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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 and 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)
    0.75 mm
  • b)
    0.80 mm
  • c)
    0.70 mm
  • d)
    0.50 mm
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divi...
L.C. = 
-ve zero error = - 5 x L.C. = -0.005 mm
∴ Measured value
= main scale reading + screw gauge reading - zero error 
= 0.5 mm + {25 x 0.001 - (-0.05)} mm  = 0.80 mm
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Most Upvoted Answer
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divi...
Given data:
- Pitch of the screw gauge = 0.5 mm
- Circular scale divisions = 50
- 45th division coincides with the main scale line when the jaws are brought in contact
- The zero of the main scale is barely visible

To find:
- Thickness of the sheet when the main scale reading is 0.5 mm and the 25th division coincides with the main scale line

Formula:
- Total reading = Main scale reading + (Circular scale reading x Least count)

Solution:
1. Least count of the screw gauge:
- The least count of the screw gauge is the smallest measurement that can be measured by it.
- The least count of the screw gauge is given by the formula: Least count = Pitch of the screw gauge / Number of circular scale divisions
- Substituting the given values, the least count is: Least count = 0.5 mm / 50 = 0.01 mm

2. Position of the zero of the main scale:
- The zero of the main scale is barely visible, so it can be assumed that the zero of the main scale coincides with the 45th division of the circular scale.
- This means that the zero of the main scale is 45 x least count away from the reference line.
- The position of the zero of the main scale is: Zero position = 45 x 0.01 mm = 0.45 mm

3. Main scale reading:
- The main scale reading is given as 0.5 mm.

4. Circular scale reading:
- The 25th division coincides with the main scale line.
- This means that there are 25 divisions between the zero position and the main scale line.
- The circular scale reading is: Circular scale reading = 25 x least count = 25 x 0.01 mm = 0.25 mm

5. Total reading:
- The total reading is given by the formula: Total reading = Main scale reading + (Circular scale reading x Least count)
- Substituting the given values, the total reading is: Total reading = 0.5 mm + (0.25 mm x 0.01 mm) = 0.5 mm + 0.0025 mm = 0.5025 mm

Conclusion:
Therefore, the thickness of the sheet is 0.5025 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 when the two jaws of the screw gauge and 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)0.75 mmb)0.80 mmc)0.70 mmd)0.50 mmCorrect answer is option 'B'. Can you explain this answer?
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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 and 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)0.75 mmb)0.80 mmc)0.70 mmd)0.50 mmCorrect answer is option 'B'. 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 when the two jaws of the screw gauge and 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)0.75 mmb)0.80 mmc)0.70 mmd)0.50 mmCorrect answer is option 'B'. 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 when the two jaws of the screw gauge and 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)0.75 mmb)0.80 mmc)0.70 mmd)0.50 mmCorrect answer is option 'B'. Can you explain this answer?.
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