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:

The diameter of a spherical bob is measured using a vernier callipers. $9$ divisions of the main scale,in the vernier callipers,are equal to $10$ divisions of vernier scale. One main scale division is $1\, {mm}$. The main scale reading is $10\, {mm}$ and $8^{\text{th}}$ division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of $0.04\, {cm}$,then the radius of the bob is $...... \,\times 10^{-2} \,{cm}$.

Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its $10$ Vernier Scale Divisions $(V.S.D.)$ are equal to its $9$ Main Scale Divisions $(M.S.D.)$. The least division on the $M.S.$ is $0.1 \ cm$ and the zero of $V.S.$ is at $x=0.1 \ cm$ when the jaws of the Vernier callipers are closed. If the main scale reading for the diameter is $M=5 \ cm$ and the number of the coinciding vernier division is $8$,the measured diameter after zero error correction is: (in $cm$)

$A$ screw gauge has a pitch of $1.5\; mm$ and there is no zero error. The linear scale has markings at $MSD = 1\; mm$ and there are $100$ equal divisions on the circular scale. When the diameter of a sphere is measured with this instrument,the $2\; mm$ mark is visible on the linear scale,but the $3\; mm$ mark is not visible. The $76^{th}$ division of the circular scale is in line with the linear scale. What is the diameter of the sphere in $mm$?

In a screw gauge,the fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are $50$ divisions on the circular scale,and the main scale moves by $0.5 \, mm$ on a complete rotation. For a particular observation,the reading on the main scale is $5 \, mm$ and the $20^{th}$ division of the circular scale coincides with the reference line. Calculate the true reading in $mm$.

$A$ 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 $cm$)