Advertisements
Advertisements
Question
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.
Advertisements
Solution
(i) Value of 1 m.s.d. =1 mm
10 vernier divisions = 9 m.s.d.
L.C. = Value of 1 m.s.d./number of divisions on the vernier scale
= 1 mm/10
= 0.1 mm or 0.01 cm
(ii) On bringing the jaws together, the zero of the vernier scale is ahead of zero of the main scale, the error is positive.
3rd vernier division coincides with a main scale division.
Total no. of vernier divisions = 10
Zero error = +3 × L.C.
= +3 × 0.01 cm
= +0.03 cm
APPEARS IN
RELATED QUESTIONS
What is meant by zero error of vernier callipers ? How is it determined? Draw neat diagrams to explain it. How is it taken in account to get the correct measurement?
A pendulum completes 2 oscillations in 5 s. What is its time period? If g = 9.8 m s-2, find its length.
What is the need for measuring length with vernier callipers?
When does a vernier callipers has the negative error?
Figure shows a screw gauge in which circular scale has 200 divisions. Calculate the least count and radius of the wire.

What do you understand by the following term as applied to screw gauge?
Positive zero error
Find the volume of a book of length 25 cm, breadth 18 cm and height 2 cm in m3.
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.
What is the least count in the case of the following instrument?
Spring balance
