MHT CET · Physics · Thermal Properties of Matter
A black sphere has radius ' \(R\) ' whose rate of radiation is ' \(\mathrm{E}\) ' at temperature ' \(\mathrm{T}\) '. If radius is made \(R / 3\) and temperature ' \(3 \mathrm{~T}\) ', the rate of radiation will be
- A \(\mathrm{E}\)
- B \(3 \mathrm{E}\)
- C \(6 \mathrm{E}\)
- D \(9 \mathrm{E}\)
Answer & Solution
Correct Answer
(D) \(9 \mathrm{E}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \mathrm{E}=\mathrm{eA} \sigma \mathrm{T}^4 \\ & \text { But for sphere, } A=4 \pi \mathrm{R}^2 \\ \therefore \quad & \mathrm{E}=\mathrm{e}\left(4 \pi \mathrm{R}^2\right) \sigma \mathrm{T}^4 \\ \therefore \quad & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{R}_1^2 \mathrm{~T}_1^4}{\mathrm{R}_2^2 \mathrm{~T}_2^4} \\ \therefore \quad & \frac{\mathrm{E}}{\mathrm{E}_2}=\frac{\mathrm{R}^2 \mathrm{~T}^4}{\left(\frac{\mathrm{R}}{3}\right)^2(3 \mathrm{~T})^4} \\ \therefore \quad & \mathrm{E}_2=9 \mathrm{E}\end{array}\)
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