As per the given figure,two plates A and B of | vedclass

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(B) The rate of heat flow is given by $\frac{\Delta Q}{\Delta t} = \left(\frac{1}{R}\right) \Delta T$,where $R$ is the thermal resistance.For a plate

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(B) The rate of heat flow is given by $\frac{\Delta Q}{\Delta t} = \left(\frac{1}{R}ight) \Delta T$,where $R$ is the thermal resistance.For a plate of thickness $L$,thermal conductivity $K$,and area $A$,the thermal resistance is $R = \frac{L}{KA}$.For plate $A$: $R_{1} = \frac{L_{1}}{K_{1}A} = \frac{4.0}{K(120)}$.For plate $B$: $R_{2} = \frac{L_{2}}{K_{2}A} = \frac{2.5}{(2K)(120)}$.Since the plates are in series,the equivalent thermal resistance is $R_{	ext{eq}} = R_{1} + R_{2}$.The equivalent thermal conductivity $K_{	ext{eq}}$ for a compound plate of total thickness $L_{	ext{eq}} = L_{1} + L_{2}$ and area $A$ is given by $\frac{L_{	ext{eq}}}{K_{	ext{eq}}A} = \frac{L_{1}}{K_{1}A} + \frac{L_{2}}{K_{2}A}$.Substituting the values: $\frac{4.0 + 2.5}{K_{	ext{eq}}(120)} = \frac{4.0}{K(120)} + \frac{2.5}{2K(120)}$.$\frac{6.5}{K_{	ext{eq}}} = \frac{4}{K} + \frac{1.25}{K} = \frac{5.25}{K} = \frac{21/4}{K} = \frac{21}{4K}$.Therefore,$K_{	ext{eq}} = \frac{6.5 	imes 4K}{21} = \frac{26}{21}K = \left(1 + \frac{5}{21}ight)K$.Comparing this with $\left(1 + \frac{5}{\alpha}ight)K$,we get $\alpha = 21$.

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