Advertisements
Advertisements
प्रश्न
A screw gauge has 50 divisions on its circular scale and its screw moves by 1 mm on turning it by two revolutions. When the flat end of the screw is in contact with the stud, the zero of the circular scale lies below the base line and 4th division of the circular scale is in line with the base line. Find
(i) The pitch,
(ii) The least count and
(iii) The zero error of the screw gauge
Advertisements
उत्तर
No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/2 = 0.5 mm
(ii) L.C. = Pitch/No. of divisions on the circular head
= (0.5/50) mm
= 0.01 mm
(iii) Because the zero of the circular scale lies below the base line, when the flat end of the screw is in contact with the stud, the error is positive.
No. of circular division coinciding with m.s.d. = 4
Zero error = + (4 × L.C.)
= + (4 × 0.01) mm
= + 0.04 mm
APPEARS IN
संबंधित प्रश्न
Explain the terms : Least count of a screw gauge.
How are they determined?
State one use of a screw gauge.
What do you understand by the following term as applied to screw gauge?
Zero error
Name the measuring employed to measure the diameter of a needle.
State whether the following statement is true or false by writing T/F against it.
A metre scale can measure a length of 6.346 cm.
State whether the following statement is true or false by writing T/F against it.
The metre scale, vernier calipers, and screw gauge are in decreasing order of least count.
Define the least count of an instrument.
Thickness of a cricket ball is measured by ______.
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?
