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प्रश्न
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उत्तर
1 MSD = 1 mm
10 VSD = 9 mm
1 VSD =`9/10`mm
Least count = 1 MSD - 1 VSD
=`1/10`MSD
=`1/10` x 1 mm = 0.1 mm
= 0.01 cm
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संबंधित प्रश्न
Define the term 'Vernier constant'.
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.
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
Is it possible to increase the degree of accuracy by mathematical manipulations? Support your answer by an example.
Up to how many decimal places can a common vernier callipers measure the length in cm?
What is the function of ratchet in the screw gauge?
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.
In a vernier calliper, (N + 1) divisions of the vernier scale coincide with N divisions of the main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is ______.
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.
