The pitch of screw gauge is $0.5 \mathrm{mm}$ and there are $100$ divisions on it's circular scale. The instrument reads $2$ circular divisions when nothing is put in between it jaws. In measuring the diameter of a wire, there are $8$ divisions on the main scale and ${83}^{\mathrm{rd}}$ division coincides with the reference line. Then choose the correct option (s).

In an experiment to find out the diameter of wire using screw gauge, the following observation were noted:

(a) Screw moves $0.5 \mathrm{mm}$ on main scale in one complete rotation
(b) Total divisions on circular scale $=50$
(c) Main scale reading is $2.5 \mathrm{mm}$
(d) ${45}^{\mathrm{th}}$${ }^{}$ division of circular scale is in the pitch line
(e) Instrument has $0.03 \mathrm{mm}$ negative error Then the diameter of wire is :

There are two vernier calipers both of which have $1 \mathrm{cm}$ divided into $10$ equal divisions on the main scale. The Vernier scale of one of the calipers $\left(C_1\right)$ has $10$ equal divisions that correspond to $9$ main scale divisions. The Vernier scale of the other caliper $\left(C_2\right)$ has $10$ equal divisions that correspond to $11$ main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in $\mathrm{cm}$) by calipers $C_1$ and $C_2$ respectively, are

$\mathrm{Assertion} A:$ If in five complete rotations of the circular scale, the distance travelled on the main scale of the screw gauge is $5 \mathrm{mm}$ and there are $50$ total divisions on a circular scale, then the least count is $0.001 \mathrm{cm}.$

$\mathrm{Reason} R:$
Least Count $=\frac{\text{ Pitch }}{\text{ Total divisions on circular scale }}$
In the light of the above statements, choose the most appropriate answer from the options given below.

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

Student $A$ and student $B$ used two screw gauges of equal pitch and $100$ equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is $0.322 \mathrm{cm}$. The absolute value of the difference between the final circular scale readings observed by the students $A$ and $B$ is _______.

[Figure shows position of reference $O$ when jaws of screw gauge are closed]

Given pitch $=0.1 \mathrm{cm}$.