A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is 172 7 ∘ C and power radiated by the wire is 94.2 W. Its emissivity is 8 x where x = _______ (Given σ = 6.0 × 1 0 − 8 W m − 2 K − 4 , π = 3.14 and assume that the emissivity of wire material is same at all wavelength.)

L = 10 cm , d = 0.5 mm , T = 172 7 ∘ C = 2000 K Power, P = 94.2 W P = ε σ AT 4 94.2 = ε × ( 6 × 1 0 − 8 ) ( π dL ) ( 2000 ) 4 94.2 = ε × ( 6 × 1 0 − 8 ) ( 3.14 ) ( 0.5 ) ( 1 0 − 3 ) ( 10 × 1 0 − 2 ) ( 2000 ) 4 ε = ( 94.2 ) ( 16 ) 94.2 = 8 5

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PhysicsHeat & ThermoRadiationEasy2 minPYQ_2025

PhysicsEasynumerical

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 $σ = 6.0 \times 1 0^{- 8} W m^{- 2} K^{- 4}, π = 3.14$ and assume that the emissivity of wire material is same at all wavelength.)

Answer:

5.00

Solution:

$L = 10 cm, d = 0.5 mm, T = 172 7^{\circ} C = 2000 K$
Power, $P = 94.2 W$
$P = ε σ AT^{4}$
$94.2 = ε \times (6 \times 1 0^{- 8}) (π dL) (2000)^{4}$
$94.2 = ε \times (6 \times 1 0^{- 8}) (3.14) (0.5) (1 0^{- 3})$
$(10 \times 1 0^{- 2}) (2000)^{4}$
$ε = \frac{94.2}{( 94.2 ) ( 16 )} = \frac{5}{8}$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Radiation

⏱ 2mℹ️ Source: PYQ_2025

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