Skip to content
doubtly
Sign up
Log in to upvote 0 Log in to downvote
Topic: Heat transfer: conduction, convection, radiation· Properties of Solids and Liquids

Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity 2K while that in the middle has thermal conductivity K. The left end of the combination is maintained at temperature…

Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity \(2K\) while that in the middle has thermal conductivity \(K\). The left end of the combination is maintained at temperature \(3T\) and the right end at \(T\). The rods are thermally insulated from outside. In steady state, temperature at the left junction is \(T_1\) and that at the right junction is \(T_2\). The ratio \(T_1/T_2\) is

three rods in series: 2K, K, 2K; left end 3T, right end T; junction temps T1, T2

1. \(\frac{3}{2}\)

2. \(\frac{4}{3}\)

3. \(\frac{5}{3}\)

4. \(\frac{5}{4}\)

Doubtly Asked NEET UG 2025 Practice test neet Save WhatsApp Report
Sign in to comment

1 answer

Log in to upvote 0 Log in to downvote

Best answer

Answer: 3. \(\frac{5}{3}\)

Key idea: in steady state the same heat current flows through all three rods, so \(kA\,\Delta T/l\) is equal for each.

With equal \(A\) and \(l\):

$$2K(3T-T_1)=K(T_1-T_2)=2K(T_2-T)$$

From the first equality: \(6T-2T_1=T_1-T_2\Rightarrow 3T_1-T_2=6T\).
From the second: \(T_1-T_2=2T_2-2T\Rightarrow T_1=3T_2-2T\).

Substituting: \(9T_2-6T-T_2=6T\Rightarrow T_2=\tfrac32T\), and \(T_1=\tfrac52T\).

$$\frac{T_1}{T_2}=\frac{5/2}{3/2}=\frac53$$
Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
Was this answer helpful?
Sign in to comment

Your answer

Log in or create a free account to answer this question.