Two plates of the same area and the same thickness having thermal conductivities $k_{1}$ and $k_{2}$ are placed one on top of the other. The top and bottom faces of the composite plate are maintained at different constant temperatures. The thermal conductivity of the composite plate will be

$(k_{1} + k_{2})$

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$\frac{1}{2} (k_{1} + k_{2})$

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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 D
$\frac{2 k_{1} k_{2}}{(k_{1} + k_{2})}$

Let d be the thickness and A the area of each plate. Let $T_{1}$ be the temperature of the top face of the upper plate I and $T_{2}$ that of the bottom face of the lower plate II. Let T be the temperature of the surface in contact.

In the steady state, the rate of flow of heat through plate I is the same as that through plate II, i.e.,
$\frac{Q}{t} = \frac{k_{1} A (T_{1} - T)}{d} = \frac{k_{2} A (T - T_{2})}{d}$
$\Rightarrow \frac{k_{1}}{k_{2}} = \frac{T - T_{2}}{T_{1} - T}$
$\Rightarrow \frac{k_{1}}{k_{2}} + 1 = \frac{T - T_{2}}{T_{1} - T} + 1$
$\Rightarrow \frac{k_{1} + k_{2}}{k_{2}} = \frac{T_{1} - T_{2}}{T_{1} - T}$ .....(1)
Now, the thickness of the composite plate is 2d. If k is the thermal conductivity of the composite plate, then the rate of flow of heat through it is
$\frac{Q}{t} = \frac{k A (T_{1} - T_{2})}{2 d}$
In the steady state, this must be the same as the rate of flow of heat through either plate I or plate II.
$∴ \frac{k A (T_{1} - T_{2})}{2 d} = \frac{k_{1} A (T_{1} - T)}{d}$
$\Rightarrow \frac{T_{1} - T_{2}}{T_{1} - T} = \frac{2 k_{1}}{k}$ .....(2)
Using (2) in (1), we get $k = \frac{2 k_{1} k_{2}}{(k_{1} + k_{2})}$ , which is choice (d).

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