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The filament of an evacuated light bulb has a length 10 cm, diameter 0.2 mm and emissivity 0.2. Calculate the power it radiates at 2000K $(σ = 5.67 \times 10^{- 8} W / m^{2} K^{4})$

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15.5 W

21.5 W

8.9 W

11.4 W

answer is D.

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Detailed Solution

$E = σ A e T^{4}$

$E = σ (2 π r L) e T^{4} = 11.4 w a t t$

Where $σ = 5.6 \times 10^{- 8} w / m^{2} k^{4}$

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