Two rods A and B of different materials are welded together as shown in Figure. If their thermal conductivities are $k_{1}$ and $k_{2}$ , the thermal conductivity of the composite rod will be

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

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$\frac{3}{2} (k_{1} + k_{2})$

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

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

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Solution

The correct option is D
$\frac{1}{2} (k_{1} + k_{2})$

If the ends of the rods are maintained at temperatures , the rate of flow of heat through each rod is given by $Q_{1} = \frac{k_{1} A (θ_{1} - θ_{2})}{d} a n d Q_{2} = \frac{K_{1} A (θ_{1} - θ_{2})}{d}$ Where A is the cross – sectional area of each rod and d is the length of each rod. Now the cross sectional area _ of the composite rod is 2A. If k is the thermal conductivity of the composite rod, the rate of heat flow is
$Q = \frac{k (2 A) (θ_{1} - θ_{2}))}{d} b u t Q = Q_{1} + Q_{2} . t h e r e f o r e \frac{(2 A) (θ_{1} - θ_{2})}{d} = \frac{K_{1} A (θ_{1} - θ_{2})}{d} + \frac{K_{1} A (θ_{1} - θ_{2})}{d} O r k = \frac{1}{2} (k_{1} + k_{2})$

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