In a vernier callipers,each $cm$ on the main scale is divided into $20$ equal parts. If the $10^{th}$ vernier scale division coincides with the $9^{th}$ main scale division,then the value of the vernier constant will be $\dots \; \times 10^{-2} \; mm$.

Length of $9 \ MSD$ of a vernier caliper are equal to $10 \ VSD$ and $1 \ MSD = 1 \ mm$. For measuring the length of a rod,the reading on the main scale is $6.4 \ cm$ and the $8^{th}$ division on the vernier scale is in line with a marking on the main scale. If there is no zero error,find the length of the rod: (in $cm$)

$A$ specially designed Vernier calliper has the main scale least count of $1 \,mm$. On the Vernier scale,there are $10$ equal divisions and they match with $11$ main scale divisions. Then,the least count of the Vernier calliper is ........... $mm$.

$A$ screw gauge gives the following reading 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 of the circular scale. The diameter of the wire from the above data is: (in $cm$)

The diameter of a wire is measured with a screw gauge having a least count of $0.01\;mm$. Which of the following correctly expresses the diameter?

$A$ vernier callipers has $20$ divisions on the vernier scale, which coincide with $19$ divisions on the main scale. The least count of the instrument is $0.1 \,mm$. One main scale division is equal to $...$ $mm$.