The diameter of a wire is measured with a screw gauge having least count $0.01\;mm$. Which of the following correctly expresses the diameter?

If the screw on a screw-gauge is given six rotations, it moves by $3\; \mathrm{mm}$ on the main scale. If there are $50$ divisions on the circular scale the least count of the screw gauge 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:

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

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 $..........\,cm$

A screw gauge has $50$ divisions on its circular scale. The circular scale is $4$ units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of $0.5\, mm$ is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively