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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]
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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
- A body of mass 3 kg acted upon by a constant force is displaced by S meter
- A car moves from X to Y with a uniform speed v(u) and returns to Y with a uniform speed
- A body travelling along a straight line traversed one-third of the total distance
- In a P-N junction
- Two bodies of masses m₁ and m₂ and having velocities v₁ and v₂ respectively
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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