Two slabs A and B of different materials but of the same thickness are joined end to end to form a composite slab. The thermal conductivities of A and B are `K_(1)` and `K_(2)` respectively. A steady temperature difference of `12^(@)`C is maintained across the composite slab. If `K_(1) = K_(2)/2`, the temperature difference across slab A is

Options

A. `4^(@)`C
B. `6^(@)`C
C. `8^(@)`C
D. `10^(@)`C

Answer

C

Explanation

The given equation can be shown as

Rate of flow of heat will be equal in both the slabs
`therefore (12-x)K_(1)= K_(2) (x-0)`
`12-x = 2x rArr x=4^(@)`C `(therefore K_(1) = (K_(2)/2)`
The temperature difference across slab
`A = (12-x) = (12-4) = 8^(@)`C