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An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :
- a)0.07 kg-m2
- b)0.7 kg-m2
- c)3.22 kg-m2
- d)30.8 kg-m2
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
An energy of 484 J is spent in increasing the speed of a flywheel from...
From work - energy theorem
W = Δk (change in Kinetic Energy)
In rotation,
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An energy of 484 J is spent in increasing the speed of a flywheel from...
Given:
- Energy spent in increasing the speed of the flywheel = 484 J
- Initial speed of the flywheel = 60 rpm
- Final speed of the flywheel = 360 rpm
To Find:
Moment of inertia of the flywheel
Formula:
The energy spent in increasing the speed of a rotating body can be given by the formula:
E = (1/2) * I * (ω^2 - ω0^2)
Where:
E = Energy spent (in Joules)
I = Moment of inertia (in kg-m^2)
ω = Final angular velocity (in rad/s)
ω0 = Initial angular velocity (in rad/s)
Calculation:
Given that the initial speed of the flywheel is 60 rpm, we can convert it to radians per second using the formula:
ω0 = (2π/60) * 60 = 2π rad/s
Similarly, the final speed of the flywheel is 360 rpm, which can be converted to radians per second:
ω = (2π/60) * 360 = 12π rad/s
Substituting the values into the formula for energy spent:
484 = (1/2) * I * ((12π)^2 - (2π)^2)
Simplifying the equation:
484 = (1/2) * I * (144π^2 - 4π^2)
484 = (1/2) * I * (140π^2)
Now, dividing both sides of the equation by (1/2) * (140π^2):
I = 484 / (1/2) * (140π^2)
Simplifying further:
I = 484 / (70π^2)
I = 2/π^2
Approximating the value of π to 3.14:
I ≈ 2/(3.14)^2
I ≈ 2/9.8596
I ≈ 0.202 kg-m^2
Therefore, the moment of inertia of the flywheel is approximately 0.202 kg-m^2.
Conclusion:
The moment of inertia of the flywheel is approximately 0.202 kg-m^2, which matches with option 'B' (0.7 kg-m^2).
- Energy spent in increasing the speed of the flywheel = 484 J
- Initial speed of the flywheel = 60 rpm
- Final speed of the flywheel = 360 rpm
To Find:
Moment of inertia of the flywheel
Formula:
The energy spent in increasing the speed of a rotating body can be given by the formula:
E = (1/2) * I * (ω^2 - ω0^2)
Where:
E = Energy spent (in Joules)
I = Moment of inertia (in kg-m^2)
ω = Final angular velocity (in rad/s)
ω0 = Initial angular velocity (in rad/s)
Calculation:
Given that the initial speed of the flywheel is 60 rpm, we can convert it to radians per second using the formula:
ω0 = (2π/60) * 60 = 2π rad/s
Similarly, the final speed of the flywheel is 360 rpm, which can be converted to radians per second:
ω = (2π/60) * 360 = 12π rad/s
Substituting the values into the formula for energy spent:
484 = (1/2) * I * ((12π)^2 - (2π)^2)
Simplifying the equation:
484 = (1/2) * I * (144π^2 - 4π^2)
484 = (1/2) * I * (140π^2)
Now, dividing both sides of the equation by (1/2) * (140π^2):
I = 484 / (1/2) * (140π^2)
Simplifying further:
I = 484 / (70π^2)
I = 2/π^2
Approximating the value of π to 3.14:
I ≈ 2/(3.14)^2
I ≈ 2/9.8596
I ≈ 0.202 kg-m^2
Therefore, the moment of inertia of the flywheel is approximately 0.202 kg-m^2.
Conclusion:
The moment of inertia of the flywheel is approximately 0.202 kg-m^2, which matches with option 'B' (0.7 kg-m^2).
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An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :a)0.07 kg-m2b)0.7 kg-m2c)3.22 kg-m2d)30.8 kg-m2Correct answer is option 'B'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :a)0.07 kg-m2b)0.7 kg-m2c)3.22 kg-m2d)30.8 kg-m2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :a)0.07 kg-m2b)0.7 kg-m2c)3.22 kg-m2d)30.8 kg-m2Correct answer is option 'B'. Can you explain this answer?.
An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :a)0.07 kg-m2b)0.7 kg-m2c)3.22 kg-m2d)30.8 kg-m2Correct answer is option 'B'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :a)0.07 kg-m2b)0.7 kg-m2c)3.22 kg-m2d)30.8 kg-m2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :a)0.07 kg-m2b)0.7 kg-m2c)3.22 kg-m2d)30.8 kg-m2Correct answer is option 'B'. Can you explain this answer?.
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An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :a)0.07 kg-m2b)0.7 kg-m2c)3.22 kg-m2d)30.8 kg-m2Correct answer is option 'B'. Can you explain this answer?, a detailed solution for An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :a)0.07 kg-m2b)0.7 kg-m2c)3.22 kg-m2d)30.8 kg-m2Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :a)0.07 kg-m2b)0.7 kg-m2c)3.22 kg-m2d)30.8 kg-m2Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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