The temperature of the two outer surfaces of a composite slab consisting of two materials having coefficient of thermal conductivity K and 2K and thickness x and 4x respectively are `T_2` and `T_1(T_2gtT_1)`. The rate of heat transfer through the slab in steady state is `((AK(T_2 -T_1))/x)f`. where, f is equal to
.

Options

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

Answer

D

Explanation

Equation of thermal conducity of the given
combination `K_(eq) = (l_(1)+l_(2))/(l_(1)/K_(1)+l_(2)/K_(2)) = (x+4x)/((x)/(K)+(4x)/(2K)) = 5/3K`.
Hence rate of flow of heat through the given
combination is `Q/t = (k_(eq)*A(T_(2)-T_(1)))/((x+4x))`
`=(5/3 KA(T_(2)-T_(1)))/(5x) = (1/3 KA(T_(2)-T_(1)))/(x)` on comparing it
with given equation we get `f = 1/3`.