In an experiment to find out the diameter of a wire using a screw gauge,the following observations were noted:
$(a)$ The 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 $2.5\,mm$.
$(d)$ The $45^{\text{th}}$ division of the circular scale is on the pitch line.
$(e)$ The instrument has a $0.03\,mm$ negative zero error.
Then the diameter of the wire is $...........\,mm$.

$A$ screw gauge has a least count of $0.01\, mm$ and there are $50$ divisions on its circular scale. The pitch of the screw gauge is $........\, mm$.

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	$(b)$ $0.001\,cm$
	$(c)$ $0.0001\,cm$

One main scale division of a vernier callipers is $a \ cm$ and $n^{\text{th}}$ division of the vernier scale coincides with $(n-1)^{\text{th}}$ division of the main scale. The least count of the callipers in $mm$ is

In a screw gauge,there are $100$ divisions on the circular scale and the main scale moves by $0.5\,mm$ on a complete rotation of the circular scale. The zero of the circular scale lies $6$ divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs,$4$ linear scale divisions are clearly visible while the $46^{\text{th}}$ division of the circular scale coincides with the reference line. The diameter of the wire is $...........\times 10^{-2}\,mm$.

$A$ screw gauge gives the following reading 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 of the circular scale. The diameter of the wire from the above data is: (in $cm$)