L=10 cm, d=0.5 mm, T=1727∘C=2000 K
Power, P=94.2 W
P=εσAT4
94.2=ε×(6×10−8)(πdL)(2000)4
94.2=ε×(6×10−8)(3.14)(0.5)(10−3)
(10×10−2)(2000)4
ε=(94.2)(16)94.2=85
JEE Main 2025 — Physics Thermodynamics
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 8x 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.)
Held on 7 Apr 2025 · Verified 6 Jul 2026.
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