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प्रश्न
Figure shows a screw gauge in which circular scale has 200 divisions. Calculate the least count and radius of the wire.
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उत्तर
No. of circular scale divisions = 200
Pitch = 1 mm
Least count (L.C.) = `"Pitch"/"No. of circular divisions"`
= `(1 "mm")/200`
= 0.005 mm
L.C. = 0.0005 cm
Main scale reading = 5 mm = 0.5 cm
(C.S.D.) circular scale reading = 34 divisions
Observed diameter of wire = Main scale reading + L.C. × C.S.D.
= 0.5 + 0.0005 × 34
= 0.5 + 0.0170
= 0.5170 cm
Redius of wire = `"diameter"/2`
= `0.5170/2`
= 0.2585 cm
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संबंधित प्रश्न
How can the least count of a screw gauge be decreased?
(a) A vernier scale has 10 divisions. It slides over the main scale, whose pitch is 1.0 mm. If the number of divisions on the left hand of zero of the vernier scale on the main scale is 56 and the 8th vernier scale division coincides with the main scale, calculate the length in centimetres.
(b) If the above instrument has a negative error of 0.07 cm, calculate the corrected length.
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 screw gauge?
Negative zero error
What is the least count in the case of the following instrument?
vernier calipers
The main scale reading while measuring the thickness of a rubber ball using Vernier caliper is 7 cm and the Vernier scale coincidence is 6. Find the radius of the ball.
