Question

A Vernier has $10.$ division and they are equal to-$9$ divisions of main scale in length. If the main scale is calibrated in mm, what is its least count?
A. $0.1mm$
B. $0.1cm$
C. $0.9mm$
D. $0.9cm$

Answer 1

Hint: Least count of Vernier caliper is the difference between the smallest reading of the main scale and the smallest reading of the Vernier scale.
Formula used:
$LC = 1 MSD - 1 VSD$

Complete step by step answer:
 A Vernier calliper is an instrument used to measure very small length, diameter, radius etc.
The minimum reading that Vernier callipers can take is called the least count of Vernier callipers.
There are two scales in Vernier calliper.
(1) Maini Scale: It gives reading in centimetre
(2) Vernier Scale: Slides over the main scale and gives reading even less than$1$millimetre.
Least count of Vernier callipers is the difference between smallest reading in main scale and smallest reading in Vernier scale.
i.e.$L.C = 1 MSD - 1 VSD$ . . . (1)
where,$ L.C$ is least count
$MSD$ is main scale reading
$VSD$ Vernier scale reading
Now, it is given in the question that,
$10$ divisions of Vernier are equal to $9$ divisions of main scale.
I.e. $10 VSD = 9 MSD$
dividing both the sides by $10$, we get
$\dfrac{{10VSD}}{{10}} = \dfrac{{9MSD}}{{10}}$
$ \Rightarrow 1VSD = 0.9MSD$
By putting this value in equation (1), we get
$\therefore L.C. = 1MSD - 0.9MSD$
$ \Rightarrow L.C. = 0.1MSD$
According to the question, main scaler is calibrated in $mm.$
$\therefore L.C. = 0.1mm$

So, the correct answer is “Option A”.

Note:
If you have seen Vernier callipers then you already know the other such questions are better understood if we have seen or used the instruments. So, always try to learn and understand about the instruments you use in the lab.