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)` .

Three rods of material `x` and three of material `y` are connected as

1.	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)` .
Answer» Correct Answer - B::C::D<br>Thermal resistance, `R = l/(KA)`<br> `R_x/R_y = K_y/ K_x ` (as `l_x = l_y` and `A_x = A_y`)<br> `=0.46/0.92`<br> `=1/2`<br> So, if `R_x = R` then` R_y = 2R`<br> CEDB forms a balanced wheatstone bridge, i.e. `T_C - T_D` and no heat flows through CD.<br> `:. 1/R_(BE) = 1/(R+R) + 1/(2R + 2R)`<br> or `R_(BE) = 4/3 R`<br> The total resistance between A and E will be<br> `R_(AE) = R_(AB) + R_(BE) = 2R + 4/3R = 10/3R`<br> `:.` Heat current between A and E is<br> `H = ((DeltaT)_(AE))/R_(AE) = (60-10)/((10//3)R) = 15/R`<br> Now, if `T_B` is the temperature at B,<br> `H_(AB) = ((DeltaT)_(AB))/R_(AB)`<br> or `15/R = (60 - T_(B))/(2R)`<br> or `T_(B) = 30^@C`<br> Further,<br> `H_(AB) = H_(BC) + H_(BD)`<br> or `15/R = (30-T_(C))/R + (30-T_(D))/(2R) [T_C = T_D = T(say)]`<br> or `15 = (30-T) + (30-T)/2`<br> Solving this, we get `T=20^@C`<br> or `T_C = T_D = 20^@C`

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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)` .
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