Physics · Heat Transfer

JEE Advanced 2018 — Paper 1 — Question 14

Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T1=300 KT_{1}=300 \mathrm{~K} and T2=100KT_{2}=100 K, as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K1K_{1} and K2K_{2} respectively. If the temperature at the junction of the two cylinders in the steady state is 200 K , then K1/K2=K_{1} / K_{2}= _____\_\_\_\_\_ .

Question figure

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

In steady state, heat current in both material is same

K1(300−200)AL=K2(200−100)4AL\frac{K_{1}(300-200) A}{L}=\frac{K_{2}(200-100) 4 A}{L}

⇒K1K2=4\Rightarrow \frac{K_{1}}{K_{2}}=4

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 1
Subject
Physics
Chapter
Heat Transfer
Topic
Advanced Problems on Conduction
Two conducting cylinders of equal length but different radii are… | JEE Advanced 2018 PYQ with Solution · DhiX AI