| ⇦ | 
| ⇨ |
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 body of mass 0.5 kg travels in a straight line with velocity v=a x³/², where
- A body is hangfing from a rigid support by an inextensible string of length
- If a small amount of antimony is added to germanium crystal
- Which of the following parameters does not characterize the thermodynamic state
- When a radioactive substance emits an ∝ particle, its position in the periodic table
Topics: Properties of Bulk Matter
(130)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A body of mass 0.5 kg travels in a straight line with velocity v=a x³/², where
- A body is hangfing from a rigid support by an inextensible string of length
- If a small amount of antimony is added to germanium crystal
- Which of the following parameters does not characterize the thermodynamic state
- When a radioactive substance emits an ∝ particle, its position in the periodic table
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