In the question number of 45, the equivalent thermal conductivity of t

1.	In the question number of 45, the equivalent thermal conductivity of the compound bar is
Answer» `(K_(1)K_(2))/(K_(1)+K_(2))` `(2K_(1)K_(2))/(K_(1)+K_(2))` `(K_(1))/(K_(1)+K_(2))` `(K_(2))/(K_(1)+K_(2))` Solution :Let K be thermal CONDUCTIVITY of the temperature BAR. Heat current through the compound bar of length `2L` is `H = (KA(T_(1) - T_(2)))/(2L)` At steady state, `H = H_(1) = H_(2)` `:. (KA(T_(1) - T_(2)))/(2L) = (K_(1)A(T_(1) - T_(0)))/(L) "â€¦..."(ii)` Substituting the value of `T_(0)` Eq (i) in (ii), we GET `(K(T_(1)-T_(2)))/(2)= K_(1)[T_(1) - ((K_(1)T_(1) + K_(2)T_(2))/((K_(1)+K_(2))))] = (K_(1)K_(2)(T_(1)-T_(2)))/((K_(1)+K_(2)))` `rArr = (2K_(1)K_(2))/(K_(1)+K_(2))`

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