Assertion: The equivalent thermal conductivit | vedclass

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(C) For two plates of equal thickness $d$ and thermal conductivities $K_1$ and $K_2$ placed in series,the total thermal resistance is $R_{eq} = R_1 +

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If the Assertion is correct but Reason is incorrect.

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(C) For two plates of equal thickness $d$ and thermal conductivities $K_1$ and $K_2$ placed in series,the total thermal resistance is $R_{eq} = R_1 + R_2$.Since $R = \frac{d}{KA}$,we have $\frac{2d}{K_{eq}A} = \frac{d}{K_1A} + \frac{d}{K_2A}$.Simplifying this,we get $\frac{2}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2}$,which implies $K_{eq} = \frac{2K_1K_2}{K_1 + K_2}$.This is the harmonic mean of $K_1$ and $K_2$. The harmonic mean of two numbers is always less than the larger number and greater than the smaller number,but specifically,the equivalent conductivity $K_{eq}$ satisfies $K_{min} Wait,let's re-evaluate: If $K_1 = 10$ and $K_2 = 2$,then $K_{eq} = \frac{2(10)(2)}{10+2} = \frac{40}{12} \approx 3.33$.Here $3.33 > 2$ (the smaller value). Thus,the Assertion is incorrect.

Assertion: The equivalent thermal conductivity of two plates of the same thickness in contact is less than the smaller value of thermal conductivity.
Reason: For two plates of equal thickness in contact,the equivalent thermal conductivity is given by: $\frac{2}{K} = \frac{1}{K_1} + \frac{1}{K_2}$

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Assertion: The equivalent thermal conductivity of two plates of the same thickness in contact is less than the smaller value of thermal conductivity.Reason: For two plates of equal thickness in contact,the equivalent thermal conductivity is given by: $\frac{2}{K} = \frac{1}{K_1} + \frac{1}{K_2}$