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QPages 98-102: Q20: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q20 Solution: A hole of radius is made centrally in a uniform circular disc…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the annular disc as a flat slab and use the plane‑wall formula $Q=KAdDelta T/L$ with $A=pi(r_2^2-r_1^2)d$ and $L=r_2-r_1$. In radial conduction the cross‑sectional area changes with radius, so the correct differential area is $2pi r d$, which after integration gives $Q=2pi K d( heta_1- heta_2)/ln(r_2/r_1)$. Using the wrong area or length leads to an incorrect heat‑flow expression.

A hole of radius r1r_1 is made centrally in a uniform circular disc of thickness d and radius r2r_2. The inner surface a cylinder of length d and radius r1r_1) is maintained at a temperature θ1\theta_1 and the outer surface a cylinder of length d and radius r2r_2) is maintained at a temperature θ2\theta_2 θ1>θ2\theta_1 > \theta_2). The thermal conductivity of the material of the disc is K. Calculate the heat flowing per unit time through the disc.

Step-by-Step Verified Solution

**Answer:** Heat flow = 2�Kd(_____)/ln(r_/r_). **Step-by-Step Explanation:** For a cylindrical shell of inner radius r_, outer radius r_, length d:
Qt=2πKd(θ1−θ2)ln⁡(r2/r1)\frac{Q}{t} = \frac{2\pi K d (\theta_1 - \theta_2)}{\ln(r_2/r_1)}

All Questions in this Chapter

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