MHT CET · Physics · Thermal Properties of Matter
A black sphere has radius \(R\) whose rate of radiation is E at temperature T . If radius is made half and temperature 4 T , the rate of radiation will be
- A 64 E
- B 32 E
- C 16 E
- D 8 E
Answer & Solution
Correct Answer
(A) 64 E
Step-by-step Solution
Detailed explanation
\(\mathrm{E}=\mathrm{e} \mathrm{~A} \sigma \mathrm{~T}^4\)
But for sphere, \(A=4 \pi R^2\)
\(\begin{array}{ll}
\therefore & \quad \mathrm{E}=\mathrm{e}\left(4 \pi \mathrm{R}^2\right) \sigma \mathrm{T}^4 \\
\therefore & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{R}_1^2 \mathrm{~T}_1^4}{\mathrm{R}_2^2 \mathrm{~T}_2^4} \\
\therefore & \frac{\mathrm{E}}{\mathrm{E}_2}=\frac{\mathrm{R}^2 \mathrm{~T}^4}{\left(\frac{\mathrm{R}}{2}\right)^2(4 \mathrm{~T})^4} \\
\therefore & \mathrm{E}_2=64 \mathrm{E}
\end{array}\)
But for sphere, \(A=4 \pi R^2\)
\(\begin{array}{ll}
\therefore & \quad \mathrm{E}=\mathrm{e}\left(4 \pi \mathrm{R}^2\right) \sigma \mathrm{T}^4 \\
\therefore & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{R}_1^2 \mathrm{~T}_1^4}{\mathrm{R}_2^2 \mathrm{~T}_2^4} \\
\therefore & \frac{\mathrm{E}}{\mathrm{E}_2}=\frac{\mathrm{R}^2 \mathrm{~T}^4}{\left(\frac{\mathrm{R}}{2}\right)^2(4 \mathrm{~T})^4} \\
\therefore & \mathrm{E}_2=64 \mathrm{E}
\end{array}\)
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