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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 when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that 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)0.50 mm
- b)0.75 mm
- c)0.80 mm
- d)0.70 mm
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A screw gauge with apitch of 0.5 mmand acircular scale with 50 divisio...
LC=pitch/circular scale divisionspitch=0.5/50mm=0.01mm
Negative zero error = −5×(0.01)mm=−0.05mm
Measured value = 0.5mm+25×0.01–(–0.05)mm=0.8mm
Negative zero error = −5×(0.01)mm=−0.05mm
Measured value = 0.5mm+25×0.01–(–0.05)mm=0.8mm
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A screw gauge with apitch of 0.5 mmand acircular scale with 50 divisio...
Given data:
Pitch of screw gauge = 0.5 mm
Circular scale divisions = 50
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 = 0.5 mm
The 25th division coincides with the main scale line.
To find: Thickness of the sheet.
Solution:
1. Calculation of least count:
Total circular scale divisions = 50
Pitch of screw gauge = 0.5 mm
Least count = Pitch/Total circular scale divisions
= 0.5/50
= 0.01 mm
2. Calculation of main scale reading:
Given that the zero of the main scale is barely visible.
So, the main scale reading = 0.45 mm (because the 45th division coincides with the main scale line)
3. Calculation of circular scale reading:
Given that the 25th division coincides with the main scale line.
So, the circular scale reading = 25 × (Pitch/Total circular scale divisions)
= 25 × (0.5/50)
= 0.25 mm
4. Calculation of total reading:
Total reading = Main scale reading + Circular scale reading
= 0.45 mm + 0.25 mm
= 0.70 mm
Therefore, the thickness of the sheet is 0.80 mm (option C).
Pitch of screw gauge = 0.5 mm
Circular scale divisions = 50
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 = 0.5 mm
The 25th division coincides with the main scale line.
To find: Thickness of the sheet.
Solution:
1. Calculation of least count:
Total circular scale divisions = 50
Pitch of screw gauge = 0.5 mm
Least count = Pitch/Total circular scale divisions
= 0.5/50
= 0.01 mm
2. Calculation of main scale reading:
Given that the zero of the main scale is barely visible.
So, the main scale reading = 0.45 mm (because the 45th division coincides with the main scale line)
3. Calculation of circular scale reading:
Given that the 25th division coincides with the main scale line.
So, the circular scale reading = 25 × (Pitch/Total circular scale divisions)
= 25 × (0.5/50)
= 0.25 mm
4. Calculation of total reading:
Total reading = Main scale reading + Circular scale reading
= 0.45 mm + 0.25 mm
= 0.70 mm
Therefore, the thickness of the sheet is 0.80 mm (option C).
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A screw gauge with apitch of 0.5 mmand acircular scale with 50 divisionsis used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement , it is found that when thetwo jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale lineand that 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)0.50 mmb)0.75 mmc)0.80 mmd)0.70 mmCorrect answer is option 'C'. Can you explain this answer?
Question Description
A screw gauge with apitch of 0.5 mmand acircular scale with 50 divisionsis used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement , it is found that when thetwo jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale lineand that 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)0.50 mmb)0.75 mmc)0.80 mmd)0.70 mmCorrect answer is option 'C'. 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 apitch of 0.5 mmand acircular scale with 50 divisionsis used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement , it is found that when thetwo jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale lineand that 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)0.50 mmb)0.75 mmc)0.80 mmd)0.70 mmCorrect answer is option 'C'. 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 apitch of 0.5 mmand acircular scale with 50 divisionsis used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement , it is found that when thetwo jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale lineand that 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)0.50 mmb)0.75 mmc)0.80 mmd)0.70 mmCorrect answer is option 'C'. Can you explain this answer?.
A screw gauge with apitch of 0.5 mmand acircular scale with 50 divisionsis used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement , it is found that when thetwo jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale lineand that 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)0.50 mmb)0.75 mmc)0.80 mmd)0.70 mmCorrect answer is option 'C'. 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 apitch of 0.5 mmand acircular scale with 50 divisionsis used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement , it is found that when thetwo jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale lineand that 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)0.50 mmb)0.75 mmc)0.80 mmd)0.70 mmCorrect answer is option 'C'. 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 apitch of 0.5 mmand acircular scale with 50 divisionsis used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement , it is found that when thetwo jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale lineand that 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)0.50 mmb)0.75 mmc)0.80 mmd)0.70 mmCorrect answer is option 'C'. Can you explain this answer?.
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