Two slabs A and B, each though similar, in all other respects, have different materials. They are placed one above the other in perfect contact and a steady difference of temperature of `36^(@)C` is maintained across the combination. If the thermal conductivity of A is twice that of B, what is the temperature of interface ?

Options

A. `16^(@)C`
B. `12^(@)C`
C. `28^(@)C`
D. `24^(@)C`.

Answer

D

Explanation

In steady state, rate of flow of heat is same
`H_(1)=H_(2)`.
`(k_(1)ADeltatheta_(1))/(d)=(k_(2)ADeltatheta_(2))/(d)`
`(Deltatheta_(1))/(Deltatheta_(2))=(k_(2))/(k_(1))=1/2`
`Deltatheta_(2)=2Deltatheta_(1)`.
Also `Deltatheta_(1)+Deltatheta_(2)=36^(@)C`
`:." "3Deltatheta_(1)=36^(@)C`
`Deltatheta_(1)=12^(@)C`
`rArr36-T=12rArrT=24^(@)C`
Thus correct choice is (d).