A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is $1727^{\circ} \mathrm{C}$ and power radiated by the wire is 94.2 W . Its emissivity is $\frac{x}{8}$ where $x=$ _________.

(Given $\sigma=6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}, \pi=3.14$ and assume that the emissivity of wire material is same at all wavelength.)

<p>To determine the emissivity ($\varepsilon$) of the wire used in the bulb, we start with the given parameters and the formula for power radiated by a black body:</p> <p><p>Length of the wire ($L$): 10 cm = 0.1 m</p></p> <p><p>Diameter of the wire ($d$): 0.5 mm = 0.0005 m</p></p> <p><p>Temperature of the wire ($T$): $1727^{\circ} \mathrm{C} = 2000 \mathrm{K}$</p></p> <p><p>Power radiated ($P$): 94.2 W</p></p> <p>The power radiated by the wire can be expressed as:</p> <p>$ P = \varepsilon \sigma A T^4 $</p> <p>Where:</p> <p><p>$\sigma$ is the Stefan-Boltzmann constant ($6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}$)</p></p> <p><p>$A$ is the surface area of the wire, calculated using the formula for the surface area of a cylinder: $A = \pi d L$</p></p> <p>Substituting the given values into the formula:</p> <p>$ 94.2 = \varepsilon \times (6 \times 10^{-8}) \times (3.14 \times 0.0005 \times 0.1) \times (2000)^4 $</p> <p>Simplifying the expression inside the parentheses:</p> <p>$ 94.2 = \varepsilon \times (6 \times 10^{-8}) \times 3.14 \times 0.0005 \times 0.1 \times (2000)^4 $</p> <p>On solving for $\varepsilon$:</p> <p>$ \varepsilon = \frac{94.2}{[(6 \times 10^{-8}) \times 3.14 \times 0.0005 \times 0.1 \times (2000)^4]} $</p> <p>Given that $\varepsilon$ is $\frac{x}{8}$, evaluating this expression provides the value of $x = 5$. Thus, the emissivity $\varepsilon$ is $\frac{5}{8}$.</p>