$A$ 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 $cm$)

In a vernier callipers,$50$ vernier scale divisions are equal to $48$ main scale divisions. If one main scale division $= 0.05 \ mm$,then the least count of the vernier callipers is . . . . . . $mm$.

The main scale of a vernier caliper reads in millimeters and its vernier scale is divided into $8$ divisions,which coincide with $5$ divisions of the main scale. When the two jaws of the instrument touch each other,the zero of the vernier scale coincides with the zero of the main scale. $A$ rod is placed between the two jaws. It is observed that the zero of the vernier scale lies just to the left of the $36^{th}$ division of the main scale and the fourth division of the vernier scale coincides with a main scale division. The measured value is .......... $cm$.

In a screw gauge,when the circular scale is given five complete rotations,it moves linearly by $2.5 \text{ mm}$. If the circular scale has $100$ divisions,the least count of the screw gauge is . . . . . . $\text{mm}$.

$A$ 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 $mm$?

The least count of a vernier caliper is $\frac{1}{20N} \text{ cm}$. The value of one division on the main scale is $1 \text{ mm}$. Then the number of divisions of the main scale that coincide with $N$ divisions of the vernier scale is: