If two conducting slabs of thickness `d_1` and `d_2`, and thermal conductivity `K_1` and `K_2` are placed together face to face as shown in figure in the steady state temperature of outer surfaces are `theta_1` and `theta_2`. The temperature of common surface 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))`