In a vernier callipers 10 VSD = 9 MSD and zero mark of vernier scale lies to the right of zero mark of main scale and to the left of first smallest division with $4^{th}$ mark of vernier division coinciding with a mark on main scale division when jaws of the callipers are in contact with each other. Then zero error of the vernier callipers is (take 1 smallest division on main scale as 1mm)

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- 0.2 mm

+ 0.4 mm

- 0.04 mm

+ 0.02 mm

answer is B.

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Detailed Solution

Here the zero error is positive error

Least count of the vernier scale $= (1 - \frac{9}{10}) mm = 0.1 mm$

$∴$ Zero error = 4 x 0.1 mm = + 0.4 mm

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