Question

In an experiment the angles required to be measured using an instrument. 29 divisions of the main scale exactly coincide with 30 divisions of the vernier scale. If the smallest division of the main scale is half-a-degree $ \left( {0.5^\circ } \right) $, then the least count of the instrument is:
(A) One minute
(B) Half minute
(C) One degree
(D) Half degree

Answer 1

Hint
From the given values in the question, we can find the value of one vernier scale division in terms of the main scale divisions. Then the least count will be the difference between one main scale and one vernier scale. Then by multiplying the least count to the smallest division of the main scale, we get the least count.
Formula Used: In the solution, we will be using the following formula,
 $\Rightarrow L.C. = 1MSD - 1VSD $
where $ L.C. $ is the least count,
 $ MSD $ is the main scale division and $ VSD $ is the vernier scale division.

Complete step by step answer
In the question we are given that 29 divisions of the main scale exactly coincide with 30 divisions of the vernier scale. So we can write that,
 $\Rightarrow 30VSD = 29MSD $
Now we can divide both the sides of the above equation by 30. Then we can find the value of 1 vernier scale division. So we get,
 $\Rightarrow 1VSD = \dfrac{{29}}{{30}}MSD $
Now the least count of the instrument is given by the formula,
 $\Rightarrow L.C. = 1MSD - 1VSD $
We can substitute the value of 1 vernier scale division in the above formula, and hence we get,
 $\Rightarrow L.C. = 1MSD - \dfrac{{29}}{{30}}MSD $
Therefore we can take the 1MSD common in the RHS of the equation,
 $\Rightarrow L.C. = \left( {1 - \dfrac{{29}}{{30}}} \right)MSD $
On calculating we get,
 $\Rightarrow L.C. = \left( {\dfrac{{30 - 29}}{{30}}} \right)MSD $
Now in the question we are given that the value of 1 smallest main scale division is half a degree. Hence we have the value of 1 main scale division as 30 minutes as 1 degree is 60 minutes.
Therefore on substituting we get,
 $\Rightarrow L.C. = \dfrac{1}{{30}} \times 30{\text{mins}} $
On cancelling the 30, we have the least count of the instrument as
 $\Rightarrow L.C. = 1{\text{min}} $
Hence the least count is one minute. So the correct answer is option A.

Note
The smallest value that can be measured using the measuring instrument is called the least count of that instrument. It is also related to the precision of an instrument, as the instrument having the smallest least count value can measure smaller changes relative to the instrument which has a larger value of the least count.