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 :-

Question

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 :- (A) 0.02 cm (B) 0.1 mm (C) 0.025 cm (D) 0.01 mm

Tardigrade Admin · Accepted Answer

Main scale division ( MSD )=(1/10)=0.1 cm Vernier scale division ( VSD )=(2/25)=0.08 cm ∴ Least count = MSD - VSD =0.02 cm

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

0.02 cm

0.1 mm

0.025 cm

0.01 mm

Main scale division $(MS D) = \frac{1}{10} = 0.1 c m$
Vernier scale division $(V S D) = \frac{2}{25} = 0.08 c m$
$∴$ Least count $= MS D - V S D = 0.02 c m$

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