Question:
A spherical perfect black body of radius 10 cm is maintained at \( 727^{\circ}C \). The total power radiated from it is (approximately) Stefan-Boltzmann constant, \( \sigma=5.67\times10^{-8} \, Wm^{-2}K^{-4} \)
A spherical perfect black body of radius 10 cm is maintained at \( 727^{\circ}C \). The total power radiated from it is (approximately) Stefan-Boltzmann constant, \( \sigma=5.67\times10^{-8} \, Wm^{-2}K^{-4} \)
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In radiation problems, always combine powers of 10 first (like \( (10^3)^4 = 10^{12} \)). It reduces calculation errors and saves time in exams.
Updated On: Jun 8, 2026
- \( 7120 \, W \)
- \( 7270 \, W \)
- \( 1000 \, W \)
- \( 7000 \, W \)
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The Correct Option is A
Solution and Explanation
Concept:
According to Stefan–Boltzmann law, power radiated by a black body is:
\[
P = \sigma A T^4
\]
where \(A = 4\pi R^2\) and \(T\) is in Kelvin.
Step 1: Convert given quantities to SI units.
• Temperature: \[ T = 727^\circ C + 273 = 1000 \, K \]
• Radius: \[ R = 10 \, cm = 0.1 \, m \]
• Surface area: \[ A = 4\pi R^2 = 4\pi (0.1)^2 = 0.04\pi \, m^2 \]
Step 2: Apply Stefan–Boltzmann law.
\[ P = 5.67 \times 10^{-8} \times 0.04\pi \times (10^3)^4 \] Since: \[ (10^3)^4 = 10^{12} \] \[ P = 5.67 \times 0.04\pi \times 10^4 \] Using \( \pi \approx 3.14 \): \[ 0.04\pi \approx 0.1256 \] \[ P = 5.67 \times 0.1256 \times 10^4 \] \[ = 0.712 \times 10^4 \] \[ = 7120 \, W \]
Step 3: Final answer.
\[ \boxed{7120 \, W} \]
Step 1: Convert given quantities to SI units.
• Temperature: \[ T = 727^\circ C + 273 = 1000 \, K \]
• Radius: \[ R = 10 \, cm = 0.1 \, m \]
• Surface area: \[ A = 4\pi R^2 = 4\pi (0.1)^2 = 0.04\pi \, m^2 \]
Step 2: Apply Stefan–Boltzmann law.
\[ P = 5.67 \times 10^{-8} \times 0.04\pi \times (10^3)^4 \] Since: \[ (10^3)^4 = 10^{12} \] \[ P = 5.67 \times 0.04\pi \times 10^4 \] Using \( \pi \approx 3.14 \): \[ 0.04\pi \approx 0.1256 \] \[ P = 5.67 \times 0.1256 \times 10^4 \] \[ = 0.712 \times 10^4 \] \[ = 7120 \, W \]
Step 3: Final answer.
\[ \boxed{7120 \, W} \]
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