Discussion Forum

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 the same width. Heat Q flows only from left to right through the blocks. Then, in steady-state

Answer:

Explanation:

 Thermal  resistance

 $R= \frac{l}{KA}$

 $\therefore$   $R_{A}= \frac{ L}{(2K)(4L \omega)}=\frac{1}{ 8K \omega}$ 

    (here , $\omega$= width)

 $R_{B}= \frac{ 4L}{3K(L \omega)}= \frac{4}{ 3K \omega}$

 $R_{C}= \frac{4L}{(4K)(2L \omega)}=\frac{1}{2K \omega}$

 $R_{D}=\frac{4L}{(5K)(L \omega)}=\frac{4}{5 K \omega}$

 $R_{E}= \frac{L}{(6K)(L \omega)}=\frac{1}{6 K \omega}$

 $R_{A}:R_{B}:R_{C}:R_{D}:R_{E}$

 =15:160:60:96:12

So, Let us write , $R_{A}=15 R$,$R_{B}=160 R$ etc and draw a simple electrical circuit as shown in figure.

 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

 $\therefore$     $H_{B}:H_{C}:H_{D}= \frac{1}{R_{B}}:\frac{1}{R_{C}}:\frac{1}{R_{D}}$

   = $\frac{1}{160}:\frac{1}{60}:\frac{1}{96}$

=9:24-:15

  $\therefore$     $H_{B}=\left(\frac{9}{9+24+15}\right)H= \frac{3}{16} H$

     $H_{C}=\left(\frac{24}{9+24+15}\right)H= \frac{1}{2} H$

 and    $H_{D}=\left(\frac{15}{9+24+15}\right)H= \frac{5}{16} H$

 $H_{C}=H_{B}+H_{D}$

 $\therefore$ option (d) is correct 

 Temperature  difference (let us call it T)= (Heat current ) x (Thermal resistance) 

 $T_{A}= H_{A} R_{A}= (H) (15 R)  =15 HR$

 $T_{B}=H_{B}R_{B}==\left(\frac{3}{16}H\right) (160 R)=30HR$

 $T_{C}=H_{C}R_{C}==\left(\frac{1}{2}H\right) (60 R)=30HR$

 $T_{D}=H_{D}R_{D}==\left(\frac{5}{16}H\right) (96 R)=30HR$

 $T_{E}=H_{E}R_{E}=(H) (12 R)=12 HR$

 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 the question is simple.
(ii) In heat transfer, questions are mainly asked from conduction and radiation topic.