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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?
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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).
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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A spherical black body with a radius of 12 cm radiates 450 watt power...
∴ q ∝ R2T4
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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 bea)225b)450c)1000d)1800Correct answer is option 'D'. Can you explain this answer?
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