A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is 1727 degrees C and power radiated by the wire is 94.2 W. Its emissivity is $$\frac{x}{8}$$ where x = ______.
(Given $$\sigma = 6.0 \times 10^{-8}$$ W m$$^{-2}$$ K$$^{-4}$$, $$\pi = 3.14$$ and assume that the emissivity of wire material is same at all wavelength. )

The power radiated from a hot surface is given by the Stefan-Boltzmann law
$$P = \varepsilon \,\sigma \,A \,T^{4}$$
where $$\varepsilon$$ is the emissivity, $$\sigma$$ is the Stefan-Boltzmann constant, $$A$$ is the surface area and $$T$$ is the absolute temperature.

Given data:
Length of wire, $$L = 10 \,\text{cm} = 0.10 \,\text{m}$$
Diameter of wire, $$d = 0.5 \,\text{mm} = 0.0005 \,\text{m}$$
Temperature, $$T = 1727^{\circ}\text{C} = 1727 + 273 = 2000 \,\text{K}$$
Power radiated, $$P = 94.2 \,\text{W}$$
Stefan-Boltzmann constant, $$\sigma = 6.0 \times 10^{-8} \,\text{W m}^{-2}\text{K}^{-4}$$

For a cylindrical wire, surface area (neglecting the ends)
$$A = \pi \,d\,L$$
Substituting the values,
$$A = 3.14 \times 0.0005 \,\text{m} \times 0.10 \,\text{m}$$
$$A = 3.14 \times 5.0 \times 10^{-5} \,\text{m}^{2}$$
$$A = 1.57 \times 10^{-4} \,\text{m}^{2}$$

Fourth power of temperature:
$$T^{4} = (2000\,\text{K})^{4} = 2^{4} \times 10^{12} = 16 \times 10^{12} = 1.6 \times 10^{13}$$

Put these numbers into the Stefan-Boltzmann law and solve for $$\varepsilon$$:
$$\varepsilon = \frac{P}{\sigma\,A\,T^{4}}$$
$$\varepsilon = \frac{94.2}{\left(6.0 \times 10^{-8}\right)\!\left(1.57 \times 10^{-4}\right)\!\left(1.6 \times 10^{13}\right)}$$

First multiply the constants in the denominator:
$$6.0 \times 1.57 = 9.42$$
$$9.42 \times 1.6 = 15.072$$
Power of ten: $$10^{-8}\times 10^{-4}\times 10^{13}=10^{1}$$
Hence,
$$\sigma A T^{4} = 15.072 \times 10^{1} = 150.72$$

Therefore,
$$\varepsilon = \frac{94.2}{150.72} = 0.625$$

Expressing $$\varepsilon$$ as a fraction:
$$0.625 = \frac{5}{8}$$

The problem states $$\varepsilon = \frac{x}{8}$$, so
$$x = 5$$

Final answer: $$x = 5$$