A wall consists of alternating blocks with length d and coefficient of thermal conductivity. The cross-sectional area of the blocks is the same. The equivalent coefficient of thermal conductivity of the wall between left and right is

K_{1} + K_{2}

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\frac{(K_{1} + K_{2})}{2}

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\frac{K_{1} K_{2}}{K_{1} + K_{2}}

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\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}

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Solution

The correct option is B $\frac{(K_{1} + K_{2})}{2}$
These will be in parallel combination because the left ends are at same temperature and the right ends are at
another same temperature.
In parallel new coefficient $\frac{K_{1} A_{1} + K_{2} A_{2}}{A_{1} + A_{2}}$ .......(1)
Cross sectional Area $A_{1} = A_{2} = A$ , for all rods.
for any two rods having same coefficient, $K_{1}$ the resultant is also $K_{1}$ from formula-1
so the above combination will reduce to a combination having just two rods one with $K_{1}$ and another with $K_{2}$
So net coefficient of conductivity will be $K = \frac{K_{1} + K_{2}}{2}$ from formula-1
Option B is correct.

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