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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: - The potential energy of a simple harmonic oscillator when the particle is half way
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Topics: Properties of Bulk Matter
(130)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The potential energy of a simple harmonic oscillator when the particle is half way
- The potential energy of a 1 kg particle free to move along the x-axis is given by
- Two bodies A and B having temperature 327° C and 427° C are radiating heat
- The refracting angle of a prism is A and refractive index of the material of the prism
- The electromagnetic waves detected using a thermopile and used in physical therapy
Topics: Properties of Bulk Matter (130)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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