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A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius is halved and the temperature doubled, the power radiated in watt would be
  • a)
    225
  • b)
    450
  • c)
    1000
  • d)
    1800
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A spherical black body with a radius of 12 cm radiates 450 watt power...
Solution:
Given,
Radius of the black body, r1 = 12 cm
Power radiated, P1 = 450 W
Temperature, T1 = 500 K
New radius, r2 = r1/2 = 6 cm
New temperature, T2 = 2T1 = 1000 K
We need to find the power radiated, P2

Using Stefan-Boltzmann law, the power radiated by a black body is given by:
P = εσAT^4
where ε is the emissivity of the black body (which is 1 for a perfect black body), σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4), A is the surface area of the black body, and T is the absolute temperature in Kelvin.

1. Surface area of the black body
The surface area of the black body is proportional to the square of the radius. Therefore, we have:
A1/A2 = (r1/r2)^2
A1/A2 = (12/6)^2 = 4
A2 = A1/4
A2 = 4πr2^2
A2 = 4π(6)^2
A2 = 452.39 cm^2

2. Power radiated at temperature T1
Using the given values of r1, T1, and A1, we can find the power radiated:
P1 = εσA1T1^4
P1 = (1)(5.67 x 10^-8)(4π(12)^2)(500)^4
P1 = 449.88 W (approx)

3. Power radiated at temperature T2
Using the new values of r2, T2, and A2, we can find the power radiated:
P2 = εσA2T2^4
P2 = (1)(5.67 x 10^-8)(452.39)(1000)^4
P2 = 1802.9 W (approx)

Therefore, the power radiated by the black body when its radius is halved and temperature is doubled is 1800 W (option D).
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Community Answer
A spherical black body with a radius of 12 cm radiates 450 watt power...
∴ q ∝ R2T4
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