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

Options

A. `1`
B. `1//2`
C. `2//3`
D. `1//3`

Answer

D

Explanation

`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)`