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प्रश्न
The pitch of a screw gauge is 0.5 mm and the head scale is divided in 100 parts. What is the least count of a screw gauge?
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उत्तर
Pitch of a screw gauge = 0.5 mm
No. of divisions on the circular scale = 100
L.C. = `(0.5/100)` mm
= 0.005 mm or 0.0005 cm
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संबंधित प्रश्न
The final result of the calculation in an experiment is 125,347,200. Express the number in term of significant place when accuracy is between 1 and 100
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Figure shows the position of vernier scale while measuring the external length of a wooden cylinder.
- What is the length recorded by the main scale?
- Which reading of vernier scale coincides with the main scale?
- Calculate the length.
(a) A vernier scale has 20 divisions. It slides over the main scale, whose pitch is 0.5 mm. If the number of divisions on the left hand of the zero of vernier on the main scale is 38 and the 18th vernier scale division coincides with the main scale, calculate the diameter of the sphere, held in the jaws of vernier callipers.
(b) If the vernier has a negative error of 0.04 cm, calculate the corrected radius of the sphere.
A micrometre screw gauge has a negative zero error of 8 divisions. While measuring the diameter of a wire the reading on the main scale is 3 divisions and the 24th circular scale division coincides with baseline.
If the number of divisions on the main scale are 20 to a centimetre and circular scale has 50 divisions, calculate
- pitch
- observed diameter.
- least count
- corrected diameter.
Explain the term least count of a screw gauge. How are they determined?
