| ⇦ | 
| ⇨ |
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: - Three unequal resistors in parallel are equivalent to a resistance 1 Ω.
- A point source emits sound equally in all directions in a non-absorbing medium.
- A particle executes simple harmonic oscillations with an amplitude
- The acceleration of an electron in an electric field of magnitude 50 V/cm,
- For the radioactive nuclei that undergo either α or β decay, which one
Topics: Properties of Bulk Matter
(130)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Three unequal resistors in parallel are equivalent to a resistance 1 Ω.
- A point source emits sound equally in all directions in a non-absorbing medium.
- A particle executes simple harmonic oscillations with an amplitude
- The acceleration of an electron in an electric field of magnitude 50 V/cm,
- For the radioactive nuclei that undergo either α or β decay, which one
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