In a dark room with ambient temperature , a black body is kept at a temperature T. Keeping the temperature of the black body constant (at T), sunrays are allowed to fall on the black body through a hole in the roof of the dark room. Assuming that there is no change in the ambient temperature of the room, which of the following statement(s) is/are correct?
- (A)The quantity of radiation absorbed by the black body in unit time will increase.✓ Correct
- (B)Since emissivity = absorptivity, hence the quantity of radiation emitted by black body in unit time will increase.✓ Correct
- (C)Black body radiates more energy in unit time in the visible spectrum.✓ Correct
- (D)The reflected energy in unit time by the black body remains same.✓ Correct
Answer: (A), (B), (C), (D)
Solution
When sunrays fall on the black body, it absorbs additional radiation from the sun, so the quantity of radiation absorbed increases (option A correct). By Kirchhoff's law, for a black body, emissivity equals absorptivity. Since the black body's temperature T is maintained constant and it now absorbs more radiation, it must emit more to maintain thermal equilibrium, so emission increases (option B correct). With increased emission at the same temperature T, more energy is radiated across all wavelengths including the visible spectrum (option C correct). A black body by definition has zero reflectivity, so reflected energy remains zero (same) regardless of incident radiation (option D correct).
JEE Advanced · 2006 · 09 Apr 2006 · One or more correct options