Two bars of same length and same cross-sectional area but of different thermal conductivites `K_(1)` and `K_(2)` are joined end to end as shown in the figure. One end of the compound bar it is at temperature `T_(1)` and the opposite end at temperature `T_(2)` (where`T_(1) gt T_(2)`). The temperature of the junction is

Options

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

Answer

A

Explanation

Let L and A be length and area of cross-section of each bar respectively.
`:.` Heat current through the bar 1 is
`H_(1) = (K_(1)A(T_(1) - T_(0)))/(L)`
Here `T_(0)` is junction temperature.
Heat current through the bar 2 is
`H_(2) = (K_(2)A(T_(0) - T_(2)))/(L)`
At steady state, `H_(1) = H_(2)`
`:. (K_(1)A(T_(1)-T_(0)))/(L) = (K_(2)A(T_(0) - T_(2)))/(L)`
`K_(1)(T_(1) - T_(0)) = K_(2) (T_(0) - T_(2))`
`K_(1)T_(1) - K_(1)T_(0) = K_(2)T_(0) - K_(2)T_(2)`
`K_(1)T_(0) + K_(2)T_(0) = K_(1)T_(1) + K_(2)T_(2)`
`T_(0)(K_(1)+K_(2)) = K_(1)T_(1) + K_(2)T_(2)`
`T_(0) = (K_(1)T_(1) + K_(2)T_(2))/((K_(1) + K_(2)))"......"(i)`