SATHEE: Heat and Thermodynamics 3 Question 24

Heat and Thermodynamics 3 Question 24

24. 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 (2011)

(a) heat flow through $A$ and $E$ slabs are same

(b) heat flow through slab $E$ is maximum

(d) heat flow through $C$ = heat flow through $B +$ heat flow through $D$

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Answer:

Correct Answer: 24. (a,c,d)

Solution:

Thermal resistance $R = \frac{l}{K A}$

$\begin{aligned} R_{A} = \frac{L}{(2 K) (4 L w)} (Here w = width) \\ = \frac{1}{8 K w}, \\ R_{B} = \frac{4 L}{3 K (L w)} = \frac{4}{3 K w} \\ R_{C} = \frac{4 L}{(4 K) (2 L w)} = \frac{1}{2 K w} \\ R_{D} = \frac{4 L}{(5 K) (L w)} = \frac{4}{5 K w} \\ R_{E} = \frac{L}{(6 K) (L w)} = \frac{1}{6 K w} \\ R_{A} : R_{B} : R_{C} : R_{D} : R_{E} \\ = 15 : 160 : 60 : 96 : 12 \end{aligned}$

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.

$H_{A} = H_{E} = H$

In parallel current distributes in inverse ratio of resistance.

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

$\begin{matrix} = \frac{1}{160} : \frac{1}{60} : \frac{1}{96} \\ = 9 : 24 : 15 \\ ∴ H_{B} = (\frac{9}{9 + 24 + 15}) H = \frac{3}{16} H \\ H_{C} = (\frac{24}{9 + 24 + 15}) H = \frac{1}{2} H \end{matrix}$

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

$H_{C} = H_{B} + H_{D}$

Temperature difference (let us call it $T$ )

$= ($ Heat current $) \times ($ Thermal resistance $)$

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

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

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

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

$T_{E} = H_{E} R_{E} = (H) (12 R) = 12 H R$

Here, $T_{E}$ is minimum. Therefore option (c) is also correct.

$∴$ Correct options are (a), (c) and (d).

Heat and Thermodynamics 3 Question 24

Answer:

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Heat and Thermodynamics 3 Question 24

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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