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The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.
The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :
  • a)
    5. 755 m
  • b)
    5. 725 mm
  • c)
    5. 740 m
  • d)
    5. 950 mm
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The pitch and the number of divisions, on the circular scale, for a gi...

LC = 0.5 × 10–2 mm
+ve error = 3 × 0.5 × 10–2 mm
= 1.5 × 10–2 mm = 0.015 mm
Reading = MSR + CSR – (+ve error)
= 5.5 mm + (48 × 0.5 × 10–2) – 0.015
= 5.5 + 0.24 – 0.015 = 5.725 mm
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Most Upvoted Answer
The pitch and the number of divisions, on the circular scale, for a gi...
Given data:
- Pitch (p) = 0.5 mm
- Number of divisions on the circular scale (N) = 100
- Zero of the circular scale lies 3 divisions below the mean line when fully tightened without any object.
- Main scale reading (MSR) = 5.5 mm
- Circular scale reading (CSR) = 48

To find:
Thickness of the thin sheet

Formula:
Total distance moved by the screw = MSR + (CSR/N) - Zero Error

Calculation:
- Zero Error = 3 divisions * (p/N) = 3 * (0.5/100) = 0.015 mm
- Total distance moved by the screw = 5.5 mm + (48/100) mm - 0.015 mm = 5.5 mm + 0.48 mm - 0.015 mm = 5.965 mm

Therefore, the thickness of the thin sheet is 5.965 mm.

Answer:
The correct option is 5.725 mm.
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The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :a)5. 755 mb)5. 725 mmc)5. 740 md)5. 950 mmCorrect answer is option 'B'. Can you explain this answer?
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The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :a)5. 755 mb)5. 725 mmc)5. 740 md)5. 950 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 The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :a)5. 755 mb)5. 725 mmc)5. 740 md)5. 950 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 The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :a)5. 755 mb)5. 725 mmc)5. 740 md)5. 950 mmCorrect answer is option 'B'. Can you explain this answer?.
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