Measuring Instruments And Least Count
A screw gauge of pitch 0.5 mm has 50 circular divisions and a positive zero error of three divisions; while measuring a wire it reads two main-scale divisions and twenty-eight circular divisions, so what is the corrected diameter?
Select the correct option:
Solution
1.25 mm
A screw gauge measures small thicknesses by converting rotation into linear advance, and its least count equals the pitch divided by the number of circular divisions. Here least count = 0.5 mm / 50 = 0.01 mm, while each main-scale division equals the pitch of 0.5 mm. The observed reading is the main-scale contribution plus the circular contribution: two main divisions give 2 × 0.5 = 1.00 mm, and twenty-eight circular divisions give 28 × 0.01 = 0.28 mm, so the observed diameter is 1.28 mm. A positive zero error means the instrument reads too high by three divisions, that is +3 × 0.01 = 0.03 mm, which must be subtracted: corrected diameter = 1.28 - 0.03 = 1.25 mm. The choice 1.31 mm wrongly adds the zero-error correction. The choice 1.28 mm ignores the zero error entirely. The choice 1.22 mm subtracts too much by doubling the correction. This procedure follows the NCERT treatment of zero-error correction. As a check, a positive zero error must always reduce the final reading, consistent with our subtraction.
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About This Question
- Subject
- physics
- Chapter
- physics and measurement
- Topic
- measuring instruments and least count
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1.25 mm
A screw gauge measures small thicknesses by converting rotation into linear advance, and its least count equals the pitch divided by the number of circular divisions. Here least count = 0.5 mm / 50 = 0.01 mm, while each main-scale division equals the pitch of 0.5 mm. The observed reading is the main-scale contribution plus the circular contribution: two main divisions give 2 × 0.5 = 1.00 mm, and twenty-eight circular divisions give 28 × 0.01 = 0.28 mm, so the observed diameter is 1.28 mm. A positive zero error means the instrument reads too high by three divisions, that is +3 × 0.01 = 0.03 mm, which must be subtracted: corrected diameter = 1.28 - 0.03 = 1.25 mm. The choice 1.31 mm wrongly adds the zero-error correction. The choice 1.28 mm ignores the zero error entirely. The choice 1.22 mm subtracts too much by doubling the correction. This procedure follows the NCERT treatment of zero-error correction. As a check, a positive zero error must always reduce the final reading, consistent with our subtraction.
This medium difficulty physics question is from the chapter physics and measurement, covering the topic of measuring instruments and least count. It appeared in the 2025 exam.
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