In a vernier callipers,$20$ $VSD$ coincide with $16$ $MSD$ (each division of length $1 \text{ mm}$). The least count of the vernier callipers is: (in $\text{ cm}$)

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

In a screw gauge,the zero of the circular scale lies $3$ divisions above the horizontal pitch line when their metallic studs are brought in contact. Using this instrument,the thickness of a sheet is measured. If the pitch scale reading is $1 \ mm$ and the circular scale reading is $51$,then the correct thickness of the sheet is . . . . . . $mm$. [Assume least count is $0.01 \ mm$]

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 . . . . . . $\text{mm}$.

$A$ student measured the diameter of a small steel ball using a screw gauge of least count $0.001 \, cm$. The main scale reading is $5 \, mm$ and the circular scale division coinciding with the reference level is $25$. If the screw gauge has a zero error of $-0.004 \, cm$,the correct diameter of the ball is: (in $, cm$)