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The temperature of the two outer surface...

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thicknesses x and 4x, respectively are `T_(2)` and `T_(1)(T_(2)gtT_(1))`. The rate of heat of heat transfer through the slab, in ? steady state is `[(A(T_(2)-T_(1))/(x)]f`, with f equal to :-

A

`1`

B

`1//2`

C

`2//3`

D

`1//3`

Text Solution

Verified by Experts

The correct Answer is:
D

`R_(1)`= Resistance fo left part = `(x)/(KA)`
`R_(2)` = Resistance of right part = `(4x)/(2KA) = (2x)/(KA)`
Total Resistance `R_(1) + R_(2) = (x)/(RA) + (2x)/(KA) = (3x)/(KA)`
Thermal current =`(T_(2)-T_(1))/(R) = (K(T_(2) - T_(1))A)/(3x)`
Comparing with the given result f = `(1)/(3)`
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