JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Heat & Thermodynamics

A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is

172 7^{\circ}

C and power radiated by the wire is 94.2 W. Its emissivity is

\frac{X}{8}

where

x =

____ (Given

σ = 6.0 \times 1 0^{- 8}

W m

^{- 2}

^{- 4}

π = 3.14

and assume that the emissivity of wire material is same at all wavelength.)

Answer: 5

Solution

Power radiated from the curved surface follows the Stefan--Boltzmann law

P = ε σ A T^{4}

, with surface area

A = π d L

. Converting the data,

L = 0.1

d = 0.5 \times 1 0^{- 3}

m and

T = 172 7^{\circ} C = 2000

A = π d L = 3.14 \times (0.5 \times 1 0^{- 3}) \times 0.1 = 1.57 \times 1 0^{- 4} m^{2},

σ A T^{4} = (6.0 \times 1 0^{- 8}) (1.57 \times 1 0^{- 4}) (2000)^{4} = 150.72 W .

Hence

ε = \frac{P}{σ A T ^{4}} = \frac{94.2}{150.72} = \frac{5}{8} .

Comparing with

ε = \frac{x}{8}

, the value of

x

5

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Integer answer