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A batch of 10 cutting tools could produce 500 components while working at 50 rpm with a tool feed of 0.25 mm/rev and depth of cut of 1 mm. A similar batch of 10 tools of the same specification could produce 122 components while working at 80 rpm with a feed of 0.25 mm/rev and 1 mm depth of cut. How many components can be produced with one cutting tool at 60 rpm?
[ME 2003]
- a)29
- b)31
- c)37
- d)42
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A batch of 10 cutting tools could produce 500 components while working...
Velocity of first cutting tool,
v1 = 50 × 0.25 = 12.5 mm/min
Velocity of second cutting tool,
v2 = 0.25 × 80= 20 mm/min
Since tool life = number of component produce × tool constant
∴ Tool life for first tool = T1 = 500 × k
Took life for second tool = T21 = 122 × k
⇒ n = 0.3332 Hence number of components produced by one cutting tool at 60 rpm is:
v1 = 50 × 0.25 = 12.5 mm/min
Velocity of second cutting tool,
v2 = 0.25 × 80= 20 mm/min
Since tool life = number of component produce × tool constant
∴ Tool life for first tool = T1 = 500 × k
Took life for second tool = T21 = 122 × k
⇒ n = 0.3332 Hence number of components produced by one cutting tool at 60 rpm is:
Most Upvoted Answer
A batch of 10 cutting tools could produce 500 components while working...
To solve this problem, we can use the concept of productivity in machining. Productivity is defined as the number of components produced per unit time.
Given data:
Batch 1:
- Number of cutting tools = 10
- Number of components produced = 500
- RPM = 50
- Feed = 0.25 mm/rev
- Depth of cut = 1 mm
Batch 2:
- Number of cutting tools = 10
- Number of components produced = 122
- RPM = 80
- Feed = 0.25 mm/rev
- Depth of cut = 1 mm
We need to find the number of components that can be produced with one cutting tool at 60 rpm.
Let's calculate the productivity for both batches using the formula:
Productivity = (Number of components produced) / (Number of cutting tools * RPM)
Productivity for Batch 1:
= 500 / (10 * 50)
= 1 component/(tool*min)
Productivity for Batch 2:
= 122 / (10 * 80)
= 0.1525 component/(tool*min)
From the given data, we can observe that the productivity of Batch 2 is lower than that of Batch 1. This indicates that the cutting tools in Batch 2 are not performing as efficiently as those in Batch 1.
To find the number of components that can be produced with one cutting tool at 60 rpm, we can use the concept of proportionality. Since the productivity is directly proportional to the number of components produced, we can set up the following proportion:
Productivity1 / Productivity2 = Number of components produced1 / Number of components produced2
Substituting the values:
1 / 0.1525 = 500 / x
Simplifying the equation:
x = (500 * 0.1525) / 1
x = 76.25 components
Therefore, one cutting tool at 60 rpm can produce approximately 76.25 components. Since the number of components must be a whole number, we can round it to the nearest whole number, which is 76.
Thus, the correct answer is option 'A', 76 components.
Given data:
Batch 1:
- Number of cutting tools = 10
- Number of components produced = 500
- RPM = 50
- Feed = 0.25 mm/rev
- Depth of cut = 1 mm
Batch 2:
- Number of cutting tools = 10
- Number of components produced = 122
- RPM = 80
- Feed = 0.25 mm/rev
- Depth of cut = 1 mm
We need to find the number of components that can be produced with one cutting tool at 60 rpm.
Let's calculate the productivity for both batches using the formula:
Productivity = (Number of components produced) / (Number of cutting tools * RPM)
Productivity for Batch 1:
= 500 / (10 * 50)
= 1 component/(tool*min)
Productivity for Batch 2:
= 122 / (10 * 80)
= 0.1525 component/(tool*min)
From the given data, we can observe that the productivity of Batch 2 is lower than that of Batch 1. This indicates that the cutting tools in Batch 2 are not performing as efficiently as those in Batch 1.
To find the number of components that can be produced with one cutting tool at 60 rpm, we can use the concept of proportionality. Since the productivity is directly proportional to the number of components produced, we can set up the following proportion:
Productivity1 / Productivity2 = Number of components produced1 / Number of components produced2
Substituting the values:
1 / 0.1525 = 500 / x
Simplifying the equation:
x = (500 * 0.1525) / 1
x = 76.25 components
Therefore, one cutting tool at 60 rpm can produce approximately 76.25 components. Since the number of components must be a whole number, we can round it to the nearest whole number, which is 76.
Thus, the correct answer is option 'A', 76 components.
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A batch of 10 cutting tools could produce 500 components while working at 50 rpm with a tool feed of 0.25 mm/rev and depth of cut of 1 mm. A similar batch of 10 tools of the same specification could produce 122 components while working at 80 rpm with a feed of 0.25 mm/rev and 1 mm depth of cut. How many components can be produced with one cutting tool at 60 rpm?[ME 2003]a)29b)31c)37d)42Correct answer is option 'A'. Can you explain this answer?
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A batch of 10 cutting tools could produce 500 components while working at 50 rpm with a tool feed of 0.25 mm/rev and depth of cut of 1 mm. A similar batch of 10 tools of the same specification could produce 122 components while working at 80 rpm with a feed of 0.25 mm/rev and 1 mm depth of cut. How many components can be produced with one cutting tool at 60 rpm?[ME 2003]a)29b)31c)37d)42Correct answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A batch of 10 cutting tools could produce 500 components while working at 50 rpm with a tool feed of 0.25 mm/rev and depth of cut of 1 mm. A similar batch of 10 tools of the same specification could produce 122 components while working at 80 rpm with a feed of 0.25 mm/rev and 1 mm depth of cut. How many components can be produced with one cutting tool at 60 rpm?[ME 2003]a)29b)31c)37d)42Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A batch of 10 cutting tools could produce 500 components while working at 50 rpm with a tool feed of 0.25 mm/rev and depth of cut of 1 mm. A similar batch of 10 tools of the same specification could produce 122 components while working at 80 rpm with a feed of 0.25 mm/rev and 1 mm depth of cut. How many components can be produced with one cutting tool at 60 rpm?[ME 2003]a)29b)31c)37d)42Correct answer is option 'A'. Can you explain this answer?.
A batch of 10 cutting tools could produce 500 components while working at 50 rpm with a tool feed of 0.25 mm/rev and depth of cut of 1 mm. A similar batch of 10 tools of the same specification could produce 122 components while working at 80 rpm with a feed of 0.25 mm/rev and 1 mm depth of cut. How many components can be produced with one cutting tool at 60 rpm?[ME 2003]a)29b)31c)37d)42Correct answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A batch of 10 cutting tools could produce 500 components while working at 50 rpm with a tool feed of 0.25 mm/rev and depth of cut of 1 mm. A similar batch of 10 tools of the same specification could produce 122 components while working at 80 rpm with a feed of 0.25 mm/rev and 1 mm depth of cut. How many components can be produced with one cutting tool at 60 rpm?[ME 2003]a)29b)31c)37d)42Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A batch of 10 cutting tools could produce 500 components while working at 50 rpm with a tool feed of 0.25 mm/rev and depth of cut of 1 mm. A similar batch of 10 tools of the same specification could produce 122 components while working at 80 rpm with a feed of 0.25 mm/rev and 1 mm depth of cut. How many components can be produced with one cutting tool at 60 rpm?[ME 2003]a)29b)31c)37d)42Correct answer is option 'A'. Can you explain this answer?.
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