A wall has two layers A and B, each made of different material. Both the layers have the same thickness. The thermal conductivity of the material of A is twice that of B . Under thermal equilibrium, the temperature difference across the wall is `36^@C.` The temperature difference across the layer A is

Options

A. `12 ^(@)C`
B. `18 ^(@)C`
C. `6^(@)C`
D. `24 ^(@)C`

Answer

A

Explanation

Here, `K_(A) = 2K_(B), T_(A) - T_(B) = 36^(@)C`
Let T is the temperature of the junction,
As `((DeltaT)/(Deltat))_(A) = ((DeltaT)/(Deltat))_(B)`
`:. (K_(A) A(T_(A) - T))/(x) = (K_(B)A(T-T_(B)))/(x)`
`2K_(B)(T_(A) - T) = K_(B)(T- T_(B))`
`2(T_(A) - T) = T - T_(B)`
Add `(T_(A) - T)` on both sides, we get
`3(T_(A) - T') = T_(A) - T + t - T_(B)`
`3(T_(A) - T') = T_(A) - T_(B)`
`T_(A) - T = (T_(A) - T_(B))/(3) = 36/3 = 12 .^(@)C`
`:.` Temperature difference across the layer
`A = T_(A) - T = 12 .^(@)C`