$A$ 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 in $cm$?

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 ........... $\times 10^{-2} \,cm$.

The figure below shows a particular position of the Vernier calipers on a centimetre scale. In this position,the value of $x$ shown in the figure is .......... $cm$ (figure is not to scale).

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 $, cm$)

The vernier calliper used for measurement has a positive zero error of $0.3 \ mm$. While taking measurement for the internal diameter of a vessel,it was observed that the zero of the vernier scale lies between $9.5 \ cm$ and $9.6 \ cm$ of the main scale and the $6^{th}$ division of the vernier scale coincides with a main scale division. If the least count of the vernier calliper is $0.01 \ cm$,the correct value of the diameter will be: (in $cm$)

$A$ 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?