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NEET Physics QUESTION #1028
Question 1
Three identical heat conducting rods in series: outer rods have conductivity \(2K\), middle rod has conductivity \(K\). Left end at \(3T\), right end at \(T\). The ratio \(T_1/T_2\) (left junction to right junction temperature) is:
  • \(\dfrac{5}{3}\)✔️
  • \(\dfrac{5}{4}\)
  • \(\dfrac{3}{2}\)
  • \(\dfrac{4}{3}\)
Correct Answer Explanation
In steady state, heat flow rate \(Q\) is same through all rods. Let each rod have length \(L\) and area \(A\). \(\frac{2KA(3T-T_1)}{L} = \frac{KA(T_1-T_2)}{L} = \frac{2KA(T_2-T)}{L}\). From 1st and 3rd: \(2(3T-T_1)=2(T_2-T)\Rightarrow 3T-T_1=T_2-T\Rightarrow T_1+T_2=4T\). From 1st and 2nd: \(2(3T-T_1)=T_1-T_2\Rightarrow 6T-2T_1=T_1-T_2\Rightarrow 3T_1-T_2=6T\). Solving: \(T_1=\frac{5T}{2}, T_2=\frac{3T}{2}\). Ratio \(=\frac{5/2}{3/2}=\frac{5}{3}\).