Question

Each division on the main scale is $1mm$. Which of the following vernier scales give vernier constant equal to$0.01mm$?
A. \[9mm\] divided into $10$ divisions.
B. \[90mm\] divided into 100 divisions.
C. \[99mm\] divided into 100 divisions.
D. \[9mm\] divided into 100 divisions.

Answer 1

Hint: Vernier scale is an instrument used to measure the length accurately and reduce human error. A vernier scale has two scales ,one is the main scale above which a vernier scale slides. The vernier constant of a vernier scale is defined by the difference between main scale division and vernier scale division. It is also known as the least count of vernier calliper.

Formulas used:
$\text{Least count=}\dfrac{\text{value}\ \text{of}\ \text{main}\ \text{scale}\ \text{division}}{\text{number}\ \text{of}\ \text{marks}\ \text{of}\ \text{vernier}\ \text{scale}}$
\[\begin{align}
  & \text{Vernier constant = one main scale division }-\text{ one vernier scale division}\text{.} \\
 & \text{i}\text{.e}\text{. LC=1MSD}-\text{1VSD} \\
\end{align}\]

Complete step by step answer:
Vernier constant or least count is given by the difference between the main scale division i.e.
$LC=1MSD-1VSD$.
Given each division on main scale is $1mm$. i.e. $1MSD=1mm$
Vernier constant or least count is $0.01mm$. According to formula.
 $\begin{align}
  & LC=1MSD-1VSD \\
 & \Rightarrow 1VSD=1MSD-LC=0.1mm-0.01mm=0.99mm \\
\end{align}$
So the vernier scale of these callipers should have $1VSD=0.99mm$.
For option A. \[9mm\] divided into $10$ divisions. So $10$ vernier scale division is equal to \[9mm\].
i.e. $10VSD=9mm\Rightarrow 1VSD=\dfrac{9}{10}=0.9mm$
So in option A. the vernier scale division is $0.9mm$. So it is not the correct option.
For option B. \[90mm\] divided into 100 divisions. So $100$ vernier scale division is equal to \[90mm\].
i.e. $100VSD=90mm\Rightarrow 1VSD=\dfrac{90}{100}=0.9mm$
So in option B. the vernier scale division is $0.9mm$. So it is not the correct option.
For option C. \[99mm\] divided into 100 divisions.So $100$ vernier scale division is equal to\[99mm\].
i.e. $100VSD=99mm\Rightarrow 1VSD=\dfrac{99}{100}=0.99mm$
So in option B the vernier scale division is $0.99mm$. Which is our required vernier scale division. So option C. is correct.
For option D. \[9mm\] divided into 100 divisions. So $100$ vernier scale division is equal to \[9mm\].
i.e. $100VSD=9mm\Rightarrow 1VSD=\dfrac{9}{100}=0.09mm$
So in option D. the vernier scale division is $0.09mm$. So it is not the correct option.
So the only correct option is option C.

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

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 instrument i.e., vernier caliper has an error called zero error. It may be positive or negative. If the experimental value is less than the accepted value, the error is negative. If the experimental value is larger than the accepted value, the error is positive.