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:

The pitch of the screw gauge is $1\, mm$ and there are $100$ divisions on the circular scale. When nothing is put in between the jaws,the zero of the circular scale lies $8$ divisions below the reference line. When a wire is placed between the jaws,the first linear scale division is clearly visible while $72^{nd}$ division on the circular scale coincides with the reference line. The radius of the wire is.........$mm$

In a vernier calliper,when both jaws touch each other,the zero of the vernier scale shifts towards the left and its $4^{\text{th}}$ division coincides exactly with a certain division on the main scale. If $50$ vernier scale divisions $(VSD)$ are equal to $49$ main scale divisions $(MSD)$ and the zero error in the instrument is $0.04 \text{ mm}$,then how many main scale divisions are there in $1 \text{ cm}$?

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:

The vernier scale used for measurement has a positive zero error of $0.2\, mm$. If while taking a measurement it was noted that the '$0$' on the vernier scale lies between $8.5\, cm$ and $8.6\, cm$ and the vernier coincidence is $6$,then the correct value of measurement is ............. $cm$. (Least count $= 0.01\, cm$)

An experiment is performed to find the refractive index of glass using a travelling microscope. In this experiment,distances are measured by