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

B, C, D

Explanation

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