Advertisements
Advertisements
Question
In a vernier callipers, 19 main scale divisions coincide with 20 vernier scale divisions. If the main scale has 20 divisions in a centimetre, calculate
- pitch
- L.C. of vernier callipers.
Advertisements
Solution
Main scale divisions of vernier callipers in one centimetre = 20 Unit
Pitch = `"Unit of main scale"/"Number of divisions in the unit"`
= `1/20` cm
= 0.05 cm
Least count = `"Pitch"/"No. of vernier scale divisions"`
= `0.05/20`
= 0.0025 cm
APPEARS IN
RELATED QUESTIONS
Explain the terms : Least count of a screw gauge.
How are they determined?
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.
The main scale of vernier callipers has 10 divisions in a centimetre and 10 vernier scale divisions coincide with 9 main scale divisions. Calculate
- pitch
- L.C. of vernier callipers.
Which part of vernier callipers is used to measure the internal diameter of a hollow cylinder?
Figure shows a screw gauge in which circular scale has 200 divisions. Calculate the least count and radius of the wire.
How do you account for negative zero error, for calculating the correct diameter of wires?
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.
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.
