Question

The total number of division of a vernier scale in slide calipers is 20.The value of the smallest division in main scale is 1 mm. What is the vernier constant?

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Answer 1

Solution:

The total number of Vernier Divisions = 20, so it will be equal to (20-1) = 19 Main Scale divisions.

So, we can say :

$20 \: VSD = 19 \: MSD$

$\implies 1\: VSD = \dfrac{19}{20} \: MSD$

Now, Vernier Constant is :

$VC = 1 \: MSD - 1 \: VSD$

$\implies V C = 1 \: MSD - \dfrac{19}{20} \:MSD$

$\implies V C = \dfrac{20 - 19}{20} \:MSD$

$\implies V C = \dfrac{1}{20} \:MSD$

$\implies V C = \dfrac{1}{20} \times 1 \: mm$

$\implies V C = 0.05 \: mm$

So, Vernier Constant is 0.05 mm.

Answer 2

Explanation:

Solution:

The total number of Vernier Divisions = 20, so it will be equal to (20-1) = 19 Main Scale divisions.

So, we can say :

20 \: VSD = 19 \: MSD20VSD=19MSD

\implies 1\: VSD = \dfrac{19}{20} \: MSD⟹1VSD=

20

19

MSD

Now, Vernier Constant is :

VC = 1 \: MSD - 1 \: VSDVC=1MSD−1VSD

\implies V C = 1 \: MSD - \dfrac{19}{20} \:MSD⟹VC=1MSD−

20

19

MSD

\implies V C = \dfrac{20 - 19}{20} \:MSD⟹VC=

20

20−19

MSD

\implies V C = \dfrac{1}{20} \:MSD⟹VC=

20

1

MSD

\implies V C = \dfrac{1}{20} \times 1 \: mm⟹VC=

20

1

×1mm

\implies V C = 0.05 \: mm⟹VC=0.05mm

So, Vernier Constant is 0.05 mm.