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.

Hint :
In order to solve the above problem, first we have to calculate the length of 1 division of the Vernier scale and main scale.
After then putting the values in least count expression, which is given as
Least count $ = $ value of one main scale division $ - $ value of one Vernier scale division

Complete step by step solution :
Given that the main scale of Vernier callipers is graduated in mm.
So, the value of one main scale division is $ = $ 1 mm
Also given that
10 division of Vernier scale $ = $ 9 mm
So, 1 division of Vernier scale $ = \dfrac{9}{{10}}mm$
We know that
Least count $ = $ value of one main scale division $ - $ value of one Vernier scale division
$L.C. = 1 - \dfrac{9}{{10}}$
$L.C. = \dfrac{{10 - 9}}{{10}}$
$L.C. = \dfrac{1}{{10}} = 0.1mm$
$L.C. = 0.1 \times {10^{ - 1}}cm$
$L.C. = {10^{ - 1}} \times {10^{ - 1}}cm$
$L.C. = {10^{ - 2}}cm$ …..(1)
Given that the least count is ${10^{ - x}}$ cm.
Hence, on comparing with equation 1, we get
$x = 2$

Note :
In many problems of Vernier scale instruments, students may get confused between the least count of main scale and Vernier scale of Vernier callipers.
i.e., Least count of Vernier scale $ = 0.01cm$
Least count of main scale $ = 0.1cm$