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The following data are given for calculating limits of dimensions and tolerance for a hole: Tolerance unit ‘i’ (microns) = 0.45 D1/3 + 0.001D. Where D is in mm. Diameter step is 18-30. If the fundamental deviation for hole H is zero and 1T8 =25 i, the maximum and minimum limit of dimension for 25H8 hole (in mm) respectively area)24.984 and 24.967b)25.017 and 24.984c)24.033 and 25.00d)25.00 and 24.967Correct answer is option 'D'. 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 dimensions and tolerance for a hole: Tolerance unit ‘i’ (microns) = 0.45 D1/3 + 0.001D. Where D is in mm. Diameter step is 18-30. If the fundamental deviation for hole H is zero and 1T8 =25 i, the maximum and minimum limit of dimension for 25H8 hole (in mm) respectively area)24.984 and 24.967b)25.017 and 24.984c)24.033 and 25.00d)25.00 and 24.967Correct answer is option 'D'. 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 dimensions and tolerance for a hole: Tolerance unit ‘i’ (microns) = 0.45 D1/3 + 0.001D. Where D is in mm. Diameter step is 18-30. If the fundamental deviation for hole H is zero and 1T8 =25 i, the maximum and minimum limit of dimension for 25H8 hole (in mm) respectively area)24.984 and 24.967b)25.017 and 24.984c)24.033 and 25.00d)25.00 and 24.967Correct answer is option 'D'. Can you explain this answer?.

Solutions for The following data are given for calculating limits of dimensions and tolerance for a hole: Tolerance unit ‘i’ (microns) = 0.45 D1/3 + 0.001D. Where D is in mm. Diameter step is 18-30. If the fundamental deviation for hole H is zero and 1T8 =25 i, the maximum and minimum limit of dimension for 25H8 hole (in mm) respectively area)24.984 and 24.967b)25.017 and 24.984c)24.033 and 25.00d)25.00 and 24.967Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering. Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free.

Here you can find the meaning of The following data are given for calculating limits of dimensions and tolerance for a hole: Tolerance unit ‘i’ (microns) = 0.45 D1/3 + 0.001D. Where D is in mm. Diameter step is 18-30. If the fundamental deviation for hole H is zero and 1T8 =25 i, the maximum and minimum limit of dimension for 25H8 hole (in mm) respectively area)24.984 and 24.967b)25.017 and 24.984c)24.033 and 25.00d)25.00 and 24.967Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The following data are given for calculating limits of dimensions and tolerance for a hole: Tolerance unit ‘i’ (microns) = 0.45 D1/3 + 0.001D. Where D is in mm. Diameter step is 18-30. If the fundamental deviation for hole H is zero and 1T8 =25 i, the maximum and minimum limit of dimension for 25H8 hole (in mm) respectively area)24.984 and 24.967b)25.017 and 24.984c)24.033 and 25.00d)25.00 and 24.967Correct answer is option 'D'. Can you explain this answer?, a detailed solution for The following data are given for calculating limits of dimensions and tolerance for a hole: Tolerance unit ‘i’ (microns) = 0.45 D1/3 + 0.001D. Where D is in mm. Diameter step is 18-30. If the fundamental deviation for hole H is zero and 1T8 =25 i, the maximum and minimum limit of dimension for 25H8 hole (in mm) respectively area)24.984 and 24.967b)25.017 and 24.984c)24.033 and 25.00d)25.00 and 24.967Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of The following data are given for calculating limits of dimensions and tolerance for a hole: Tolerance unit ‘i’ (microns) = 0.45 D1/3 + 0.001D. Where D is in mm. Diameter step is 18-30. If the fundamental deviation for hole H is zero and 1T8 =25 i, the maximum and minimum limit of dimension for 25H8 hole (in mm) respectively area)24.984 and 24.967b)25.017 and 24.984c)24.033 and 25.00d)25.00 and 24.967Correct answer is option 'D'. 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 dimensions and tolerance for a hole: Tolerance unit ‘i’ (microns) = 0.45 D1/3 + 0.001D. Where D is in mm. Diameter step is 18-30. If the fundamental deviation for hole H is zero and 1T8 =25 i, the maximum and minimum limit of dimension for 25H8 hole (in mm) respectively area)24.984 and 24.967b)25.017 and 24.984c)24.033 and 25.00d)25.00 and 24.967Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.