Question

$1$ cm of the main scale of a vernier calliper is divided into $10$ divisions. The least count of the callipers is $0.005cm$, then the vernier scale must have:

• A. 10 divisions
• B. 20 divisions
• C. 25 divisions
• D. 50 divisions

Answer 1

The correct option is B 20 divisions

Each division on MSR=1/10=0.1 cmMSR=110=0.1 cm

Let N be the number of division that must present on vernier scale so that each division on MSR corresponds to the maximum value that vernier scale can measure.

N×0.005 cm=0.1 cmN×0.005 cm=0.1 cm

N=20N=20 Divisions