MHT CET · Physics · Thermal Properties of Matter
A star (P) behaves like a perfectly black body emitting radiant energy at temperature ' \(\mathrm{T}\) '. Another star (Q) also behaves like perfectly black body emitting radiant energy at temperature ' \(\mathrm{T} / 4\) ' and has radius eight times the radius of star \((\mathrm{P})\). The ratio of radiant energy emitted by \((\mathrm{P})\) to that by \((\mathrm{Q})\) is
- A 1:8
- B 1:1
- C 4:1
- D 1:4
Answer & Solution
Correct Answer
(B) 1:1
Step-by-step Solution
Detailed explanation
Concept: According to the Stefan-Boltzmann law energy radiated per unit area (j)
\(\mathrm{j}=\sigma \mathrm{T}^4\)
Where \(\sigma\) is the proportionality constant and \(\mathrm{T}\) is temperature radiant energy \(E=j A\), where \(A\) is the surface black body.
Given,
For \((\mathrm{a})\) blackbody \(\mathrm{P}\), radius \(=\mathrm{R}\) and temperature \(=\mathrm{T}\)
(b) blackbody \(\mathrm{Q}\), radius \(=8 \mathrm{R}\) and temperature \(=\frac{\mathrm{T}}{4}\)
\(\therefore \frac{E_{\mathrm{P}}}{\mathrm{E}_{\mathrm{Q}}}=\frac{\sigma \mathrm{T}^4 4 \pi \mathrm{R}^2}{\sigma\left(\frac{\mathrm{T}}{4}\right)^4 4 \pi(8 \mathrm{R})^2}=\frac{4^4}{8^2}=\frac{(8)^2}{8^2}=1\)
\(\mathrm{j}=\sigma \mathrm{T}^4\)
Where \(\sigma\) is the proportionality constant and \(\mathrm{T}\) is temperature radiant energy \(E=j A\), where \(A\) is the surface black body.
Given,
For \((\mathrm{a})\) blackbody \(\mathrm{P}\), radius \(=\mathrm{R}\) and temperature \(=\mathrm{T}\)
(b) blackbody \(\mathrm{Q}\), radius \(=8 \mathrm{R}\) and temperature \(=\frac{\mathrm{T}}{4}\)
\(\therefore \frac{E_{\mathrm{P}}}{\mathrm{E}_{\mathrm{Q}}}=\frac{\sigma \mathrm{T}^4 4 \pi \mathrm{R}^2}{\sigma\left(\frac{\mathrm{T}}{4}\right)^4 4 \pi(8 \mathrm{R})^2}=\frac{4^4}{8^2}=\frac{(8)^2}{8^2}=1\)
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