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Difficulty: HardMeasurement of Length and Measuring Instruments

A micrometer screw gauge with a pitch of 0.5 mm0.5\text{ mm} and 5050 divisions on its circular thimble scale has a negative zero error such that when the jaws are fully closed, the 46th46\text{th} division on the thimble scale aligns with the main scale datum line. When this instrument is used to measure the total thickness of a stack of 4040 identical sheets of paper, the main scale reads 10.0 mm10.0\text{ mm} and the thimble scale reads 1616 divisions. What is the actual thickness of a single sheet of paper?

  1. A
    0.253 mm0.253\text{ mm}
  2. 0.255 mm0.255\text{ mm}Answer
  3. C
    0.254 mm0.254\text{ mm}
  4. D
    0.266 mm0.266\text{ mm}

Answer

The actual thickness of a single sheet of paper is 0.255 mm0.255\text{ mm}.
The least count of the micrometer is 0.5 mm50=0.01 mm\frac{0.5\text{ mm}}{50} = 0.01\text{ mm}. Because the 46th division coincides with the datum line upon closure, the zero error is negative: −(50−46)×0.01 mm=−0.04 mm-(50 - 46) \times 0.01\text{ mm} = -0.04\text{ mm}. The observed reading is 10.0 mm+0.16 mm=10.16 mm10.0\text{ mm} + 0.16\text{ mm} = 10.16\text{ mm}. Correcting for zero error gives a true total thickness of 10.16 mm−(−0.04 mm)=10.20 mm10.16\text{ mm} - (-0.04\text{ mm}) = 10.20\text{ mm}. Dividing this total thickness by 40 sheets gives an individual sheet thickness of 0.255 mm0.255\text{ mm}.

Step-by-Step Solution

1
Determine the least count (LC) of the micrometer screw gauge.
LC=PitchNumber of thimble divisions=0.5 mm50=0.01 mm\text{LC} = \frac{\text{Pitch}}{\text{Number of thimble divisions}} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}.
The least count is the minimum distance measurable by one circular scale division.
2
Calculate the zero error of the instrument.
Zero error=−(50−46)×0.01 mm=−0.04 mm\text{Zero error} = -(50 - 46) \times 0.01\text{ mm} = -0.04\text{ mm}.
Since the 46th division aligns with the datum line when jaws are closed, the zero mark lies 4 divisions below the line, indicating a negative zero error.
3
Find the observed reading for the stack of 40 sheets.
Observed reading=Main scale+(Thimble reading×LC)=10.0 mm+(16×0.01 mm)=10.16 mm\text{Observed reading} = \text{Main scale} + (\text{Thimble reading} \times \text{LC}) = 10.0\text{ mm} + (16 \times 0.01\text{ mm}) = 10.16\text{ mm}.
The total observed value combines the main scale reading and the fractional thimble scale reading.
4
Calculate the true (corrected) total thickness of the stack.
True reading=Observed reading−Zero error=10.16 mm−(−0.04 mm)=10.20 mm\text{True reading} = \text{Observed reading} - \text{Zero error} = 10.16\text{ mm} - (-0.04\text{ mm}) = 10.20\text{ mm}.
Zero error correction requires subtracting the zero error (with sign) from the observed reading.
5
Calculate the thickness of a single sheet of paper.
Thickness of one sheet=10.20 mm40=0.255 mm\text{Thickness of one sheet} = \frac{10.20\text{ mm}}{40} = 0.255\text{ mm}.
Dividing the total corrected thickness by the total number of sheets gives the thickness per sheet.

Key Concept

Measurement of length using micrometer screw gauge with negative zero error correction
Estimated Time:2m 0s
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Topics

Measurement of Length and Measuring Instruments

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