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QPages 98-102: Q42: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q42 Solution: A 100 W bulb has tungsten filament of total length 1.0 m and…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently write the filament’s area as $A=2rl$ instead of the correct $A=2\pi rl$, forgetting the $pi$ factor; this underestimates the radiating area and inflates the calculated temperature. Always include $pi$ when using the lateral surface area formula for a cylinder.

A 100 W bulb has tungsten filament of total length 1.0 m and radius 4×10−5 m4 \times 10^{-5} \text{ m}. The emissivity of the filament is 0.8 and σ=6.0×10−8 W m−2 K−4\sigma = 6.0 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}. Calculate the temperature of the filament when the bulb is operating at correct wattage.

Step-by-Step Verified Solution

**Answer:** T = 2200 K. **Step-BY-Step Explanation:** Given: P = 100 W, l = 1.0 m, r = 4 _ 10__ m, e = 0.8, _ = 6.0 _ 10__. Surface area = 2�rl = 2� _ 4 _ 10__ _ 1 = 2.51 _ 10__ m_.
100=0.8×6.0×10−8×2.51×10−4×T4⇒T=2200 K100 = 0.8 \times 6.0 \times 10^{-8} \times 2.51 \times 10^{-4} \times T^4 \Rightarrow T = 2200 \text{ K}

All Questions in this Chapter

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