Question

If in a vernier caliper 10 VSD coincides with 8 MSD, then the least count of vernier caliper is (given 1 MSD = 1 mm)

Answer 1

Step 1: Given data:

In the vernier callipers

$10$ VSD= $8$ MSD

Where VSD= Vernier scale division

And, MSD= Main scale division

$1$ MSD = $1mm$

Step 2: Calculation of the least count of the vernier callipers:

The least count of vernier callipers or vernier constant is given by;

Least count= Vernier constant = $1$ MSD $-1$ VSD

Since,

$10$ VSD= $8$ MSD

$1$ VSD= $\frac{8}{10}$ MSD

$1$ VSD= $0.8$ MSD

Now,

Least count= $1$ MSD $-0.8$ MSD

= $0.2$ MSD

But, $1$ MSD= $1mm$

Therefore the least count of the vernier callipers will be

= $0.2\times 1mm$

= $0.2mm$

Thus,

The least count of the vernier callipers is $0.2mm$