A vernier calipers have 10 divisions on vernier scale coincide with 9 main scale division. When there is not any object between jaws and jaws touch each other the first main scale division coincides with first vernier scale division. If one main scale division is 1 mm then the distance moved by vernier calipers till third vernier scale division coincides with third main scale division is $x \times 10^{- 1} m m$ . Then the value of $x$ is

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answer is 2.

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Detailed Solution

10 VSD=9 MSD
$\Rightarrow L . C . = 0.1 m m$

So the distance the vernier scale should move is give by vernier scale reading -Zero error.
Distance=2 x VSD=0.2 mm

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