| ⇦ | 
| ⇨ |
The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T₁ and T₂ (T₁ > T₂). The rate of heat transfer, dQ/dt through the rod in a steady state is given by:
Options
(a) dQ/dt = k(T₁ – T₂) / LA
(b) dQ/dt = kLA (T₁ – T₂)
(c) dQ/dt = kA (T₁ – T₂) / L
(d) dQ/dt = kL (T₁ – T₂) / A
Correct Answer:
dQ/dt = kA (T₁ – T₂) / L
Explanation:
dQ / dt = kA (T₁ -T₂) / L
[(T₁ -T₂) is the temperature difference]
Related Questions: - A dynamo converts
- A magnetic field of 1600 A/m produces a magnetic flux of 2.4 x 10⁻⁵ Wb
- A proton of mass m and charge q is moving in a plane with kinetic energy E. If there exists
- If Q, E and W denote respectively the heat added, change in internal energy
- Two cities are 150 km apart. Electric power is sent from one city to another city
Topics: Properties of Bulk Matter
(130)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A dynamo converts
- A magnetic field of 1600 A/m produces a magnetic flux of 2.4 x 10⁻⁵ Wb
- A proton of mass m and charge q is moving in a plane with kinetic energy E. If there exists
- If Q, E and W denote respectively the heat added, change in internal energy
- Two cities are 150 km apart. Electric power is sent from one city to another city
Topics: Properties of Bulk Matter (130)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply