If the length of a vernier scale having $25$ divisions corresponding to $24$ main scale divisions end given that $1 M S D = 1 m m$ , then the least count of vernier calipers is

0.004 c m

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4 μ m

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0.04 m m

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0.04 m

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Solution

The correct options are
B $0.004 c m$
C $0.04 m m$
Least count, $L . C = 1 M S D - 1 V S D$ where $M S D =$ main scale division and $V S D =$ vernier scale division.
Here, $25 V S D = 24 M S D$ or $1 V S D = (24 / 25) M S D$
Thus, $L . C = 1 M S D - (24 / 25) M S D = (1 / 25) M S D$
Given, $M S D = 1 m m$ so $L . C = 0.04 \times 1 m m = 0.04 m m = 0.004 c m$

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If the length of a vernier scale having 25 divisions corresponding to 24 main scale divisions end given that 1MSD=1mm, then the least count of vernier calipers is

The correct options are B 0.004cm C 0.04mmLeast count, L.C=1MSD−1VSD where MSD= main scale division and VSD=vernier scale division.Here, 25VSD=24MSD or 1VSD=(24/25)MSDThus, L.C=1MSD−(24/25)MSD=(1/25)MSDGiven, MSD=1mm so L.C=0.04×1mm=0.04mm=0.004cm

If the length of a vernier scale having $25$ divisions corresponding to $24$ main scale divisions end given that $1 M S D = 1 m m$ , then the least count of vernier calipers is