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Saturday, 31 October

JEE Main 2025 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 7 April 2025, Shift 1, Physics Q47

Question 47 of 75 in this shift, Physics question 22 of 25, Section B.

A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is 1727∘1727^\circC and power radiated by the wire is 94.2 W. Its emissivity is x8\frac{x}{8} where xx = ___________. (Given σ=6.0×10−8 W m−2 K−4\sigma = 6.0 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}, π=3.14\pi = 3.14 and assume that the emissivity of wire material is same at all wavelength.)

Official answer

5

NTA final key.

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.