What is zero error of the screw gauge and how to identify the zero error?

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

The vernier scale used for measurement has a positive zero error of $0.2 \mathrm{mm}.$ If while taking a measurement it was noted that $0$ on the vernier scale lies between $8.5 \mathrm{cm}$ and $8.6 \mathrm{cm}$. Vernier coincidence is $6,$ then the correct value of measurement is $\mathrm{cm}.$

(least count$=0.01 \mathrm{cm}$)

$\mathrm{Assertion} A:$ 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 smallest division on the main scale of a Vernier calipers is $0.1 \mathrm{cm}$. Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is

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):