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:

Using a screw gauge with a pitch of $0.1 \ cm$ and $50$ divisions on its circular scale,the thickness of an object is measured. How should the measurement be correctly recorded (in $cm$)?

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 $mm$.

Given below are two statements:
Statement $I$: In a vernier callipers,one vernier scale division is always smaller than one main scale division.
Statement $II$: The vernier constant is given by one main scale division multiplied by the number of vernier scale division.
In the light of the above statements,choose the correct answer from the options given below.

In a screw gauge,$5$ complete rotations of the screw cause it to move a linear distance of $0.25\, cm$. There are $100$ circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of $4$ main scale divisions and $30$ circular scale divisions. Assuming negligible zero error,the thickness of the wire is (in $, cm$)

Diameter of a steel ball is measured using a Vernier callipers which has divisions of $0.1\,cm$ on its main scale $(MS)$ and $10$ divisions of its vernier scale $(VS)$ match $9$ divisions on the main scale. Three such measurements for a ball are given as:

$S$.No.	$MS\;(cm)$	$VS$ divisions
$(1)$	$0.5$	$8$
$(2)$	$0.5$	$4$
$(3)$	$0.5$	$6$

If the zero error is $-0.03\,cm,$ then the mean corrected diameter is ........... $cm$.