac{L}{\left(6K\right)\left(Lw\right)}=\frac{1}{6Kw} \\ {R}_A:R_B:R_C:R_D;R_E \\ =15:160:60:96:12 \end{gathered}$$

So, let us write,  $R_A=15R,R_B=160R$ etc and draw a simple electrical circuit as shown in figure

H = Heat current = Rate of heat flow
$H_A=H_E=H$
In parallel current distributes in inverse ratio of resistanance
 $\therefore H_B:H_C:H_D=\frac{1}{R_B}:\frac{1}{R_C}:\frac{1}{R_D}$$=\frac{1}{160}:\frac{1}{60}:\frac{1}{96} =9:24:15$

$$\begin{gathered} H_B=\left(\frac{9}{9+24+15}\right)H=\frac{3}{16}H \\ {H}_C=\left(\frac{24}{9+24+15}\right)H=\frac{1}{2}H \\ And H_D=\left(\frac{15}{9+24+15}\right)H=\frac{5}{16}H \\ {H}_C=H_B+H_D \end{gathered}$$
Temperature difference (let us call it T)
= (Heat current ) x (Thermal resistance)
$$\begin{gathered} T_A=H_A{R}_A=\left(H\right)\left(15R\right)=15HR \\ {T}_B=H_B{R}_B=\left(\frac{3}{16}H\right)\left(160R\right)=30HR \\ {T}_C=H_C{R}_C=\left(\frac{1}{2}H\right)\left(60R\right)=30HR \\ {T}_D=H_D{R}_D=\left(\frac{5}{16}H\right)\left(96R\right)=30HR \\ {T}_E=H_E{R}_E=\left(H\right)\left(12R\right)=12HR \end{gathered}$$

Here $T_E$ is minimum. Therefore option (c) is also correct.
$\therefore$Correct option are a,c and d