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QPages 98-102: Q27: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q27 Solution: Suppose the bent part of the frame of the previous problem has…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often assume the bent and straight sections have identical cross‑sectional area and length, so they set $Q_{ ext{bent}}/Q_{ ext{straight}}=k_{ ext{bent}}/k_{ ext{straight}}$. In reality $Q=kADelta T/L$; if $A$ or $L$ differ, the ratio is not just the conductivity ratio. Always verify the geometry before simplifying.

Suppose the bent part of the frame of the previous problem has a thermal conductivity of 780 J s−1 m−1 �C−1780 \text{ J s}^{-1} \text{ m}^{-1} \text{ �C}^{-1} whereas it is 390 J s−1 m−1 �C−1390 \text{ J s}^{-1} \text{ m}^{-1} \text{ �C}^{-1} for the straight part. Calculate the ratio of the rate of heat flow through the bent part to the rate of heat flow through the straight part.

Step-by-Step Verified Solution

**Answer:** Ratio = 2:1. **Step-by-Step Explanation:** Given: K_bent = 780, K_straight = 390. For same geometry: Q_bent/Q_straight = K_bent/K_straight = 780/390 = 2.

All Questions in this Chapter

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