Three rods of material `x` and three of material `y` are connected as shown in figure. All the rods are identical in length and cross sectional area. If the end `A` is maintained at `60^(@)C` and the junction `E` at `10^(@)C` , calculate the temperature of the junction `B`. The thermal conductivity of `x` is `800Wm^(-1).^(@)C^(-1)` and that of `y` is `400Wm^(-1).^(@)C^(-1)` .

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`