Three rods of material x and three rods of material y are connected as shown in figure. All the rods are of identical length and cross-sectional area. If the end A is maintained at `60^@`C and the junction E at `10^@C` , calculate temperature of junctions B, C and D. The thermal conductivity of x is 0.92 `cal//cm -s^@C` and that of y is `0.46 cal// cm-s^@C`.
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Explanation

` R_(X)prop1/K_(X), R_(Y)prop 1/K_(Y) implies R_(X)/R_(Y) = K_(Y)/K_(X)=0.46/0.92=1/2`
Let ` R_(X)=R therefore R_(Y)=2R`
The total resistance `sumR = R_(Y)` + effective resistance in the bridge
`sumR=2R+(2Rxx4R)/(2Rxx4R) = 2R + 4/3 R =10/3R & because Deltatheta = lxxR`
Further ` I_(BCE) (2R)=I_(BDE)(4R)` and `I_(BCE)+ I_(BDE) = I implies I_(BCE) = 2/3I` and`I_(BDE)= 1/3I`
For `A` and `B` ` theta_(A) - theta_(B)= 60^(@) - theta_(B) = 2RxxI ...(i)`
For `A` and `C` ` theta_(B) - theta_(C) =2/3(IxxR) ... (ii) theta_(C) -theta_(E)= 2/3 xx RxxI`
For `A` and `E` `theta_(A) -theta_(E) = 60-10 =50 implies10/3(RxxI)=50` ..(iii)` therefore RxxI=15`
` therefore theta_(A)-theta_(B)- 2xx15=30, theta_(B)=60-30=30^(@)C, theta_(B)-theta_(C) = ((2)/(3)) xx 15 =10`
`therefore theta_(C)=30-10=20^(@)C` Obviously, ` theta_(C)=theta_(D) = 20^(@)C`