A vernier has 10. divisions and they are equal to 9 divisions of main scale in length. If the main scale is calibrated in mm, what is its least count?

0.1 mm

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0.1 cm

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0.9 mm

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0.9 cm

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Solution

The correct option is A 0.1 mm
Least count can calculated using the formula.
$L e a s t C o u n t = L e n g t h o f 1 M S D - L e n g t h o f I V S D$
Where $M S D$ and $V S D$ refer to Main Scale Division and Vernier Scale Division respectively. Here 10 divisions on main scale coincide with 9 divisions on vernier scale.
So, $1 V S D$ is equivalent to $0.9$ main scale divisions.
Using the formula for Least Count, we get $L C = 1 M S D - 1 V S D = 1 M S D - 0.9 M S D = 0.1 M S D = 0.1 \times 1 m m = 0.1 m m$
As Given, $1 M S D = 1 m m$

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