A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is 1727°C and power radiated by the wire is 94.2 W. Its emissivity is $ \frac{x}{8} $, where $ x = \ldots $
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Correct Answer: 5
Approach Solution - 1
- Length \( L = 0.1\, \text{m} \)
- Diameter \( d = 0.0005\, \text{m} \Rightarrow r = 0.00025\, \text{m} \)
- Temperature \( T = 1727^\circ C = 2000\, \text{K} \)
- Power radiated \( P = 94.2\, \text{W} \)
Approach Solution -2
Given values: \[ L = 10 \, \text{cm}, \, d = 0.5 \, \text{mm}, \, T = 1727^\circ \text{C} = 2000 \, \text{K}, \, \text{Power}, P = 94.2 \, \text{W} \] The formula for the power is: \[ P = \epsilon \sigma A T^4 \] Substituting the known values: \[ 94.2 = \epsilon \times (6 \times 10^{-8}) \times (3.14) \times (0.5) \times (10^{-3}) \times (2000)^4 \] Simplifying further: \[ 94.2 = \epsilon \times (6 \times 10^{-8}) \times (3.14) \times (0.5) \times (10^{-3}) \times (10 \times 10^{-2}) \times (2000)^4 \] Solving for \( \epsilon \): \[ \epsilon = \frac{94.2}{(94.2) \times (16)} = \frac{5}{8} \] \[ \boxed{\epsilon = \frac{5}{8}} \]
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