In a vernier callipers, 20 VSD coincide with 16 MSD (each division of length 1 mm). The least count of the vernier callipers is: ____.

In a vernier caliper the least count is the difference between one main scale division (MSD) and one vernier scale division (VSD).

Given:

• Length of one MSD = 1 mm

• 20 VSD coincide with 16 MSD

First find the length represented by one VSD.

$length\backslash \; of\backslash \; 1\backslash \; VSD\; =\; \backslash frac\{16\backslash \; MSD\}\{20\backslash \; VSD\}\; =\; \backslash frac\{16\; \backslash times\; 1\backslash ;mm\}\{20\}\; =\; \backslash frac\{16\}\{20\}\backslash ;mm\; =\; 0.8\backslash ;mm$

Now the least count (LC) is

$LC\; =\; 1\backslash ;MSD\; -\; 1\backslash ;VSD\; =\; 1\backslash ;mm\; -\; 0.8\backslash ;mm\; =\; 0.2\backslash ;mm$

Convert millimetres to centimetres (1 mm = 0.1 cm):

$0.2\backslash ;mm\; =\; 0.2\; \backslash times\; 0.1\backslash ;cm\; =\; 0.02\backslash ;cm$

Therefore, the least count of the vernier calipers is 0.02 cm.