Rods `X` and `Y` of identical dimensions but of different materials are joined as shown in The length of each part is the same It the temperature of end `A` and `F` be main tained at `100^(@)C` and `20^(@)C` respectively then find the temperatures of the junctions `B` and `E` (Thermal conductivity of `X` is double that of `Y`) .

Explanation

Givne `K_(X) =2K_(Y)`
`R_(x) =(L)/(K_(x)A) and R_(y) = (L)/(K_(Y)A),i.e,R_(Y)=2R_(X)`
If `R_(X) =R ,R_(Y) =2R`
The corresponding thermal network is as shown
`(100-20)/(R+(4R)/(3)+2R)= (100-theta_(B))/(R)rArrtheta_(B)=81.54^(@)C`
`(100-20)/(R+(4R)/(3)+2R)= (theta_(E)-20)/(2R)rArrtheta_(E)=56.92^(@)C`
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