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Heat and Thermodynamics question

2020 · Shift 1 · Q49
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  5. /2020 · Shift 1 · Q49

Heat and Thermodynamics question

2020 · Shift 1 · Q49

JEE AdvancedPhysicsHeat and ThermodynamicsMultiple correct+4 / −2
The filament of a light bulb has surface area 64 mm2 . The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then (Take Stefan-Boltzmann constant = 5.67 ×\times× 10−8 Wm−2K−4 , Wien’s displacement constant = 2.90 ×\times× 10−3 m-K, Planck’s constant = 6.63 ×\times× 10−34 Js, speed of light in vacuum = 3.00 ×\times× 108 ms−1)
  1. A
    power radiated by the filament is in the range 642 W to 645 W
  2. B
    radiated power entering into one eye of the observer is in the range 3.15 ×\times× 10−8 W to 3.25 ×\times× 10−8 W
  3. C
    the wavelength corresponding to the maximum intensity of light is 1160 nm
  4. D
    taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 ×\times× 1011 to 2.85 ×\times× 1011
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