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

Options

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

Answer

B

Explanation

Suppose thickness of each wall is `x` then
`(Q/t)_("combination") = (Q/t)_(A)`
`rArr (K_(S)A(theta_(1)-theta_(2)))/(2x) = (2KA(theta_(1)-theta))/(x)`
`:' K_(S) = (2xx2KxxK)/((2K+K)) = 4/3 K and (theta_(1)-theta_(2)) = 36^(@)`
`rArr (4/3 KA xx36)/(2x) = (2KA(theta_(1)-theta))/(x)`

Hence temperature difference across wall `A` is
`(theta_(1)-theta) = 12^(@)C`.