Two materials having coefficients of thermal conductivity ‘3 K’ and ‘K’ and thickness ‘d’ and ‘3 d’, respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are `theta_(2)` and `theta_(1)` respectively, `(theta_(2) gt theta_(1))` The temperature at the interface is:

Options

A. `(theta_(1))/(6)+(5theta_(2))/(6)`
B. `(theta_(1))/(10)+(9theta_(2))/(10)`
C. `(theta_(2)+theta_(1))/(2)`
D. `(theta_(1))/(3)+(2theta_(2))/(3)`

Answer

B

Explanation

`d(dQ)/(dt)=(3KA(theta_(2)-theta))/(d)=(KA(theta-theta_(1)))/(3d)`
`theta=(theta_(1))/(10)+(9theta_(2))/(10)`