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JEE Advanced 2020Paper 1
Exam
JEE Advanced
Year
2020
Paper
Paper 1
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Advanced 2020 · Paper 1 · Physics · Thermal Properties of Matter

The filament of a light bulb has surface area 64 mm ^ 2 . The...

The filament of a light bulb has surface area 64mm2.64{mm}^{2} . The filament can be considered as a black body at temperature 2500K2500K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100m100m. Assume the pupil of the eyes of the observer to be circular with radius 3mm3mm. Then:

(Take Stefan-Boltzmann constant =5.67×10−8Wm−2K−4,= 5 . 67 \times {10}^{- 8}W{m}^{- 2}{K}^{- 4}, Wiens' displacement constant =2.90×10−3mK= 2 .90 \times {10}^{- 3}mK, Planck's constant =6.60×10−34Js= 6 .60 \times {10}^{- 34}Js, speed of light in vacuum =3.00×108ms−1= 3 .00 \times {10}^{8}m{s}^{- 1})

Options

  1. Apower radiated by the filament is in the range 642W642W to 645W645W
  2. Bradiated power entering into one eye of the observer is in the range 3.15×10−8W3 . 15 \times {10}^{- 8}W to 3.25×10−8W3 . 25 \times {10}^{- 8}W
  3. Cthe wavelength corresponding to the maximum intensity of light is 1160nm1160nm
  4. Dtaking the average wavelength of emitted radiation to be 1740nm,1740nm, the total number of photons entering per second into one eye of the observer is in the range 2.75×10112 . 75 \times {10}^{11} to 2.85×10112 . 85 \times {10}^{11}

Correct answer

B. radiated power entering into one eye of the observer is in the range 3.15×10−8W3 . 15 \times {10}^{- 8}W to 3.25×10−8W3 . 25 \times {10}^{- 8}W

Step-by-step solution

P=\sigmaAeT4P = \sigmaAe{T}^{4}

P=5.6×10−8×64×10−6×1×(2500)4P = 5 . 6 \times {10}^{- 8} \times 64 \times {10}^{- 6} \times 1 \times ( 2500{)}^{4}

P=14175×10−14×108×104P = 14175 \times {10}^{- 14} \times {10}^{8} \times {10}^{4}

(a) P=141.75WP = 141 . 75W

(b) \sigmaAeT44π(100)2×π(3×10−3)2=141.75×9×10−64×104\dfrac{\sigmaAe{T}^{4}}{4\pi ( 100{)}^{2}} \times \pi{\left(3 \times {10}^{- 3}\right)}^{2} = 141 . 75 \times \dfrac{9 \times {10}^{- 6}}{4 \times {10}^{4}}

318.937×10−10318 . 937 \times {10}^{- 10}

3.18937×108W3 . 18937 \times {10}^{8}W

(c) \lambdaT=b\lambdaT = b

λ=2.93×10−62500=1160nm\lambda = \dfrac{2 . 93 \times {10}^{- 6}}{2500} = 1160nm

(d) 3.18937×10−8=(nsec⁡)hcλ3 . 18937 \times {10}^{- 8} = \left(\dfrac{n}\sec\right)\dfrac{hc}{\lambda}

3.18937×10−8λ\lambdae=n=279.00×10−8×10−910−34×1083 . 18937 \times \dfrac{{10}^{- 8}\lambda}{\lambdae} = n = 279 . 00 \times \dfrac{{10}^{- 8} \times {10}^{- 9}}{{10}^{- 34} \times {10}^{8}}

n=279×10−17×1034×10−8n = 279 \times {10}^{- 17} \times {10}^{34} \times {10}^{- 8}

n=2.79×1011n = 2 . 79 \times {10}^{11}

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