In an experiment the angles are required to be measured using an instrument. 29 divisions of the main scale exactly coincide with the 30 divisions of the Vernier scale. If the smallest division of the main scale is half a degree then the least count of the instrument is
A. One minute
B. Half a minute
C. One degree
D. Half degree

Hint: In this question, the relation between the main scale division and Vernier scale division is already given. The value of 1 Main Scale Division is also given. So, by using the two parameters, we can find out the least count using the formula for the least count
Formula used:
For solving this question, we will be using the formula for the least count in the terms of VSD and MSD, i.e.,
Least count = 1MSD − 1VSD

Complete answer:
Before we start solving the question that is given to us, let us take a look at all the parameters that have been given to us in the above question
29 divisions of main scale coincide with 30 divisions of Vernier scale
1 MSD = ${{0.5}^{\circ }}$
So,
We have
$\Rightarrow 1V.S.D=\dfrac{29}{30}M.S.D$
Also,
As we know that
Least count = 1MSD − 1VSD
So,
By using the given parameters in the above formula
We have
 Least count = $1MSD-\dfrac{29}{30}MSD$
\[\Rightarrow Least\text{ }count=\dfrac{1}{30}MSD\]
\[\Rightarrow Least\text{ }count=\dfrac{1}{30}\times {{0.5}^{\circ }}\]
\[\Rightarrow Least\text{ }count=\dfrac{1}{30}\times 30\min \]
\[\Rightarrow Least\text{ }count=1\min \]
So, If the smallest division of the main scale is half a degree then the least count of the instrument is 1 min.

So, the correct answer to this question is Option – A, i.e., One Minute .

Note:
A Vernier scale is a visual aid that uses mechanical interpolation to take an exact measurement reading between two graduation marks on a linear scale, thus increasing precision and reducing measurement uncertainty by using Vernier acuity to minimise human error in estimation.