The vernier scale of a travelling microscope has $50$ divisions which coincide with $49$ main scale divisions. If each main scale division is $0.5\, mm$, calculate the minimum inaccuracy in the measurement of distance.

In a vernier callipers, each $cm$ on the main scale is divided into $20$ equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be $\dots \; \times 10^{-2} \;mm$

The diameter of a cylinder is measured using a vernier callipers with no zero error. It is found that the zero of the vernier scale lies between $5.10 \ cm$ and $5.15 \ cm$ of the main scale. The vernier scale has $50$ division equivalent to $2.45 \ cm$. The $24^{\text {th }}$ division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is :

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 main scale corresponds to $100$ divisions of the circular scale. The diameter of wire from the above data is:

Two full turns of the circular scale of screw gauge cover a distance of $1\,mm$ on scale. The total number of divisions on circular scale is $50$. Further, it is found that 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 number of circular scale division in line, with the main scale is $35$. The diameter of the wire is ………. $mm$