A vernier callipers has $1$ mm 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 C $0.2 m m$
Given that,
$20 V S D = 16 M S D ∴ 1 V S D = \frac{16}{20} M S D$
$∴$ LC=Length of 1 MSD-length of $1 V S D$
$= 1 m m - (\frac{16}{20}) 1 m m = 0.2 m m$

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A Vernier caliper has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which matches with 16 main scale divisions. For this Vernier caliper, the least count is:

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A vernier callipers has 1 mm 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

The correct option is C 0.2mmGiven that,20VSD=16MSD∴1VSD=1620MSD∴ LC=Length of 1 MSD-length of 1VSD=1mm−(1620)1mm=0.2mm

A vernier callipers has $1$ mm 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

The correct option is C $0.2 m m$
Given that,
$20 V S D = 16 M S D ∴ 1 V S D = \frac{16}{20} M S D$
$∴$ LC=Length of 1 MSD-length of $1 V S D$
$= 1 m m - (\frac{16}{20}) 1 m m = 0.2 m m$