Question

In a Vernier caliper, N divisions of vernier scale coincide with (N-1) divisions of main scale (in which one division represents 1 mm). The least count of the instrument (in cm) should be :
A. $N$
B. $N - 1$
C. $\dfrac{1}{{10N}}$
D. $\dfrac{1}{{N - 1}}$

Answer 1

Hint: Generally for normal measurements daily we will be using metre scale. If we want to measure things which are very minute we use vernier calipers and for things smaller than that we will be using screw gauge. Here we calculate what is 1 main scale division(1MSD) and what is one vernier scale division and we subtract 1VSD from 1MSD to get the least count.

Formula used:
Least count(LC) = 1MSD – 1VSD
1 cm = 10 mm

Complete answer:
It is given that N divisions of vernier scale coincide with N – 1 divisions of main scale initially when both jaws are meeting
From this information we can understand that
$\eqalign{
  & N(VSD) = N - 1(MSD) \cr
  & 1VSD = \dfrac{{N - 1}}{N}(MSD) \cr
  & 1VSD = 1 - \dfrac{1}{N}(MSD) \cr} $
We got the value of 1 VSD
Now Least count(LC) = 1MSD – 1VSD
Least count will be
$\eqalign{
  & 1(MSD) - [1 - \dfrac{1}{N}](MSD) \cr
  & \dfrac{1}{N}(MSD) \cr
  & \cr} $
1MSD is given is 1mm
But we have to find out least count in cm
We have
$\eqalign{
  & 1cm = 10mm \cr
  & 1mm = \dfrac{1}{{10}}cm \cr} $
So
$\dfrac{1}{N}(MSD) = \dfrac{1}{N}mm = \dfrac{1}{N} \times \dfrac{1}{{10}}cm = \dfrac{1}{{10N}}cm$

So, the correct answer is “Option C”.

Additional Information:
Least count of meter scale would be generally 1mm and least count of vernier calipers would be generally 0.1mm and least count of screw gauge would be 0.01. Just as vernier calipers have two scales, the screw gauge also has two scales that are circular scale and pitch scale.

Note:
In some cases there might be zero errors for vernier calipers. These are also called systematic errors and these are one directional errors only. These will increase the actual value or decrease the actual value. Error which increases is called positive zero error and can be corrected by subtraction of error at end and the other is negative zero error and can be corrected by error addition at the end.