The main scale of a vernier calliper has $n$ divisions per $cm$. $n$ divisions of the vernier scale coincide with $(n-1)$ divisions of the main scale. The least count of the vernier calliper is,

In a vernier callipers,$(N+1)$ divisions of vernier scale coincide with $N$ divisions of main scale. If $1 \text{ MSD}$ represents $0.1 \text{ mm}$,the vernier constant (in $\text{cm}$) is:

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 $cm$)

The figure below shows a particular position of the Vernier calipers on a centimetre scale. In this position,the value of $x$ shown in the figure is .......... $cm$ (figure is not to scale).

In an experiment to find out the diameter of a wire using a screw gauge,the following observations were noted:
$(A)$ Screw moves $0.5 \ mm$ on the main scale in one complete rotation.
$(B)$ Total divisions on the circular scale $= 50$.
$(C)$ Main scale reading is $2.5 \ mm$.
$(D)$ $45^{\text{th}}$ division of the circular scale is in the reference line.
$(E)$ Instrument has $0.03 \ mm$ negative zero error.
Then the diameter of the wire is: (in $mm$)

Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5 \text{ mm}$. The circular scale has $100$ divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.

Measurement condition	Main scale reading	Circular scale reading
Two arms of gauge touching each other without wire	$0$ division	$4$ division
Attempt-$1$: With wire	$4$ division	$20$ division
Attempt-$2$: With wire	$4$ division	$16$ division

What are the diameter and cross-sectional area of the wire measured using the screw gauge?