Question

A Vernier caliper has \[1mm\]marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16 main scale divisions. For this Vernier calipers, the least count is-
\[\begin{array}{*{35}{l}}
   A.\text{ }0.02mm \\
   B.\text{ }0.05mm \\
   C.\text{ }0.1mm \\
   D.\text{ }0.2mm \\
\end{array}\]

Answer 1

Hint: Vernier callipers is an instrument used to measure the length accurately reducing human error. A vernier calliper has two scales one is the main scale above which a vernier scale slides. The least count of a vernier calliper is the smallest measurement that can be made by the vernier callipers and is given by the difference between main scale division and vernier scale division.

Formulas used:
 $\text{Least count}=\dfrac{\text{smallest reading or division of}\ \text{main}\ \text{scale}}{\text{number}\ \text{of}\ \text{marks}\ \text{or divisions of}\ \text{vernier}\ \text{scale}}$
It may be also defined as
 \[\begin{align}
  & \text{Least count = one main scale division }-\text{ one vernier scale division}\text{.} \\
 & \text{i}\text{.e}\text{. LC=1MSD}-\text{1VSD} \\
\end{align}\]

Complete answer:
Given that 20 divisions of vernier scale is equal to 16 divisions of vernier scale.
i.e
 $\begin{align}
  & 20VSD=16MSD \\
 & \Rightarrow 1VSD=\dfrac{16}{20}MSD \\
\end{align}$
The Least count of the vernier calipers is given by the difference between one main scale division and one vernier scale division. So,
\[\begin{align}
  & \text{LC=1MSD}-\text{1VSD=1MSD}-\dfrac{16}{20}\text{MSD}\left( \because 1VSD=\dfrac{16}{20}MSD \right) \\
 & \Rightarrow \text{LC}=\left( 1-\dfrac{16}{20} \right)\text{MSD}=\left( \dfrac{20-16}{20} \right)\text{MSD} \\
 & \Rightarrow \text{LC}=\dfrac{4}{20}\text{MSD=0}\text{.2}\times \text{1}mm=0.2mm\left( \because 1MSD=1mm \right) \\
 & \Rightarrow \text{LC=0}\text{.2}mm \\
\end{align}\]

So the correct option is D.

Additional Information:
Least count can also be defined by the minimum measurement that can be measured correctly by an instrument. And is given by
$\text{Least count}=\dfrac{\text{smallest reading or division of}\ \text{main}\ \text{scale}}{\text{number}\ \text{of}\ \text{marks}\ \text{or divisions of}\ \text{vernier}\ \text{scale}}$.

Note:
Note that the instrument which has the lowest least count can give more accurate results. You can also measure volumes of some solids using vernier calipers.
Zero error: If the zero of the vernier scale doesn't coincide with the zero of the main scale, then the vernier caliper has an error called zero error. It may be positive or negative. The error is negative if the experimental value is less than the accepted value and the error is positive if the experimental value is more than the accepted value.
To remove zero error we have to subtract the zero error calculated from the experimental value to get the value without zero error.