MHT CET · Physics · Thermal Properties of Matter
Rate of radiation by a black body is 'R' at temperature 'T'. Another body has same area but emissivity is 0.2 and temperature 3 T . Its rate of radiation is
- A \((162) \mathrm{R}\)
- B \((81) \mathrm{R}\)
- C \((16.2) \mathrm{R}\)
- D \((8.1) \mathrm{R}\)
Answer & Solution
Correct Answer
(C) \((16.2) \mathrm{R}\)
Step-by-step Solution
Detailed explanation
Rate of radiation for blackbody,
\(\mathrm{R}=\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_1=\mathrm{eA} \sigma \mathrm{~T}^4=\mathrm{A} \sigma \mathrm{~T}^4 \quad \ldots(\because \mathrm{e}=1)\)
For another body,
\(\begin{aligned}
\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_2 & =\mathrm{e}^{\prime} \mathrm{A}^{\prime} \sigma\left(\mathrm{T}^{\prime}\right)^4 \\
& =0.2 \mathrm{~A} \sigma(3 \mathrm{~T})^4 \\
& =16.2 \mathrm{~A} \sigma \mathrm{~T}^4 \\
& =16.2\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_1=16. 2 \mathrm{R}
\end{aligned}\)
\(\mathrm{R}=\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_1=\mathrm{eA} \sigma \mathrm{~T}^4=\mathrm{A} \sigma \mathrm{~T}^4 \quad \ldots(\because \mathrm{e}=1)\)
For another body,
\(\begin{aligned}
\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_2 & =\mathrm{e}^{\prime} \mathrm{A}^{\prime} \sigma\left(\mathrm{T}^{\prime}\right)^4 \\
& =0.2 \mathrm{~A} \sigma(3 \mathrm{~T})^4 \\
& =16.2 \mathrm{~A} \sigma \mathrm{~T}^4 \\
& =16.2\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_1=16. 2 \mathrm{R}
\end{aligned}\)
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