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The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit i (microns) = 0.45 (D1/3) + 0.001 D. The unit of D is mm. Diameter step is 18 – 30. If the fundamental deviation for hole H is zero and IT8 = 26i, the maximum and minimum limits of dimension for a 25H8 hole (in mm) are
[PI 2008]
- a)24.984 and 24.967
- b)25.017 and 24.984
- c)25.033 and 25.00
- d)25.00, 24.967
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The following data are given for calculating limits of dimension and t...
IT8 = 26i = 26 × 1.307 = 33.98 ≈ 34mm = 0.034mm
25H8 = 250.00 + 0.034mm.
Most Upvoted Answer
The following data are given for calculating limits of dimension and t...
Based on the given data, we have the following formulas for calculating limits of dimension and tolerances for a hole:
Upper limit of hole diameter = D + (Tolerance unit i)
Lower limit of hole diameter = D - (Tolerance unit i)
Where:
- D is the nominal hole diameter in mm.
- Tolerance unit i is calculated using the formula: 0.45 (D^(1/3)) + 0.001 D.
So, to calculate the limits of dimension and tolerances for a hole with a diameter step of 18 mm, we plug in the value of D into the formulas.
For the upper limit of hole diameter:
Upper limit = D + (0.45 (D^(1/3)) + 0.001 D)
For the lower limit of hole diameter:
Lower limit = D - (0.45 (D^(1/3)) + 0.001 D)
Note: The given data does not specify a specific value for D. You need to provide a specific value for D in order to calculate the limits of dimension and tolerances.
Upper limit of hole diameter = D + (Tolerance unit i)
Lower limit of hole diameter = D - (Tolerance unit i)
Where:
- D is the nominal hole diameter in mm.
- Tolerance unit i is calculated using the formula: 0.45 (D^(1/3)) + 0.001 D.
So, to calculate the limits of dimension and tolerances for a hole with a diameter step of 18 mm, we plug in the value of D into the formulas.
For the upper limit of hole diameter:
Upper limit = D + (0.45 (D^(1/3)) + 0.001 D)
For the lower limit of hole diameter:
Lower limit = D - (0.45 (D^(1/3)) + 0.001 D)
Note: The given data does not specify a specific value for D. You need to provide a specific value for D in order to calculate the limits of dimension and tolerances.
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Community Answer
The following data are given for calculating limits of dimension and t...
IT8 = 26i = 26 × 1.307 = 33.98 ≈ 34mm = 0.034mm
25H8 = 250.00 + 0.034mm.
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The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit i (microns) = 0.45 (D1/3) + 0.001 D. The unit of D is mm. Diameter step is 18 – 30. If the fundamental deviation for hole H is zero and IT8 = 26i, the maximum and minimum limits of dimension for a 25H8 hole (in mm) are[PI 2008]a)24.984 and 24.967b)25.017 and 24.984c)25.033 and 25.00d)25.00, 24.967Correct answer is option 'C'. Can you explain this answer?
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The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit i (microns) = 0.45 (D1/3) + 0.001 D. The unit of D is mm. Diameter step is 18 – 30. If the fundamental deviation for hole H is zero and IT8 = 26i, the maximum and minimum limits of dimension for a 25H8 hole (in mm) are[PI 2008]a)24.984 and 24.967b)25.017 and 24.984c)25.033 and 25.00d)25.00, 24.967Correct answer is option 'C'. 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 The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit i (microns) = 0.45 (D1/3) + 0.001 D. The unit of D is mm. Diameter step is 18 – 30. If the fundamental deviation for hole H is zero and IT8 = 26i, the maximum and minimum limits of dimension for a 25H8 hole (in mm) are[PI 2008]a)24.984 and 24.967b)25.017 and 24.984c)25.033 and 25.00d)25.00, 24.967Correct answer is option 'C'. 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 The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit i (microns) = 0.45 (D1/3) + 0.001 D. The unit of D is mm. Diameter step is 18 – 30. If the fundamental deviation for hole H is zero and IT8 = 26i, the maximum and minimum limits of dimension for a 25H8 hole (in mm) are[PI 2008]a)24.984 and 24.967b)25.017 and 24.984c)25.033 and 25.00d)25.00, 24.967Correct answer is option 'C'. Can you explain this answer?.
The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit i (microns) = 0.45 (D1/3) + 0.001 D. The unit of D is mm. Diameter step is 18 – 30. If the fundamental deviation for hole H is zero and IT8 = 26i, the maximum and minimum limits of dimension for a 25H8 hole (in mm) are[PI 2008]a)24.984 and 24.967b)25.017 and 24.984c)25.033 and 25.00d)25.00, 24.967Correct answer is option 'C'. 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 The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit i (microns) = 0.45 (D1/3) + 0.001 D. The unit of D is mm. Diameter step is 18 – 30. If the fundamental deviation for hole H is zero and IT8 = 26i, the maximum and minimum limits of dimension for a 25H8 hole (in mm) are[PI 2008]a)24.984 and 24.967b)25.017 and 24.984c)25.033 and 25.00d)25.00, 24.967Correct answer is option 'C'. 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 The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit i (microns) = 0.45 (D1/3) + 0.001 D. The unit of D is mm. Diameter step is 18 – 30. If the fundamental deviation for hole H is zero and IT8 = 26i, the maximum and minimum limits of dimension for a 25H8 hole (in mm) are[PI 2008]a)24.984 and 24.967b)25.017 and 24.984c)25.033 and 25.00d)25.00, 24.967Correct answer is option 'C'. Can you explain this answer?.
Solutions for The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit i (microns) = 0.45 (D1/3) + 0.001 D. The unit of D is mm. Diameter step is 18 – 30. If the fundamental deviation for hole H is zero and IT8 = 26i, the maximum and minimum limits of dimension for a 25H8 hole (in mm) are[PI 2008]a)24.984 and 24.967b)25.017 and 24.984c)25.033 and 25.00d)25.00, 24.967Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit i (microns) = 0.45 (D1/3) + 0.001 D. The unit of D is mm. Diameter step is 18 – 30. If the fundamental deviation for hole H is zero and IT8 = 26i, the maximum and minimum limits of dimension for a 25H8 hole (in mm) are[PI 2008]a)24.984 and 24.967b)25.017 and 24.984c)25.033 and 25.00d)25.00, 24.967Correct answer is option 'C'. Can you explain this answer?, a detailed solution for The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit i (microns) = 0.45 (D1/3) + 0.001 D. The unit of D is mm. Diameter step is 18 – 30. If the fundamental deviation for hole H is zero and IT8 = 26i, the maximum and minimum limits of dimension for a 25H8 hole (in mm) are[PI 2008]a)24.984 and 24.967b)25.017 and 24.984c)25.033 and 25.00d)25.00, 24.967Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit i (microns) = 0.45 (D1/3) + 0.001 D. The unit of D is mm. Diameter step is 18 – 30. If the fundamental deviation for hole H is zero and IT8 = 26i, the maximum and minimum limits of dimension for a 25H8 hole (in mm) are[PI 2008]a)24.984 and 24.967b)25.017 and 24.984c)25.033 and 25.00d)25.00, 24.967Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit i (microns) = 0.45 (D1/3) + 0.001 D. The unit of D is mm. Diameter step is 18 – 30. If the fundamental deviation for hole H is zero and IT8 = 26i, the maximum and minimum limits of dimension for a 25H8 hole (in mm) are[PI 2008]a)24.984 and 24.967b)25.017 and 24.984c)25.033 and 25.00d)25.00, 24.967Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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