A composite block is made of slabs A, B, C, D and E of different...

Thermal resistance $$\begin{aligned}& R=\frac{l}{K A}\\ & \therefore \quad R_A=\frac{L}{(2 K)(4 L w)}=\frac{1}{8 K w}\\ & \text{ (Here, }w=\text{ width) }\\ & R_B=\frac{4 L}{3 K(L w)}=\frac{4}{3 K w}\\ & R_C=\frac{4 L}{(4 K)(2 L w)}=\frac{1}{2 K w}\\ & R_D=\frac{4 L}{(5 K)(L w)}=\frac{4}{5 K w}\\ & R_E=\frac{L}{(6 K)(L w)}=\frac{1}{6 K w}\\ & R_A: R_B: R_C: R_D: R_E \\ &=15: 160: 60: 96: 12 \end{aligned}$$So, let us write, $R_A=15 R, R_B=160 R$ etc and draw a simple electrical circuit as shown in figure. $<img src="/api/proxy-image?url=https$ \\ $H=$ Heat current = Rate of heat flow. H_A=H_E=H [let] $\therefore$ Option (a) is correct. In parallel, current distributes in inverse ratio of resistance. $$\begin{aligned}& \therefore H_B: H_C: H_D=\frac{1}{R_B}: \frac{1}{R_C}: \frac{1}{R_D}\\ & =\frac{1}{160}: \frac{1}{60}: \frac{1}{96}\\ & =9: 24: 15 \\ & \therefore \quad H_B=\left(\frac{9}{9+24+15}\right)H=\frac{3}{16}H \\ & H_C=\left(\frac{24}{9+24+15}\right)H=\frac{1}{2}H \\ & \text{ and } \quadH_D=\left(\frac{15}{9+24+15}\right)H=\frac{5}{16}H \\ & \end{aligned}$$H_C=H_B+H_D $\therefore$ Option (d) is correct. Temperature difference (let us call it $T$ ) $=($ Heat current $) \times($ Thermal resistance $)T_A=H_A R_A=(H)(15 R)=15 \mathrm{HR}T_B=H_B R_B=\left(\frac{3}{16} H\right)(160 R)=30 H RT_C=H_C R_C=\left(\frac{1}{2} H\right)(60 R)=30 H RT_D=H_D R_D=\left(\frac{5}{16} H\right)(96 R)=30 H RT_E=H_E R_E=(H)(12 R)=12 \mathrm{HR}$ Here, $T_E$ is minimum. Therefore, option (c) is also correct. $\therefore$ Correct options are (a), (c) and (d). Analysis of Question (i) From calculation point of view, question is difficult otherwise question is simple. (ii) In heat transfer, questions are mainly asked from conduction and radiation topic.