Two walls of thickness `d_(1) and d_(2)` and thermal conductivites `K_(1) and K_(2)` are in contact. In the steady state, if the temperature at the outer surfaces are `T_(1) and T_(2)` the temperature at the common wall is

Options

A. `(K_(1) theta_(1) d_(1)+ K_(2) theta_(2) d_(2))/( K_(1)d_(1)+K_(2)d_(2))`
B. `(K_(1) theta_(1)+K_(2) theta_(2))/(K_(1)+K_(2))`
C. `(K_(1) theta_(1)+K_(2) theta_(2))/(theta_(1)+theta_(2))`
D. `(K_(1) theta_(1)d_(2)+K_(2) theta_(2)d_(1))/(K_(1) d_(2)+K_(2)d_(1))`

Answer

D

Explanation

`theta=theta_(1)-I_(omega)R_(th1)=theta_(1)-((theta_(1)-theta_(2))/(R_(Th1)-R_(Th2)))=theta_(1)-((theta_(2)-theta_(1))/((K_(1)A)/(d_(1))+(K_(2)A)/(d_(2)))xx(K_(1)A)/(d_(1)))`
`=theta_(1)-((theta_(2)-theta_(1))/((K_(1))/(d_(1))+(K_(2))/(d_(2)))xx(K_(1))/(d_(1)))=(K_(1)theta_(1)d_(2)+K_(2)theta_(2)d_(1))/(K_(1)d_(2)+K_(2)d_(1))`