Question

In vernier callipers N divisions of vernier scale coincide with N-1 divisions of main scale . The least count of the instrument is

Answer 1

Bonjour!

"To find the least count"

If Nth division of vernier scale coincides with (N - )the division of main scale.

Therefore,
$= > N \: VSD = (N - 1) \: MSD \\ \\ = > 1 \: VSD = (\frac{N - 1}{N}) MSD$

So,
Least count (L.C.)
$= 1 \: MSD - 1 \: VSD \\ \\ = 1 \: MSD - ( \frac{N - 1}{N} )MSD \\ \\ = 1 \: MSD - 1 \: MSD + \frac{1}{N} MSD \\ \\ = \frac{1}{N} \: MSD$

Hope this helps...:)

Answer 2

Answer: The least count is $1 / n MSD .$

Given data states that the nth division on Vernier scale interacts with $(n - 1)^{th}$ division of Main scale, least count has to be calculated.

Least count = 'Main scale division (MSD)' - 'Vernier scale division (VSD)'.

Therefore,

                      $n VSD = n - 1 MSD \\=> (n - 1 / n ) MSD = 1 VSD \\=>  1 VSD = ( n - 1 / n )$

=> Least Count =$1 MSD - 1 VSD \\= 1 MSD  - ( n - 1  / n ) MSD \\= 1 / n MSD .$