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JEE Advanced Physics Thermal Properties of Matter 2020 JEE Advanced 2020 (Paper 1)

JEE Advanced Physics Question (2020) — Solution

Question

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 × 10 - 8   W   m - 2   K - 4 , Wiens' displacement constant = 2 .90 × 10 − 3   m   K , Planck's constant = 6 .60 × 10 − 34   J   s , speed of light in vacuum = 3 .00 × 10 8   m   s − 1 )

Options

  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 × 10 - 8   W to  3 . 25 × 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 × 10 11 to 2 . 8

Answer

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 × 10 11 to 2 . 8

Step-by-step solution

P = σ A e T 4 P = 5 . 6 × 10 - 8 × 64 × 10 - 6 × 1 × ( 2500 ) 4 P = 14175 × 10 - 14 × 10 8 × 10 4 (a)  P = 141 . 75   W (b)  σ A e T 4 4 π ( 100 ) 2 × π 3 × 10 - 3 2 = 141 . 75 × 9 × 10 - 6 4 × 10 4 318 . 937 × 10 - 10 3 . 18937 × 10 8   W (c)  λ T = b λ = 2 . 93 × 10 - 6 2500 = 1160   nm (d)  3 . 18937 × 10 - 8 = n sec h c λ 3 . 18937 × 10 - 8 λ λ e = n = 279 . 00 × 10 - 8 × 10 - 9 10 - 34 × 10 8 n = 279 × 10 - 17 × 10 34 × 10 - 8 n = 2 . 79 × 10 11

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Related: Physics — Thermal Properties of Matter · All PYQ Banks