JEE Advanced · Physics · 12. Thermal Properties
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
- A heat flow through \(A\) 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 & Solution
Correct Answer
(A) heat flow through \(A\) and \(E\) slabs are same
Step-by-step Solution
Detailed explanation
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.
\(H=\) 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.
\(
\begin{aligned}
& \therefore H_B: H_C: H_D=\frac{1}{R_B}: \frac{1}{R_C}: \frac{1}{R_D} \\
& =\frac{1}{160}: \frac{1}{60}: \frac{1}{96} \\
& =9: 24: 15 \\
& \therefore \quad H_B=\left(\frac{9}{9+24+15}\right) H=\frac{3}{16} H \\
& H_C=\left(\frac{24}{9+24+15}\right) H=\frac{1}{2} H \\
& \text { and } \quad H_D=\left(\frac{15}{9+24+15}\right) H=\frac{5}{16} H \\
&
\end{aligned}
\)
\(
H_C=H_B+H_D
\)
\(\therefore\) Option (d) is correct.
Temperature difference (let us call it \(T\) )
\(=(\) Heat current \() \times(\) Thermal resistance \()\)
\(T_A=H_A R_A=(H)(15 R)=15 \mathrm{HR}\)
\(T_B=H_B R_B=\left(\frac{3}{16} H\right)(160 R)=30 H R\)
\(T_C=H_C R_C=\left(\frac{1}{2} H\right)(60 R)=30 H R\)
\(T_D=H_D R_D=\left(\frac{5}{16} H\right)(96 R)=30 H R\)
\(T_E=H_E R_E=(H)(12 R)=12 \mathrm{HR}\)
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.
\(
\begin{aligned}
& R=\frac{l}{K A} \\
& \therefore \quad R_A=\frac{L}{(2 K)(4 L w)}=\frac{1}{8 K w} \\
& \text { (Here, } w=\text { width) } \\
& R_B=\frac{4 L}{3 K(L w)}=\frac{4}{3 K w} \\
& R_C=\frac{4 L}{(4 K)(2 L w)}=\frac{1}{2 K w} \\
& R_D=\frac{4 L}{(5 K)(L w)}=\frac{4}{5 K w} \\
& R_E=\frac{L}{(6 K)(L w)}=\frac{1}{6 K w} \\
& R_A: R_B: R_C: R_D: R_E \\
&=15: 160: 60: 96: 12
\end{aligned}
\)
So, let us write, \(R_A=15 R, R_B=160 R\) etc and draw a simple electrical circuit as shown in figure.
\(H=\) 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.
\(
\begin{aligned}
& \therefore H_B: H_C: H_D=\frac{1}{R_B}: \frac{1}{R_C}: \frac{1}{R_D} \\
& =\frac{1}{160}: \frac{1}{60}: \frac{1}{96} \\
& =9: 24: 15 \\
& \therefore \quad H_B=\left(\frac{9}{9+24+15}\right) H=\frac{3}{16} H \\
& H_C=\left(\frac{24}{9+24+15}\right) H=\frac{1}{2} H \\
& \text { and } \quad H_D=\left(\frac{15}{9+24+15}\right) H=\frac{5}{16} H \\
&
\end{aligned}
\)
\(
H_C=H_B+H_D
\)
\(\therefore\) Option (d) is correct.
Temperature difference (let us call it \(T\) )
\(=(\) Heat current \() \times(\) Thermal resistance \()\)
\(T_A=H_A R_A=(H)(15 R)=15 \mathrm{HR}\)
\(T_B=H_B R_B=\left(\frac{3}{16} H\right)(160 R)=30 H R\)
\(T_C=H_C R_C=\left(\frac{1}{2} H\right)(60 R)=30 H R\)
\(T_D=H_D R_D=\left(\frac{5}{16} H\right)(96 R)=30 H R\)
\(T_E=H_E R_E=(H)(12 R)=12 \mathrm{HR}\)
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.
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