past year iit thermo doubt

yup... that i had done... but... getting mixed up somewhere..

please give the steps.... how to proceed...

do the question in this way..take the resistance of rod Y = R ..so for X it will be 2R
and the middle X resistance will get cancelled due to wheatstone..so we have a circuit in which 2R + 2R = 4R and R +R = 2R
2R and 4R in parallel and then R in series
net resisitance 7R/3 ..temp difference = 50..thermal current 50*3 / 7R

so for temperature at B we have
50 - x = (50*3/7R) * R ....using this property V_A - V_B = iR
so x = 29°C
similar procedure can be done for the temp difference B and D..in that case the thermal current will be 50/7R.. so temp at D will be 16°C
and C and D will have the same temp due to wheatstone property
dunno why i am not getting the exact answers

@Govind, R_X = 50 R_Y. Look at the units ... in one there is per meter while the other one is per cm