Question:
Two spherical black bodies A and B made of the same material having masses 80 kg and 10 kg are maintained at temperatures $27^\circ C$ and $327^\circ C$ respectively. If P is the power radiated by body B, then the power radiated by body A is
Two spherical black bodies A and B made of the same material having masses 80 kg and 10 kg are maintained at temperatures $27^\circ C$ and $327^\circ C$ respectively. If P is the power radiated by body B, then the power radiated by body A is
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Radiated power:
\[
P\propto AT^4
\]
Always convert temperatures into Kelvin before applying Stefan's law.
Updated On: Jun 17, 2026
- $4P$
- $\frac{P}{4}$
- $16P$
- $\frac{P}{16}$
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The Correct Option is A
Solution and Explanation
Concept:
According to Stefan's law,
\[
P=e\sigma AT^4
\]
For spheres made of the same material,
\[
m\propto r^3
\]
and
\[
A\propto r^2
\]
Step 1: Find ratio of surface areas.
\[ \frac{m_A}{m_B} = \frac{80}{10} = 8 \] Hence, \[ \frac{r_A}{r_B} = \sqrt[3]{8} = 2 \] Therefore, \[ \frac{A_A}{A_B} = \left(\frac{r_A}{r_B}\right)^2 = 4 \]
Step 2: Convert temperatures into Kelvin.
\[ T_A=27+273=300K \] \[ T_B=327+273=600K \] Thus, \[ \left(\frac{T_A}{T_B}\right)^4 = \left(\frac{300}{600}\right)^4 = \frac1{16} \]
Step 3: Calculate power ratio.
\[ \frac{P_A}{P_B} = \frac{A_A}{A_B} \left(\frac{T_A}{T_B}\right)^4 \] \[ = 4\times\frac1{16} = \frac14 \] Therefore, \[ P_A=\frac{P}{4} \] \[ \boxed{\frac{P}{4}} \]
Step 1: Find ratio of surface areas.
\[ \frac{m_A}{m_B} = \frac{80}{10} = 8 \] Hence, \[ \frac{r_A}{r_B} = \sqrt[3]{8} = 2 \] Therefore, \[ \frac{A_A}{A_B} = \left(\frac{r_A}{r_B}\right)^2 = 4 \]
Step 2: Convert temperatures into Kelvin.
\[ T_A=27+273=300K \] \[ T_B=327+273=600K \] Thus, \[ \left(\frac{T_A}{T_B}\right)^4 = \left(\frac{300}{600}\right)^4 = \frac1{16} \]
Step 3: Calculate power ratio.
\[ \frac{P_A}{P_B} = \frac{A_A}{A_B} \left(\frac{T_A}{T_B}\right)^4 \] \[ = 4\times\frac1{16} = \frac14 \] Therefore, \[ P_A=\frac{P}{4} \] \[ \boxed{\frac{P}{4}} \]
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