WBJEE · Physics · Thermal Properties of Matter
Two black bodies \(A\) and \(B\) have equal surface areas and are maintained at temperatures \(27^{\circ} \mathrm{C}\) and \(177^{\circ} \mathrm{C}\) respectively. What will be the ratio of the thermal energy radiated per second by \(A\) to that by \(B ?\)
- A \(4: 9\)
- B \(2: 3\)
- C \(16: 81\)
- D \(27:177\)
Answer & Solution
Correct Answer
(C) \(16: 81\)
Step-by-step Solution
Detailed explanation
According to the question, Area of both bodies \(A\) and \(B=A\) Temperature of body \(A=27^{\circ} \mathrm{C}=27+273 \mathrm{K}\) Temperature of body \(B=177^{\circ} \mathrm{C}=177+273 \mathrm{K}\) Now, by Stefan-Boltzmann law, thermal energy radiated per second by a body…
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