Question

Two identical rods are made of different materials whose thermal conductivities are `K_(1) and K_(2)`. They are placed end to end between two heat reservoirs at temperatures `theta_(1) and theta_(2)`. The temperature of the junction of the rod is

A. `(theta_(1) + theta_(2))/2`
B. `(K_(1)theta_(1) + K_(2) theta_(2))/ (K_(1) + K_(2)`
C. `(K_(1)theta_(2) + K_(2)theta_(1))/(K_(1) + K_(2))`
D. `(K_(1)theta_(1)+ K_(2)theta_(2))/([K_(1)-K_(2)])`

Answer 1

Correct Answer - B
Heat currents in the two rods will be same.
`H_(1) = H_(2)`
or `(theta_(1)-theta)/(l//K_(1)A)=(theta-theta_(2))/(l//K_(2)A) (theta` = temperature of junction)
or `K_(1)(theta_(1)-theta)=K_(2)(theta-theta_(2)`
`therefore theta=(K_(1)theta_(1)+K_(2)theta_(2))/(K_(1)+K_(2))`