Two slabs are of the thicknesses `d_(1)` and `d_(2)`. Their thermal conductivities are `K_(1)` and `K_(2)` respectively They are in series. The free ends of the combination of these two slabs are kept at temperatures `theta_(1)` and `theta_(2)`. Assume `theta_(1)gttheta_(2)`. The temperature `theta` of their common junction is

Options

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

Answer

C

Explanation

For the first slab
Heat current, `H_(1)=(K_(1)(theta_(1)-theta)A)/(d_(1))`
For the second slab Heat current , `H_(2)=(K_(2)(theta-theta_(2))A)/(d_(2))`
As the slabs are connected in seriews same anount of heat current flows through the slabs
`:.,.H_(1)=H_(2)`
`(K_(1)(theta_(1)-theta)A)/(d_(1))=(K_(2)(theta-theta_(2))A)/(d_(2))`
`K_(1)theta_(1)d_(2)-K_(1)thetad_(2)=K_(2)thetad_(1)-K_(2)theta_(2)d_(1)`
`theta(K_(2)d_(1)+K_(1)d_(2))=K_(1)theta_(1)d_(2)+K_(2)theta_(2)d_(1)`
`theta(K_(1)theta_(1)d_(2)+K_(2)theta_(2)d_(1))/(K_(2)d_(1)+K_(1)d_(2))`