A vernier calipers has $1 m m$ marks on the main scale. It has $20$ equal divisions on the Vernier scale which match with $16$ main scale divisions. For this Vernier calipers, the least count is

$0.02 m m$

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$0.05 m m$

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$0.1 m m$

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$0.2 m m$

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Solution

The correct option is D
$0.2 m m$

Step 1. Given data:
Main scale reading $M S D = 1 m m$
Number of divisions on vernier caliper corresponding $16 M S D$
Step 2. Finding the Least count of vernier caliper:
The least count of vernier caliper is the difference between one main scale division and one vernier scale division,
$L C = 1 M S D - 1 V S D$
It is given that , $20 V S D = 16 M S D$
$1 V S D = \frac{16 M S D}{20}$
$\Rightarrow L C = 1 m m - \frac{16 m m}{20}$
$\Rightarrow L C = 0.2 m m$
Hence, option D is correct

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A vernier calipers has $1 m m$ marks on the main scale. It has $20$ equal divisions on the Vernier scale which match with $16$ main scale divisions. For this Vernier calipers, the least count is