COMEDK · Physics · 12. Thermal Properties of Matter
Two black bodies \(\mathrm{P}\) and \(\mathrm{Q}\) have equal surface areas and are kept at temperatures \(127^{\circ} \mathrm{C}\) and \(27^{\circ} \mathrm{C}\) respectively. The ratio of thermal power radiated by A to that by B is
- A \(256: 81\)
- B \(177: 127\)
- C \(127: 177\)
- D \(81: 256\)
Answer & Solution
Correct Answer
(A) \(256: 81\)
Step-by-step Solution
Detailed explanation
The thermal power radiated by a black body is given by the Stefan-Boltzmann law: \(P = \sigma A T^{4}\), where \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the surface area, and \(T\) is the absolute temperature in Kelvin. Given the temperatures in Celsius:…
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