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Consider an incandescent bulb having a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. The bulb is powered at constant voltage. JEE Adv Paper-1 of 2016

Question: Disprove that temperature distribution of filament is uniform.

Answer: Since the evaporation is non uniform, the area of filament is non uniform. Therefore the temperature at different places on the metal is different.

My question is what exactly does it mean for the area of a filament to be non uniform?

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  • $\begingroup$ You can easilly prove that a small non-uniformity will grow over time. $\endgroup$
    – fraxinus
    Commented Jul 16, 2021 at 10:16
  • $\begingroup$ How would you achieve that @fraxinus? $\endgroup$
    – Brian
    Commented Jul 17, 2021 at 3:34
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    $\begingroup$ If a part of the filament has higher resistance per length, it gets more power per length (all parts are connected in series so the current is equal for them). More power per length = higher temperature = more evaporation = even more resistance because after a while the evaporation gets you less tungsten per length. $\endgroup$
    – fraxinus
    Commented Jul 17, 2021 at 11:09

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Imagine a long cylinder of radius $r$ mm, it has a uniform surface area per mm, i.e

$$ 2 \pi r \times 1$$ in square mm

If we removed an approximately hemispherical shape, say of radius 0.2mm at a particular point, due to non uniform evaporation - then the surface area of the 1mm length that contained that point would increase.

Since the rest of the wire was unaffected the surface area of the filament becomes non uniform.

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