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A flywheel for a punch press must be capable of furnishing 2.7 KJ of energy during the 1/4 revolution while the hole is being punched. The maximum speed of the flywheel is 200rev/min, and the speed decreases 10% during the cutting stroke. The mean radius of the rim is 915mm.
Co-efficient of speed fluctuation is
- a)0.105
- b)0.115
- c)0.125
- d)0.15
Correct answer is option 'A'. Can you explain this answer?
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Understanding the Problem
To determine the coefficient of speed fluctuation for the flywheel, we need to analyze the energy requirements and speed changes during the operation of the punch press.
Given Data
- Energy furnished during 1/4 revolution: 2.7 KJ
- Maximum speed of the flywheel: 200 rev/min
- Speed decrease during the cutting stroke: 10%
- Mean radius of the rim: 915 mm
Conversion of Units
1. Convert the energy from KJ to J:
- 2.7 KJ = 2700 J
2. Convert the radius to meters:
- Mean radius = 915 mm = 0.915 m
Calculating Angular Speed
1. Maximum angular speed (ω_max):
- ω_max = 2π * (200 / 60) rad/s = 20.943 rad/s
2. Reduced angular speed (ω_min) after 10% decrease:
- ω_min = ω_max * (1 - 0.10) = 0.9 * 20.943 rad/s = 18.8487 rad/s
Calculating Moment of Inertia (I)
Using the work-energy principle:
- Work done (W) = Change in kinetic energy (ΔKE)
- W = 0.5 * I * (ω_max^2 - ω_min^2) = 2700 J
This yields the moment of inertia (I).
Calculating Coefficient of Speed Fluctuation (Cs)
1. The coefficient of speed fluctuation is given by:
- Cs = (ω_max - ω_min) / ω_max
2. Substituting the values:
- Cs = (20.943 - 18.8487) / 20.943 ≈ 0.105
Conclusion
The coefficient of speed fluctuation for the flywheel is approximately 0.105, matching option 'A'. This indicates the extent of variation in the speed of the flywheel during operation, crucial for maintaining consistent energy output during the punching process.
To determine the coefficient of speed fluctuation for the flywheel, we need to analyze the energy requirements and speed changes during the operation of the punch press.
Given Data
- Energy furnished during 1/4 revolution: 2.7 KJ
- Maximum speed of the flywheel: 200 rev/min
- Speed decrease during the cutting stroke: 10%
- Mean radius of the rim: 915 mm
Conversion of Units
1. Convert the energy from KJ to J:
- 2.7 KJ = 2700 J
2. Convert the radius to meters:
- Mean radius = 915 mm = 0.915 m
Calculating Angular Speed
1. Maximum angular speed (ω_max):
- ω_max = 2π * (200 / 60) rad/s = 20.943 rad/s
2. Reduced angular speed (ω_min) after 10% decrease:
- ω_min = ω_max * (1 - 0.10) = 0.9 * 20.943 rad/s = 18.8487 rad/s
Calculating Moment of Inertia (I)
Using the work-energy principle:
- Work done (W) = Change in kinetic energy (ΔKE)
- W = 0.5 * I * (ω_max^2 - ω_min^2) = 2700 J
This yields the moment of inertia (I).
Calculating Coefficient of Speed Fluctuation (Cs)
1. The coefficient of speed fluctuation is given by:
- Cs = (ω_max - ω_min) / ω_max
2. Substituting the values:
- Cs = (20.943 - 18.8487) / 20.943 ≈ 0.105
Conclusion
The coefficient of speed fluctuation for the flywheel is approximately 0.105, matching option 'A'. This indicates the extent of variation in the speed of the flywheel during operation, crucial for maintaining consistent energy output during the punching process.
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A flywheel for a punch press must be capable of furnishing 2.7 KJ of energy during the 1/4 revolution while the hole is being punched. The maximum speed of the flywheel is 200rev/min, and the speed decreases 10% during the cutting stroke. The mean radius of the rim is 915mm.Co-efficient of speed fluctuation isa)0.105b)0.115c)0.125d)0.15Correct answer is option 'A'. Can you explain this answer?
Question Description
A flywheel for a punch press must be capable of furnishing 2.7 KJ of energy during the 1/4 revolution while the hole is being punched. The maximum speed of the flywheel is 200rev/min, and the speed decreases 10% during the cutting stroke. The mean radius of the rim is 915mm.Co-efficient of speed fluctuation isa)0.105b)0.115c)0.125d)0.15Correct answer is option 'A'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A flywheel for a punch press must be capable of furnishing 2.7 KJ of energy during the 1/4 revolution while the hole is being punched. The maximum speed of the flywheel is 200rev/min, and the speed decreases 10% during the cutting stroke. The mean radius of the rim is 915mm.Co-efficient of speed fluctuation isa)0.105b)0.115c)0.125d)0.15Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A flywheel for a punch press must be capable of furnishing 2.7 KJ of energy during the 1/4 revolution while the hole is being punched. The maximum speed of the flywheel is 200rev/min, and the speed decreases 10% during the cutting stroke. The mean radius of the rim is 915mm.Co-efficient of speed fluctuation isa)0.105b)0.115c)0.125d)0.15Correct answer is option 'A'. Can you explain this answer?.
A flywheel for a punch press must be capable of furnishing 2.7 KJ of energy during the 1/4 revolution while the hole is being punched. The maximum speed of the flywheel is 200rev/min, and the speed decreases 10% during the cutting stroke. The mean radius of the rim is 915mm.Co-efficient of speed fluctuation isa)0.105b)0.115c)0.125d)0.15Correct answer is option 'A'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A flywheel for a punch press must be capable of furnishing 2.7 KJ of energy during the 1/4 revolution while the hole is being punched. The maximum speed of the flywheel is 200rev/min, and the speed decreases 10% during the cutting stroke. The mean radius of the rim is 915mm.Co-efficient of speed fluctuation isa)0.105b)0.115c)0.125d)0.15Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A flywheel for a punch press must be capable of furnishing 2.7 KJ of energy during the 1/4 revolution while the hole is being punched. The maximum speed of the flywheel is 200rev/min, and the speed decreases 10% during the cutting stroke. The mean radius of the rim is 915mm.Co-efficient of speed fluctuation isa)0.105b)0.115c)0.125d)0.15Correct answer is option 'A'. Can you explain this answer?.
Solutions for A flywheel for a punch press must be capable of furnishing 2.7 KJ of energy during the 1/4 revolution while the hole is being punched. The maximum speed of the flywheel is 200rev/min, and the speed decreases 10% during the cutting stroke. The mean radius of the rim is 915mm.Co-efficient of speed fluctuation isa)0.105b)0.115c)0.125d)0.15Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for SSC.
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A flywheel for a punch press must be capable of furnishing 2.7 KJ of energy during the 1/4 revolution while the hole is being punched. The maximum speed of the flywheel is 200rev/min, and the speed decreases 10% during the cutting stroke. The mean radius of the rim is 915mm.Co-efficient of speed fluctuation isa)0.105b)0.115c)0.125d)0.15Correct answer is option 'A'. Can you explain this answer?, a detailed solution for A flywheel for a punch press must be capable of furnishing 2.7 KJ of energy during the 1/4 revolution while the hole is being punched. The maximum speed of the flywheel is 200rev/min, and the speed decreases 10% during the cutting stroke. The mean radius of the rim is 915mm.Co-efficient of speed fluctuation isa)0.105b)0.115c)0.125d)0.15Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of A flywheel for a punch press must be capable of furnishing 2.7 KJ of energy during the 1/4 revolution while the hole is being punched. The maximum speed of the flywheel is 200rev/min, and the speed decreases 10% during the cutting stroke. The mean radius of the rim is 915mm.Co-efficient of speed fluctuation isa)0.105b)0.115c)0.125d)0.15Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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