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प्रश्न
The circular scale of a screw gauge has 50 divisions. Its spindle moves by 2 mm on the sleeve, when given four complete rotations calculate
- pitch
- least count
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उत्तर
Number of circular scale divisions (C.S.D.) = 50
Distance moved by screw (spindle) on sleeve = 2 mm
Number of complete rotations given = 4
(1) Pitch = `"Distance moved by screw on sleeve"/"No. of complete rotations"`
= `(2 "mm")/4`
= 0.5 mm
(2) Pitch = 0.05 cm
Least count = `"Pitch"/"No. of circular scale divisions"`
= `0.05/50`
= 0.001 cm
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संबंधित प्रश्न
Define the term 'Vernier constant'.
When a screw gauge with a least count of 0.01 mm is used to measure the diameter of a wire, the reading on the sleeve is found to be 1 mm and the reading on the thimble is found to be 27 divisions.
(i) What is the diameter of the wire in cm?
(ii) If the zero error is +0.005 cm, what is the correct diameter?
State the formula for determining least count for a vernier callipers.
A micrometre screw gauge having a positive zero error of 5 divisions is used to measure diameter of a wire, when reading on the main scale is 3rd division and the 48th circular scale division coincides with baseline. If the micrometer has 10 divisions to a centimetre on the main scale and 100 divisions on a circular scale, calculate
- Pitch of screw
- Least count of screw
- Observed diameter
- Corrected diameter.
Define the least count of an instrument.
How will you measure the least count of vernier caliper?
On a vernier calliper, 25 vernier scale divisions are equal in length to 24 main scale divisions. One main scale division is 1 mm. Find the least count of the instrument.
