Two rods X and Y identical dimesion but of differentmaterical are joined as shown in figure . Thelengthof each part is the same . If thetemperature of endA and F be maintained at100^(@)C and 20^(@)C respectively , then find the temperature of the junction B and E . Thethermalconducting of X is doublethat of Y.

Two rods X and Y identical dimesion but of differentmaterical are join

1.	Two rods X and Y identical dimesion but of differentmaterical are joined as shown in figure . Thelengthof each part is the same . If thetemperature of endA and F be maintained at100^(@)C and 20^(@)C respectively , then find the temperature of the junction B and E . Thethermalconducting of X is doublethat of Y.
Answer» Solution :GIVEN `K_(x) = 2K_(y)` If AB = EF = Land A be the crosssectionalarea of the rods ,then `R_(x) = (L)/(K_(x)A) and R_(y) = (L)/(K_(y)A) i.e.,R_(y) = 2R_(x)` Letthe temperature of junctions B and E be `T_(1)` and `T_(2)`respectively. Thecorresponding thermalnetworkis SHOWN belowin figure. For `AB , i =(100-T_(1))/(R_(x))""......(i) ` For `BCE, i_(I) = (T_(1) - T_(2))/(R_(y)) = (T_(1) - T_(2))/(4R_(x)) "".........(II)` For `BDE , (i-i_(1)) = (T_(1)- T_(2))/(2R_(x)) ""..........(iii)` and for `EF , i = (T_(2) -20)/(R_(y)) = (T_(2) - 20)/(2R_(x)) ""......(iv)` Adding Eqs. (ii) and (iii) `i = (3)/(4) ((T_(1) - T_(2)))/(R_(x)) "".....(v)` Equating Eqs.(i) and (iv) ` (100 - T_(1))/(R_(x)) = (T_(2) - 20)/(2R_(x)) "".....(vi) (or) 2T_(1) + T_(2) = 220` Equating Eqs. (i) and (iv) `(KA(T_(1) - T_(2)))/(L) (1)/(2) 7T_(1) - 3T_(2) = 400 ""....(vii)` Solving Eqs. (vi) and (vii) `T_(1) = 81.54^(@)C , T_(2) = 56.92^(@)C`