Thermal Properties of Matter

JEE Main level, numerical answer, medium

115 of 126

Question 115 of 126

A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is 1727∘C1727^{\circ} \mathrm{C} and power radiated by the wire is 94.2 W. Its emissivity is x8\frac{x}{8} where x=x=_______
(Given σ=6.0×10−8 W m−2 K−4,π=3.14\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.)

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Answer: 5

L=10 cm, d=0.5 mm, T=1727∘C=2000 K\mathrm{L}=10 \mathrm{~cm}, \mathrm{~d}=0.5 \mathrm{~mm}, \mathrm{~T}=1727^{\circ} \mathrm{C}=2000 \mathrm{~K}
Power, P=94.2 W\mathrm{P}=94.2 \mathrm{~W}
P=εσAT4\mathrm{P}=\varepsilon \sigma \mathrm{AT}^4
94.2=ε×(6×10−8)(πdL)(2000)494.2=\varepsilon \times\left(6 \times 10^{-8}\right)(\pi \mathrm{dL})(2000)^4
94.2=ε×(6×10−8)(3.14)(0.5)(10−3)94.2=\varepsilon \times\left(6 \times 10^{-8}\right)(3.14)(0.5)\left(10^{-3}\right)
(10×10−2)(2000)4\left(10 \times 10^{-2}\right)(2000)^4
ε=94.2(94.2)(16)=58\varepsilon=\frac{94.2}{(94.2)(16)}=\frac{5}{8}

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