$A$ screw gauge has a least count of $0.01\, mm$ and there are $50$ divisions on its circular scale. The pitch of the screw gauge is $........\, mm$.

$A$ 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 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 $, cm$)

In a screw gauge,when the circular scale is given five complete rotations,it moves linearly by $2.5 \text{ mm}$. If the circular scale has $100$ divisions,the least count of the screw gauge is . . . . . . $\text{mm}$.

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 $1 \text{ cm}$?

The vernier constant of Vernier callipers is $0.1 \,mm$ and it has zero error of $(-0.05) \,cm$. While measuring the diameter of a sphere,the main scale reading is $1.7 \,cm$ and the coinciding vernier division is $5$. The corrected diameter will be ........... $\times 10^{-2} \,cm$.