Rate of heat flow through a cylindrical rod is H 1 . Temperatures of ends of rod are T 1…
Physics · NEET · NTA Exams — Properties of Solids and Liquids
Rate of heat flow through a cylindrical rod is H1. Temperatures of ends of rod are T1 and T2. If all the dimensions of rod become double and temperature difference remains same and rate of heat flow becomes H2. Then:
- \(\mathrm{H}_{2}=2 \mathrm{H}_{1}\)
- \(H_{2}=\frac{H_{1}}{2}\)
- \(H_{2}=\frac{H_{1}}{4}\)
- \(H_{2}=4 H_{1}\)
Answer
(A) H _ 2 =2 H _ 1
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