A wire of length $10\ \text{cm}$ and diameter $0.5\ \text{mm}$ is used in a bulb. The temperature of the wire is $1727^\circ\text{C}$ and the power radiated by the wire is $94.2\ \text{W}$. Its emissivity is $\dfrac x8$, where $x$ = ______. (Given $\sigma = 6.0\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$, $\pi = 3.14$, and assume that the emissivity of the wire material is the same at all wavelengths.)
Numerical value type. Enter your answer.
Answer: 5
$T = 2000\ \text{K}$. The radiating surface is the curved surface of the wire:
$$A = \pi dL = 3.14\times0.5\times10^{-3}\times0.1 = 1.57\times10^{-4}\ \text{m}^2$$
$$P = e\sigma AT^4 \Rightarrow 94.2 = e\times6\times10^{-8}\times1.57\times10^{-4}\times16\times10^{12} = e\times150.72$$
$$e = 0.625 = \frac58 \Rightarrow x = 5$$
Solution by Sreeraj P, M.Sc Physics