The smallest division on the main scale of a vernier callipers is $1 \ mm$,and $10$ vernier divisions coincide with $9$ main scale divisions. While measuring the diameter of a sphere,the zero mark of the vernier scale lies between $2.0 \ cm$ and $2.1 \ cm$ of the main scale,and the fifth division of the vernier scale coincides with a main scale division. The diameter of the sphere is:

Consider a Vernier callipers in which each $1 \ cm$ on the main scale is divided into $8$ equal divisions and a screw gauge with $100$ divisions on its circular scale. In the Vernier callipers,$5$ divisions of the Vernier scale coincide with $4$ divisions on the main scale and in the screw gauge,one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
$(A)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.01 \ mm$.
$(B)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.005 \ mm$.
$(C)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.01 \ mm$.
$(D)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.005 \ mm$.

Match the instruments in Column-$I$ with their respective least counts in Column-$II$.

Column-$I$	Column-$II$
$(1)$ Traveling Microscope	$(a)$ $0.01\,cm$
$(2)$ Screw Gauge	$(b)$ $0.001\,cm$
	$(c)$ $0.0001\,cm$

The circular divisions of a screw gauge are $50$. It moves $0.5 \ mm$ on the main scale in one rotation. When the diameter of a wire is measured,the main scale reading is $3.5 \ mm$ and the circular scale reading is $32$. If the zero error (positive) in the screw gauge is $0.06 \ mm$,then the diameter of the wire is:

The smallest division on the main scale of a Vernier calipers is $0.1 \text{ cm}$. 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 (in $\text{ cm}$)

$A$ screw gauge gives the following readings when used to measure the diameter of a wire:
Main scale reading: $0 \, mm$
Circular scale reading: $52$ divisions
Given that $1 \, mm$ on the main scale corresponds to $100$ divisions on the circular scale. The diameter of the wire from the above data is ...... $cm$.