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A red star having radius rR at a temperature TR and a white star having radius rW at a temperature TW , radiate the same total power. If these stars radiate as perfect black bodies, then
- a)rR > rW and TR > TW
- b)rR < rW and TR > TW
- c)rR > rW and TR < TW
- d)rR < rW and TR < TW
Correct answer is option 'C'. Can you explain this answer?
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A red star having radius rR at a temperature TR and a white star havin...
We can use the Stefan-Boltzmann law to relate the radiated power to the temperature and radius of the stars:
P = σA(T⁴)
where P is the power, σ is the Stefan-Boltzmann constant, A is the surface area of the star, and T is the temperature. Since the stars are perfect black bodies, all of the radiation is emitted from their surfaces, so we can use the surface area of a sphere to calculate A:
A = 4πr²
where r is the radius of the star. Since the stars radiate the same total power, we can set their radiated powers equal to each other:
σ(4πrR²)(TR⁴) = σ(4πrW²)(TW⁴)
Canceling out the constants and simplifying, we get:
rR²TR⁴ = rW²TW⁴
Taking the square root of both sides, we get:
rRTR² = rWTW²
Dividing both sides by rRTR, we get:
rW/rR = (TW/TR)²
So the ratio of the radii is equal to the square of the ratio of the temperatures. Therefore, if the temperatures are equal, the radii must be equal as well. If the white star is hotter, then it must have a smaller radius than the red star to radiate the same total power, and if the white star is cooler, then it must have a larger radius.
P = σA(T⁴)
where P is the power, σ is the Stefan-Boltzmann constant, A is the surface area of the star, and T is the temperature. Since the stars are perfect black bodies, all of the radiation is emitted from their surfaces, so we can use the surface area of a sphere to calculate A:
A = 4πr²
where r is the radius of the star. Since the stars radiate the same total power, we can set their radiated powers equal to each other:
σ(4πrR²)(TR⁴) = σ(4πrW²)(TW⁴)
Canceling out the constants and simplifying, we get:
rR²TR⁴ = rW²TW⁴
Taking the square root of both sides, we get:
rRTR² = rWTW²
Dividing both sides by rRTR, we get:
rW/rR = (TW/TR)²
So the ratio of the radii is equal to the square of the ratio of the temperatures. Therefore, if the temperatures are equal, the radii must be equal as well. If the white star is hotter, then it must have a smaller radius than the red star to radiate the same total power, and if the white star is cooler, then it must have a larger radius.
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A red star having radius rR at a temperature TR and a white star having radius rW at a temperature TW , radiate the same total power. If these stars radiate as perfect black bodies, thena)rR > rW and TR > TWb)rR < rW and TR > TWc)rR > rW and TR < TWd)rR < rW and TR< TWCorrect answer is option 'C'. Can you explain this answer?
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A red star having radius rR at a temperature TR and a white star having radius rW at a temperature TW , radiate the same total power. If these stars radiate as perfect black bodies, thena)rR > rW and TR > TWb)rR < rW and TR > TWc)rR > rW and TR < TWd)rR < rW and TR< TWCorrect answer is option 'C'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A red star having radius rR at a temperature TR and a white star having radius rW at a temperature TW , radiate the same total power. If these stars radiate as perfect black bodies, thena)rR > rW and TR > TWb)rR < rW and TR > TWc)rR > rW and TR < TWd)rR < rW and TR< TWCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A red star having radius rR at a temperature TR and a white star having radius rW at a temperature TW , radiate the same total power. If these stars radiate as perfect black bodies, thena)rR > rW and TR > TWb)rR < rW and TR > TWc)rR > rW and TR < TWd)rR < rW and TR< TWCorrect answer is option 'C'. Can you explain this answer?.
A red star having radius rR at a temperature TR and a white star having radius rW at a temperature TW , radiate the same total power. If these stars radiate as perfect black bodies, thena)rR > rW and TR > TWb)rR < rW and TR > TWc)rR > rW and TR < TWd)rR < rW and TR< TWCorrect answer is option 'C'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A red star having radius rR at a temperature TR and a white star having radius rW at a temperature TW , radiate the same total power. If these stars radiate as perfect black bodies, thena)rR > rW and TR > TWb)rR < rW and TR > TWc)rR > rW and TR < TWd)rR < rW and TR< TWCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A red star having radius rR at a temperature TR and a white star having radius rW at a temperature TW , radiate the same total power. If these stars radiate as perfect black bodies, thena)rR > rW and TR > TWb)rR < rW and TR > TWc)rR > rW and TR < TWd)rR < rW and TR< TWCorrect answer is option 'C'. Can you explain this answer?.
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