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प्रश्न
Fig., below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on the main scale when circular head is rotated once.
Find:
(i) Pitch of the screw gauge,
(ii) Least count of the screw gauge and
(iii) The diameter of the wire.
उत्तर
No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/1 = 1 mm.
(ii) L.C. = Pitch/No. of divisions on the circular head
= (1/50) mm
= 0.02 mm
(iii) Main scale reading = 4 mm
No. of circular division coinciding with m.s.d. (p) = 47
Circular scale reading = p × L.C.
= (47 × 0.02) mm
= 0.94 mm
Diameter (Total reading) = M.s.r. + circular scale reading
= (4 + 0.94) mm
= 4.94 mm
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संबंधित प्रश्न
Explain the terms : Pitch .
How are they determined?
In the vernier callipers, there are 10 divisions on the vernier scale and 1 cm on the main scale is divided into 10 parts. While measuring the length, the zero of the vernier lies just ahead of the 1.8 cm mark and the 4th division of vernier coincides with a main scale division.
(a) Find the length.
(b) If zero error of vernier callipers is -0.02 cm,
What is the correct length ?
The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions.
(i) What is the pitch of the screw gauge?
(ii) What is the least count of the screw gauge?
Define the term least count as applied to a vernier callipers.
State the correction if the positive error is 7 divisions when the least count is 0.01 cm.
Figure shows a screw gauge in which circular scale has 100 divisions. Calculate the least count and the diameter of a wire.
