In a screw gauge,$5$ complete rotations of the screw cause it to move a linear distance of $0.25\, cm$. There are $100$ circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of $4$ main scale divisions and $30$ circular scale divisions. Assuming negligible zero error,the thickness of the wire is (in $, cm$)

$N$ divisions on the main scale of a vernier calliper coincide with $(N + 1)$ divisions of the vernier scale. If each division of the main scale is $a$ units,then the least count of the instrument is:

The least count of a vernier caliper is $\frac{1}{20N} \text{ cm}$. The value of one division on the main scale is $1 \text{ mm}$. Then the number of divisions of the main scale that coincide with $N$ divisions of the vernier scale is:

Two full turns of the circular scale of a screw gauge cover a distance of $1 \ mm$ on its main scale. The total number of divisions on the circular scale is $50$. The screw gauge has a zero error of $-0.03 \ mm$. While measuring the diameter of a thin wire,a student notes the main scale reading of $3 \ mm$ and the $35^{\text{th}}$ division of the circular scale coincides with the reference line of the main scale. The diameter of the wire is: (in $mm$)

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: In a Vernier calliper,if a positive zero error exists,then while taking measurements,the reading taken will be more than the actual reading.
Reason $(R)$: The zero error in a Vernier calliper might have happened due to a manufacturing defect or due to rough handling.
In the light of the above statements,choose the correct answer from the options given below:

$A$ travelling microscope has $20$ divisions per $cm$ on the main scale while its Vernier scale has total $50$ divisions and $25$ Vernier scale divisions are equal to $24$ main scale divisions. What is the least count of the travelling microscope in $cm$?