$A$ screw gauge with a pitch of $0.5 \ mm$ and a circular scale with $50$ divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement,it is found that when the two jaws of the screw gauge are brought in contact,the $45^{th}$ division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet (in $mm$) if the main scale reading is $0.5 \ mm$ and the $25^{th}$ division coincides with the main scale line?

In a Vernier Calipers,$10$ divisions of the Vernier scale are equal to $9$ divisions of the main scale. When both jaws of the Vernier calipers touch each other,the zero of the Vernier scale is shifted to the left of the zero of the main scale,and the $4^{\text{th}}$ Vernier scale division exactly coincides with a main scale division. One main scale division is equal to $1\,mm$. While measuring the diameter of a spherical body,the body is held between the two jaws. It is observed that the zero of the Vernier scale lies between $30$ and $31$ divisions of the main scale,and the $6^{\text{th}}$ Vernier scale division exactly coincides with a main scale division. The diameter of the spherical body is $.......\,cm$.

In a screw gauge,the fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are $50$ divisions on the circular scale,and the main scale moves by $0.5 \, mm$ on a complete rotation. For a particular observation,the reading on the main scale is $5 \, mm$ and the $20^{th}$ division of the circular scale coincides with the reference line. Calculate the true reading in $mm$.

$N$ divisions on the main scale of a vernier calliper coincide with $(N + 1)$ divisions of the vernier scale. If each division of the main scale is $a$ units,then the least count of the instrument is:

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 $, cm$)

$A$ vernier callipers has $20$ divisions on the vernier scale, which coincide with $19$ divisions on the main scale. The least count of the instrument is $0.1 \,mm$. One main scale division is equal to $...$ $mm$.