As per the given figure, two plates A and B of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of plates are 4.0 cm and 2.5 cm respectively and the area of cross-section is 120 cm2 for each plate. The equivalent thermal conductivity of the compound plate is ( 1 + 5 \over \alpha ) K, then the value of \alpha will be \underline\hspace2cm .

Heat and Thermodynamics PYQ — JEE Main 2022 · 29 Jun · Shift 1 · Q61 with Solution

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Correct answer: 21

Interpret the arrangement

Since the two plates are joined together to form a compound plate and each has the same cross-sectional area, they are in series for heat flow.
Given data
- Plate $A$ $A$ :
  - thermal conductivity $k_1 = K$
  - thickness $L_1 = 4.0\,\text{cm}$
- Plate $B$ $B$ :
  - thermal conductivity $k_2 = 2K$
  - thickness $L_2 = 2.5\,\text{cm}$
- Cross-sectional area of each plate = same, so area cancels in series formula.
Formula for equivalent conductivity in series

For two slabs in series,
$k_{\text{eq}} = \frac{L_1+L_2}{\dfrac{L_1}{k_1}+\dfrac{L_2}{k_2}}$
Substitute values
$k_{\text{eq}} = \frac{4.0+2.5}{\dfrac{4.0}{K}+\dfrac{2.5}{2K}}$ $= \frac{6.5}{\dfrac{4}{K}+\dfrac{1.25}{K}} = \frac{6.5}{\dfrac{5.25}{K}}$ $k_{\text{eq}} = \frac{6.5K}{5.25} = \frac{26K}{21}$
Compare with given form

Given,
$k_{\text{eq}} = \left(1+\frac{5}{\alpha}\right)K$
So,
$1+\frac{5}{\alpha} = \frac{26}{21}$ $\frac{5}{\alpha} = \frac{26}{21}-1 = \frac{5}{21}$
Hence,
$\alpha = 21$
Final answer
$\boxed{21}$
Comparison with stored correct answer

Stored correct answer = $21$ .

My derived answer matches it.