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प्रश्न
The circular scale of a screw gauge has 100 divisions. Its spindle moves forward by 2.5 mm when given five complete turns. Calculate
- pitch
- least count of the screw gauge
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उत्तर
Number of circular scale divisions = 100
Distance moved by spindle (screw) = 2.5 mm
No. of complete rotations given = 5
(1) Pitch = `"Distance moved by screw on sleeve"/"No. of complete rotations"`
= `2.5/5`
= 0.5 mm
= 0.05 cm
(2) Least count = `"Pitch"/"No. of circular scale divisions"`
= `0.05/100` cm
= 0.0005 cm
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संबंधित प्रश्न
State one use of a screw gauge.
(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.
State the correction if the negative error is 7 divisions when the least count is 0.01 cm.
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.
What do you understand by the following term as applied to micrometre screw gauge?
Baseline
What is the zero error of a screw gauge?
The precision of vernier calipers is ______ mm.
A Vernier calliper reads a main scale reading of 2.3 cm and the 5th division of the Vernier scale coincides with a main scale division. If the least count is 0.01 cm, what is the actual reading?
