Question: The main scale of a vernier calliper has $n$ divisions per cm. $n$ divisions of the vernier scale coincide with $(n-1)$ divisions of the main scale. The least count of the vernier callipers is:

a. $\dfrac{1}{(n+1)(n-1)}\ \text{cm}$

b. $\dfrac{1}{n}\ \text{cm}$

c. $\dfrac{1}{n^{2}}\ \text{cm}$

d. $\dfrac{1}{n(n+1)}\ \text{cm}$

Answer. c. $\dfrac{1}{n^{2}}\ \text{cm}$

Solution. The main scale has $n$ divisions per cm, therefore

$
1\ \text{MSD} = \frac{1}{n}\ \text{cm}.
$

Since $n$ vernier divisions coincide with $(n-1)$ main scale divisions,

$
1\ \text{VSD} = \frac{n-1}{n}\ \text{MSD}.
$

The least count is

$
\text{LC} = 1\ \text{MSD} – 1\ \text{VSD}
= \text{MSD}\left(1-\frac{n-1}{n}\right)
= \frac{1}{n}\cdot \text{MSD}.
$

Substituting $\text{MSD}=\tfrac{1}{n}\ \text{cm}$,

$
\text{LC} = \frac{1}{n}\cdot \frac{1}{n} = \frac{1}{n^{2}}\ \text{cm}.
$

Hence, the least count of the vernier callipers is $\dfrac{1}{n^{2}}\ \text{cm}$.