Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
The rate of flow of heat through a copper rod with temperature difference $28^{\circ} \mathrm{C}$ is $1400 \mathrm{cal} \mathrm{s}^{-1}$. The thermal resistance of copper rod will be
PhysicsThermal Properties of MatterMHT CETMHT CET 2021 (22 Sep Shift 2)
Options:
  • A $0.05 \frac{{ }^{\circ} \mathrm{Cs}}{\mathrm{cal}}$
  • B $0.02 \frac{{ }^{\circ} \mathrm{Cs}}{\mathrm{cal}}$
  • C $5 \frac{{ }^{\circ} \mathrm{Cs}}{\mathrm{cal}}$
  • D $2 \frac{{ }^{\circ} \mathrm{Cs}}{\mathrm{cal}}$
Solution:
1835 Upvotes Verified Answer
The correct answer is: $0.02 \frac{{ }^{\circ} \mathrm{Cs}}{\mathrm{cal}}$
Thermal resistance $=\frac{\text { Temp.difference }}{\text { Thermalcurrent }}=\frac{28}{1400}=0.02{ }^{\circ} \mathrm{Cs} / \mathrm{cal}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.