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  4. In a vernier calliper, 10 divisions on main scale equals 1 cm 25 vernier scale divisions equals 2 cm then find least count of vernier calliper :-

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Q. In a vernier calliper, $10$ divisions on main scale equals $1\, cm \& 25$ vernier scale divisions equals $2\, cm$ then find least count of vernier calliper :-

Physical World, Units and Measurements

Solution:

Main scale division $( MSD )=\frac{1}{10}=0.1\, cm$
Vernier scale division $( VSD )=\frac{2}{25}=0.08 \,cm$
$\therefore $ Least count $= MSD - VSD =0.02 \,cm$