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

Answer

B

Explanation

As `k_(A) = 2k_(B)`
Let area of layers be a
`T_(A) -T_(B)= 36^(@)C` …(1)
`(k_(A)A(T_(A))-T_(0))/(l) =(k_(B)A(T_(0))-T_(B))/(l)`
`rArr`` 2(T_(A)-T_(0))=T_(0)-T_(B)`
`rArr``2T_(A)+T_(B)=3T_(0)` …(2)
From (1) and (2) , we get
`2T_(A) +T_(A)-36 + 3 T_(0)`
`rArr` `3T_(A) - 3 T_(0)= 36`
`rArr`` T_(A) -T_(0)= 12^(@)C`