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An evacuated bulb contains a filament that is 10 cm long and 0.2 mm in diameter. If its emissivity is 0.2, determine the radiant power emitted by the filament at 2000 K. Take σ = 5.67 × 10⁻⁸ W/m² K⁴.

1. 21.5 W
2. 15.5 W
3. 8.9 W
4. 11.4 W

Correct answer: 11.4 W

Solution

Area = 2*pi*r*L = 2*pi*(1e-4)(0.1) = 6.28e-5 m^2. P = 0.2 * 5.67e-8 * 6.28e-5 * (2000)^4 = 0.2 * 5.67e-8 * 6.28e-5 * 1.6e13 ~ 11.4 W.

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