Answer
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Hint: The least count of the vernier caliper is equal to the difference in one main scale division and one vernier scale division. So by finding out the value of one vernier scale division in terms of the main scale divisions, and substituting it in the formula for the least count, we can get the least count of the given vernier caliper.
Complete step by step answer:
According to the question, $ N $ divisions on the main scale of a vernier callipers coincide with $ \left( {N + 2} \right) $ divisions on the vernier scale. So we can write the below equation
$ N{\text{ MSD}} = \left( {N + 2} \right){\text{ VSD}} $
Dividing both sides of the above equation by $ N + 2 $ , we get
$ 1{\text{ VSD}} = \dfrac{N}{{N + 2}}{\text{ MSD}} $ .....................(1)
Now, we know that the least count of a vernier caliper is equal to the difference in one main scale division and one vernier scale division, that is,
$ LC = 1{\text{ MSD}} - 1{\text{ VSD}} $
Substituting (1) in the above equation we get
$ LC = 1{\text{ MSD}} - \dfrac{N}{{N + 2}}{\text{ MSD}} $
$ \Rightarrow LC = \left( {1 - \dfrac{N}{{N + 2}}} \right){\text{ MSD}} $
On simplifying the above expression, we get
$ LC = \dfrac{2}{{N + 2}}{\text{ MSD}} $
According to the question, one main scale division is of $ 'a' $ units. Therefore substituting $ 1{\text{MSD}} = a $ in the above equation, we get
$ LC = \dfrac{{2a}}{{N + 2}} $
Thus, the least count of the given vernier caliper is equal to $ \dfrac{{2a}}{{N + 2}} $ .
Hence, the correct answer is option A.
Note:
A vernier caliper is used in the measurement of the inner dimensions, outer dimensions, and the depths of different objects. It has three parts which are used for the measurements. These are the main jaws, which are used for measuring the outer dimensions, the smaller jaws, which are used in measuring the inner dimensions, and the stem, which are used in measuring the depths.
Complete step by step answer:
According to the question, $ N $ divisions on the main scale of a vernier callipers coincide with $ \left( {N + 2} \right) $ divisions on the vernier scale. So we can write the below equation
$ N{\text{ MSD}} = \left( {N + 2} \right){\text{ VSD}} $
Dividing both sides of the above equation by $ N + 2 $ , we get
$ 1{\text{ VSD}} = \dfrac{N}{{N + 2}}{\text{ MSD}} $ .....................(1)
Now, we know that the least count of a vernier caliper is equal to the difference in one main scale division and one vernier scale division, that is,
$ LC = 1{\text{ MSD}} - 1{\text{ VSD}} $
Substituting (1) in the above equation we get
$ LC = 1{\text{ MSD}} - \dfrac{N}{{N + 2}}{\text{ MSD}} $
$ \Rightarrow LC = \left( {1 - \dfrac{N}{{N + 2}}} \right){\text{ MSD}} $
On simplifying the above expression, we get
$ LC = \dfrac{2}{{N + 2}}{\text{ MSD}} $
According to the question, one main scale division is of $ 'a' $ units. Therefore substituting $ 1{\text{MSD}} = a $ in the above equation, we get
$ LC = \dfrac{{2a}}{{N + 2}} $
Thus, the least count of the given vernier caliper is equal to $ \dfrac{{2a}}{{N + 2}} $ .
Hence, the correct answer is option A.
Note:
A vernier caliper is used in the measurement of the inner dimensions, outer dimensions, and the depths of different objects. It has three parts which are used for the measurements. These are the main jaws, which are used for measuring the outer dimensions, the smaller jaws, which are used in measuring the inner dimensions, and the stem, which are used in measuring the depths.
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