Question

1 cm on the main scale of Vernier callipers is divided into 10 equal parts, if 10 vernier scales coincide with 8 small divisions of main scale, then the least count of the calliper is
(A) \[0.01cm\]
(B) \[0.02cm\]
(C) \[0.05cm\]
(D) \[0.005cm\]

Answer 1

Hint: The least count is directly related to the width of each division of the main scale. 10 divisions of the Vernier scale coincides with 8 divisions of the main scale, the least count is proportional to ratio of the difference to the number of divisions of the Vernier scale.

Formula used: In this solution we will be using the following formulae;
\[MR = \dfrac{L}{N}\] where \[MR\] signifies minimum reading of the main scale, \[L\] is the length of a section of the main scale and \[N\] is the number of division in that section.
\[LC = \dfrac{{MR}}{n}\] where \[LC\] is the least count (without coincidence error)\[n\] is the number of divisions on the Vernier scale

Complete Step-by-Step solution:
To calculate the least count, we find the smallest reading on the main scale. This is equal to the length of a particular section of the Vernier calliper’s main scale divided by the number of divisions of that section. In the question, we are told a 1 cm section is divided into 10 divisions, i.e.
$\Rightarrow$ \[MR = \dfrac{L}{N}\] where \[MR\] signifies minimum reading of the main scale, \[L\] is the length of a section of the main scale and \[N\] is the number of division in that section.
Hence, we have
$\Rightarrow$ \[MR = \dfrac{{1cm}}{{10}} = 0.1cm\]
Least count can be defined as
$\Rightarrow$ \[LC = \dfrac{{MR}}{n}\] where \[n\] is the number of divisions on the Vernier scale
$\Rightarrow$ \[LC = \dfrac{{0.1cm}}{{10}} = 0.01cm\]
Nonetheless, the least count will be increased due to the “coincidence error”. Only eight divisions coincide with 10 divisions, hence, coincidence error is \[10 - 8 = 2\]
Hence, true \[LC\] would be
\[LC = 0.01cm \times 2 = 0.02cm\]

Hence, the correct option is B,

Note: Alternatively, we can simply use the relation
\[LC = MR - M{r_v}\] where \[M{r_v}\] is the minimum reading of the Vernier scale.
Since only 8 coincides, then only 8 can be read, then \[M{r_v}\] would be
\[M{r_v} = \dfrac{{8mm}}{{10}} = 0.8mm = 0.08cm\]
Hence,
\[LC = MR - M{r_v} = 0.1cm - 0.08cm = 0.02cm\]