A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is 1727 and power radiated by the wire is 94.2 W. Its emissivity is x/8 where x = _________. (Given σ = 6.0 × 10⁻⁸ W m⁻² K⁻⁴, π = 3.14 and assume that the emissivity of wire material is same at all wavelength.) Numerical answer type. Work it out, then open the answer.

Answer: 5. P = eσ A T⁴ with A = π d L and T = 1727 + 273 = 2000 K. A = 3.14 × 0.5 × 10⁻³ × 0.1 = 1.57 × 10⁻⁴ m² σ A T⁴ = 6 × 10⁻⁸ × 1.57 × 10⁻⁴ × 16 × 10¹² = 150.72 W e = (94.2)/(150.72) = 0.625 = 5/8 So x = 5.

A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The… | JEE Main 2025 (7 Apr, Shift 1)

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Thermal Radiation: Black Body, Stefan & Wien

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

Numerical answer type. Work it out, then open the answer.

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Answer: 5

$P = e\sigma A T^4$ with $A = \pi d L$ and $T = 1727 + 273 = 2000$ K.
$A = 3.14 \times 0.5 \times 10^{-3} \times 0.1 = 1.57 \times 10^{-4}\ \text{m}^2$
$\sigma A T^4 = 6 \times 10^{-8} \times 1.57 \times 10^{-4} \times 16 \times 10^{12} = 150.72$ W
$e = \dfrac{94.2}{150.72} = 0.625 = \dfrac{5}{8}$
So $x = 5$ .

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