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 $, cm$)

$A$ specially designed Vernier calliper has the main scale least count of $1 \,mm$. On the Vernier scale,there are $10$ equal divisions and they match with $11$ main scale divisions. Then,the least count of the Vernier calliper is ........... $mm$.

In a vernier callipers,$10$ divisions of vernier scale coincide with $9$ divisions of main scale. One division of main scale is of $0.1 \ cm$. If in the measurement of inner diameter of a cylinder,the zero of the vernier scale lies between $1.3 \ cm$ and $1.4 \ cm$ of the main scale and the $2^{\text{nd}}$ division of the vernier scale coincides with a main scale division,then the diameter will be: (in $cm$)

In a vernier callipers,$(N+1)$ divisions of vernier scale coincide with $N$ divisions of main scale. If $1 \text{ MSD}$ represents $0.1 \text{ mm}$,the vernier constant (in $\text{cm}$) is:

$A$ student measured the diameter of a small steel ball using a screw gauge of least count $0.001 \, cm$. The main scale reading is $5 \, mm$ and the circular scale division coinciding with the reference level is $25$. If the screw gauge has a zero error of $-0.004 \, cm$,the correct diameter of the ball is: (in $, cm$)

$10$ divisions on the main scale of a Vernier calliper coincide with $11$ divisions on the Vernier scale. If each division on the main scale is of $5$ units,the least count of the instrument is :