Question

Two perfectly black spheres A and B having radii 8 cm and 2 cm are maintained at temperatures \(127^\circ C\) and \(527^\circ C\) respectively. The ratio of the energy radiated by A to that by B is

• A. 1 : 2
• B. 1 : 1
• C. 2 : 1
• D. 1 : 4
• E. 1 : 16

Hint:

Radiation power depends strongly on temperature (\(T^4\)), so always convert to Kelvin.

Answer 1

The Correct Option is B

Concept: According to Stefan–Boltzmann law: \[ P = \sigma A T^4 \] Energy radiated per unit time depends on surface area and absolute temperature.

Step 1: Convert temperatures to Kelvin. \[ T_A = 127 + 273 = 400K \] \[ T_B = 527 + 273 = 800K \]

Step 2: Surface areas. \[ A \propto R^2 \Rightarrow A_A : A_B = 8^2 : 2^2 = 64 : 4 = 16 : 1 \]

Step 3: Apply Stefan law. \[ \frac{P_A}{P_B} = \frac{A_A T_A^4}{A_B T_B^4} \]

Step 4: Substitute values. \[ = \frac{16 \times (400)^4}{1 \times (800)^4} \]

Step 5: Simplify temperature ratio. \[ \left(\frac{400}{800}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]

Step 6: Final ratio. \[ \frac{P_A}{P_B} = 16 \times \frac{1}{16} = 1 \]

Step 7: Conclusion. \[ \boxed{1:1} \]