One centimeter on the main scale of Vernier callipers is divided into ten equal parts. If 10 divisions of Vernier scale coincide with 8 small divisions of the main scale, the least count of the callip

Question

One centimeter on the main scale of Vernier callipers is divided into ten equal parts. If 10 divisions of Vernier scale coincide with 8 small divisions of the main scale, the least count of the callipers is :

TestHub · Accepted Answer

The least count of Vernier callipers is the difference between one main scale division (MSD) and one Vernier scale division (VSD).

Given:

1 MSD = 1/10 cm = 0.1 cm.

10 VSD = 8 MSD.

Therefore, 1 VSD = $\frac{8}{10}$ MSD = $\frac{8}{10} \times 0.1$ cm = 0.08 cm.

Least Count = 1 MSD - 1 VSD = 0.1 cm - 0.08 cm = 0.02 cm.

Therefore, the least count of the callipers is 0.02 cm.

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