If $50$ Vernier divisions are equal to $49$ main scale divisions of a travelling microscope and one smallest reading of main scale is $0.5 \,mm$, the Vernier constant of travelling microscope is:

The pitch and the number of divisions on the circular scale for a given screw gauge are $0.5\,mm$ and $100$ respectively. When the screw gauge is fully tightened without any object,the zero of its circular scale lies $3$ divisions below the mean line. The readings of the main scale and the circular scale for a thin sheet are $5.5\,mm$ and $48$ respectively. The thickness of this sheet is: (in $,mm$)

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 $, cm$)

Using a screw gauge with a pitch of $0.1 \ cm$ and $50$ divisions on its circular scale,the thickness of an object is measured. How should the measurement be correctly recorded (in $cm$)?

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

In an experiment with Vernier callipers of least count $0.1\,mm$,when two jaws are joined together,the zero of the Vernier scale lies to the right of the zero of the main scale and the $6^{th}$ division of the Vernier scale coincides with a main scale division. While measuring the diameter of a spherical bob,the zero of the Vernier scale lies between $3.2\,cm$ and $3.3\,cm$ marks,and the $4^{th}$ division of the Vernier scale coincides with a main scale division. The diameter of the bob is measured as $.......\,cm$.