| ⇦ |
| ⇨ |
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: - In the diffraction pattern of a single slit
- The distance travelled by a particle starting from rest and moving with an acceleration
- The radius of a nucleus of mass number A is directly proportional to
- Two point objects of masses 1.5 g and 2.5 g respectively are at a distance of 16 cm
- A source of sound of frequency 256 Hz is moving rapidly towards a wall with a velocity
Topics: Properties of Bulk Matter
(130)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In the diffraction pattern of a single slit
- The distance travelled by a particle starting from rest and moving with an acceleration
- The radius of a nucleus of mass number A is directly proportional to
- Two point objects of masses 1.5 g and 2.5 g respectively are at a distance of 16 cm
- A source of sound of frequency 256 Hz is moving rapidly towards a wall with a velocity
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