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Class 11PhysicsChapter-11

Thermodynamics & KTG

Question 1 of 1084

1HardJEE Advanced2020

The filament of a light bulb has surface area 64 mm 2 . 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 −2 K −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 10 8 ms −1 )

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