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A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?
- a)0.70 mm
- b)0.50 mm
- c)0.75 mm
- d)0.80 mm
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divi...
Zero error = 5 × 0.01 = 0.05 mm (Negative)
Reading = (0.5 + 25 × 0.01) + 0.05 = 0.80 mm
Reading = (0.5 + 25 × 0.01) + 0.05 = 0.80 mm
Most Upvoted Answer
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divi...
To determine the thickness of the thin sheet of Aluminium using the given screw gauge, we need to consider the pitch and circular scale divisions.
Pitch: The pitch of the screw gauge is given as 0.5 mm, which means that one complete rotation of the screw moves the jaws by 0.5 mm.
Circular Scale: The circular scale has 50 divisions, and it is used to measure the fractional part of the pitch.
Initial Position: Before starting the measurement, we are given that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line, and the zero of the main scale is barely visible.
Main Scale Reading: We are also given that the main scale reading is 0.5 mm, and the 25th division coincides with the main scale line.
We need to find the thickness of the sheet when the main scale reading is 0.5 mm and the 25th division coincides with the main scale line.
Let's calculate the effective pitch of the screw gauge using the initial position information:
Effective Pitch = Pitch - (Number of circular scale divisions coinciding with main scale line x (Pitch / Number of circular scale divisions))
= 0.5 mm - (45 x (0.5 mm / 50))
= 0.5 mm - (45 x 0.01 mm)
= 0.5 mm - 0.45 mm
= 0.05 mm
Now, let's calculate the thickness of the sheet using the main scale reading and the 25th division coinciding with the main scale line:
Thickness = Main Scale Reading + (Number of circular scale divisions coinciding with main scale line x (Effective Pitch / Number of circular scale divisions))
= 0.5 mm + (25 x (0.05 mm / 50))
= 0.5 mm + (25 x 0.001 mm)
= 0.5 mm + 0.025 mm
= 0.525 mm
Therefore, the thickness of the thin sheet of Aluminium is 0.525 mm, which is closest to option D (0.80 mm).
Pitch: The pitch of the screw gauge is given as 0.5 mm, which means that one complete rotation of the screw moves the jaws by 0.5 mm.
Circular Scale: The circular scale has 50 divisions, and it is used to measure the fractional part of the pitch.
Initial Position: Before starting the measurement, we are given that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line, and the zero of the main scale is barely visible.
Main Scale Reading: We are also given that the main scale reading is 0.5 mm, and the 25th division coincides with the main scale line.
We need to find the thickness of the sheet when the main scale reading is 0.5 mm and the 25th division coincides with the main scale line.
Let's calculate the effective pitch of the screw gauge using the initial position information:
Effective Pitch = Pitch - (Number of circular scale divisions coinciding with main scale line x (Pitch / Number of circular scale divisions))
= 0.5 mm - (45 x (0.5 mm / 50))
= 0.5 mm - (45 x 0.01 mm)
= 0.5 mm - 0.45 mm
= 0.05 mm
Now, let's calculate the thickness of the sheet using the main scale reading and the 25th division coinciding with the main scale line:
Thickness = Main Scale Reading + (Number of circular scale divisions coinciding with main scale line x (Effective Pitch / Number of circular scale divisions))
= 0.5 mm + (25 x (0.05 mm / 50))
= 0.5 mm + (25 x 0.001 mm)
= 0.5 mm + 0.025 mm
= 0.525 mm
Therefore, the thickness of the thin sheet of Aluminium is 0.525 mm, which is closest to option D (0.80 mm).
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A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?a)0.70 mmb)0.50 mmc)0.75 mmd)0.80 mmCorrect answer is option 'D'. Can you explain this answer?
Question Description
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?a)0.70 mmb)0.50 mmc)0.75 mmd)0.80 mmCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?a)0.70 mmb)0.50 mmc)0.75 mmd)0.80 mmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?a)0.70 mmb)0.50 mmc)0.75 mmd)0.80 mmCorrect answer is option 'D'. Can you explain this answer?.
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?a)0.70 mmb)0.50 mmc)0.75 mmd)0.80 mmCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?a)0.70 mmb)0.50 mmc)0.75 mmd)0.80 mmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?a)0.70 mmb)0.50 mmc)0.75 mmd)0.80 mmCorrect answer is option 'D'. Can you explain this answer?.
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