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 the figure. All slabs are of same with. Heat Q flows only from left to right through the blocks. Then, in steady state

• A. heat flown trough 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)

heat flow through C = heat flow through B + heat flow through D

Thermal resistance $R=R=\frac{1}{KA}$
$\therefore R_A=\frac{L}{\left(2K\right)\left(4LW\right)}$ (Here w = width)
$=\frac{1}{8K{w}^{\prime }}$
$R_B=\frac{4L}{3k\left(LW\right)}=\frac{4}{3Kw}$
$R_C=\frac{4L}{\left(4K\right)\left(Lw\right)}=\frac{1}{2Kw}$
$R_D=\frac{4L}{\left(5K\right)\left(Lw\right)}=\frac{4}{5Kw}$
$R_E=\frac{L}{\left(6K\right)\left(Lw\right)}=\frac{1}{6Kw}$
$R_A:R_B:R_C:R_D:R_E$
=15:160:60:96:12
So, let us write, $R_A=15R,R_B=160R$ etc and draq a simple electrical circuit as shown in figure

H = Heat current = Rate of heat flow.
$H_A=H_E=H$ (let)
In parallel current distributes in inverse ratio of resistance.
$\therefore H_B:H_C:H_D=\frac{1}{R_B}:\frac{1}{R_C}:\frac{1}{R_D}$
$=\frac{1}{160}:\frac{1}{60}:\frac{1}{96}$
=9:24:15
$\therefore H_B=\left(\frac{9}{9+24+15}\right)H=\frac{3}{16}H{H}_C=\left(\frac{24}{9+24+15}\right)H=\frac{1}{2}Hand{H}_D=\left(\frac{15}{9+24+15}\right)H=\frac{5}{16}H$
Temperature difference (let us call it T)
= (Heat current) $\times$ ( Thermal resistance)
$T_A=H_A{R}_A=\left(H\right)\left(15R\right)=15HR$
$T_B=H_B{R}_B=\left(\frac{3}{16}H\right)\left(160R\right)=30HR$
$T_C=H_C{R}_C=\left(\frac{1}{2}H\right)\left(60R\right)=30HR$
$T_D=H_D{R}_D=\left(\frac{5}{16}H\right)\left(96R\right)=30HR$
$T_E=H_E{R}_E=\left(H\right)\left(12R\right)=12HR$
Here, $T_E$ is minimum. Therefore option (c) is also correct.
$\therefore$ Correct options are (a), (c), and (d).