Two full turns of the circular scale of a screw gauge cover a distance of $1 \mathrm{mm}$ on its main scale. The total number of divisions on the circular scale is $50.$ Further, it is found that the screw gauge has a zero error of $-0.03 \mathrm{mm}.$ While measuring the diameter of a thin wire, a student notes the main scale reading of $3 \mathrm{mm}$ and the number of circular scale divisions, in line with the main scale, as $35.$ The diameter $\left(\mathit{\text{in}} \mathrm{\mu m}\right)$ of the wire 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):

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.

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