| ⇦ |
| ⇨ |
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: - Nuclear fission is not found in
- X-ray beam can be deflected by
- A particle executing SHM with amplitude of 0.1 m. At a certain instant,
- The height at which the weight of a body becomes 1/16 th, its weight
- Two similar springs P and Q have spring constants Kp and Kq, such that Kp>Kq.
Topics: Properties of Bulk Matter
(130)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Nuclear fission is not found in
- X-ray beam can be deflected by
- A particle executing SHM with amplitude of 0.1 m. At a certain instant,
- The height at which the weight of a body becomes 1/16 th, its weight
- Two similar springs P and Q have spring constants Kp and Kq, such that Kp>Kq.
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