Find the temperature of junction if rods are identical and are placed symmetrically
(a) !!95^0 C!!
(b) !! 90^0 C!!
(c) !!85^0 C!!
(d) !!80^0 c!!
Assume temperature at junction, !!T!!
We will calculate the heat flow from each branch. We know total out going heat will be equal to total incoming heat.
Rate of heat flow !!(DeltaQ)/(Deltat)= - (kxxAxxDeltaT)/l!!
Since both the rods are identical, therefore k,A and l will be same for all.
!!-(kxxAxx(100-T))/l+ (-(kxxAxx(100-T))/l)= (-(kxxAxx(T-70))/l)+(-(kxxAxx(T-50))/l)!!
!!=> 100-T+100-T= T-70+T-50!!
!!=> 4T = 100+100+50+70= 320!!
!!=> T = 80^0 C!!
Therefore, answer (d) is correct.