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प्रश्न
A screw has a pitch equal to 0.5 mm. What should be the number of divisions on its head so as to read correctly up to 0.001 mm with its help?
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उत्तर
Pitch of the screw gauge = 0.5 mm
L.C. of the screw gauge = 0.001 mm
No. of divisions on circular scale = Pitch / L.C.
= 0.5 / 0.001
= 500
संबंधित प्रश्न
State the purpose of ratchet in a screw gauge.
A screw gauge has a least count 0.001 cm and zero error +0.007 cm. Draw a neat diagram to represent it.
A vernier callipers has its main scale graduated in mm and 10 divisions on its vernier scale are equal in length to 9 mm. When the two jaws are in contact, the zero of the vernier scale is ahead of the zero of the main scale and the 3rd division of the vernier
scale coincides with a main scale division.
Find : (i) The least count and
(ii) The zero error of the vernier callipers.
The pitch of a screw gauge is 1 mm and the circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on the circular scale coincides with the base line. Find
(i) The least count and
(ii) The diameter of the wire.
Define the term least count as applied to a vernier callipers.
Figure shows a screw gauge in which circular scale has 100 divisions. Calculate the least count and the diameter of a wire.

Consider the following case where the zero of vernier scale and the zero of the main scale are clearly seen. If L.C. of the vernier calipers is 0.01 cm, write the zero error and zero correction of the following.

How will you measure the least count of vernier caliper?
