One main scale division of a vernier caliper is equal to $m$ units. If $n^{\text{th}}$ division of the main scale coincides with $(n+1)^{\text{th}}$ division of the vernier scale,the least count of the vernier caliper is:

$A$ screw gauge has $50$ divisions on its circular scale. The circular scale is $4$ units ahead of the pitch scale marking,prior to use. Upon one complete rotation of the circular scale,a displacement of $0.5\, mm$ is noticed on the pitch scale. The nature of zero error involved,and the least count of the screw gauge,are respectively

$10$ divisions on the main scale of a Vernier calliper coincide with $11$ divisions on the Vernier scale. If each division on the main scale is of $5$ units,the least count of the instrument is :

$A$ screw gauge gives the following readings when used to measure the diameter of a wire:
Main scale reading: $0 \, mm$
Circular scale reading: $52$ divisions
Given that $1 \, mm$ on the main scale corresponds to $100$ divisions on the circular scale. The diameter of the wire from the above data is ...... $cm$.

In a vernier callipers,$20$ $VSD$ coincide with $16$ $MSD$ (each division of length $1 \text{ mm}$). The least count of the vernier callipers is: (in $\text{ cm}$)

The least count of a screw gauge is $0.01 \ mm$. If the pitch is increased by $75\%$ and the number of divisions on the circular scale is reduced by $50\%$,the new least count will be . . . . . . $\times 10^{-3} \ mm$.