A Vernier calliper 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 Vernier scale is ahead of zero of the main scale and 3rd division of Vernier scale coincides with a main scale division. The least count is ${10}^{-x}$cm. Find the value of x.

We must first determine the length of 1 division on both the Vernier scale and the main scale in order to solve the aforementioned problem.
The values are then entered into the following least-count expression:

Least count = value of one main scale division − value of one Vernier scale division

A major scale division's value = 1mm

10 vernier scale divisions = 9mm

1 division of vernier scale = $\frac{9}{10}mm$

You may compute the least count by using,

one major scale division's value - The worth of one vernier scale division

$LC=1-\frac{9}{10}$

$LC=\frac{10-9}{10}$

$LC=\frac{1}{10}=0.1mm$

$LC=0.1\times {10}^{-1}cm$

$LC={10}^{-1}\times {10}^{-1}cm$

$LC={10}^{-2} cm$

Thus least count

 $$\begin{gathered} {10}^{-x} \\ \Rightarrow {10}^{-2} \\ \Rightarrow x=2 \end{gathered}$$

Hence option B is the correct answer.