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Difficulty: Very hardMeasurement of Length and Measuring Instruments

A micrometer screw gauge with a pitch of 0.5 mm0.5\text{ mm} and 5050 circular scale divisions is used to measure the diameter of a uniform brass sphere. When the anvil and spindle are brought into contact without the sphere, the 45th45\text{th} division on the thimble scale aligns with the main scale index line. When the sphere is clamped between the anvil and spindle, the main scale reads 3.5 mm3.5\text{ mm} and the 28th28\text{th} division on the thimble scale coincides with the index line. What is the actual diameter of the brass sphere?

  1. 3.83 mm3.83\text{ mm}Answer
  2. B
    3.78 mm3.78\text{ mm}
  3. C
    3.73 mm3.73\text{ mm}
  4. D
    3.33 mm3.33\text{ mm}

Answer

The actual diameter of the brass sphere is 3.83 mm3.83\text{ mm}.
The correct answer of 3.83 mm3.83\text{ mm} is derived by first establishing the least count (0.01 mm0.01\text{ mm}). When the jaws are closed, the 45th45\text{th} mark lies below the reference line, giving a negative zero error of −0.05 mm-0.05\text{ mm}. Adding the main scale (3.5 mm3.5\text{ mm}) to the thimble reading (0.28 mm0.28\text{ mm}) gives an observed value of 3.78 mm3.78\text{ mm}. Subtracting the negative zero error yields 3.78 mm−(−0.05 mm)=3.83 mm3.78\text{ mm} - (-0.05\text{ mm}) = 3.83\text{ mm}.

Step-by-Step Solution

1
Determine the least count (precision) of the micrometer screw gauge.
Least Count (LC)=PitchNumber of circular scale divisions=0.5 mm50=0.01 mm\text{Least Count (LC)} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}.
Least count defines the minimum measurement value represented by one thimble scale division.
2
Calculate the zero error of the instrument.
Since the 45th45\text{th} division is aligned when closed, it is 55 divisions below the zero line (45−50=−545 - 50 = -5). Thus, Zero Error (ZE)=−5×0.01 mm=−0.05 mm\text{Zero Error (ZE)} = -5 \times 0.01\text{ mm} = -0.05\text{ mm}.
When the zero mark on the thimble lies below the index line, the instrument has a negative zero error.
3
Calculate the observed reading of the sphere.
\text{Observed Reading (OR)} = 3.5\text{ mm} + (28 \times 0.01\text{ mm}) = 3.5\text{ mm} + 0.28\text{ mm} = 3.78\text{ mm}.
The total observed reading combines the main scale reading and the circular thimble reading.
4
Apply the zero error correction to find the true diameter.
\text{True Diameter} = \text{Observed Reading} - \text{Zero Error} = 3.78\text{ mm} - (-0.05\text{ mm}) = 3.78\text{ mm} + 0.05\text{ mm} = 3.83\text{ mm}.
True measurement is always obtained by subtracting the zero error (including its sign) from the observed reading.

Key Concept

Negative zero error correction in micrometer screw gauge measurements
Estimated Time:2m 0s
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Topics

Measurement of Length and Measuring Instruments

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