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

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

The pitch and the number of divisions on the circular scale for a given screw gauge are $0.5\,mm$ and $100$ respectively. When the screw gauge is fully tightened without any object,the zero of its circular scale lies $3$ divisions below the mean line. The readings of the main scale and the circular scale for a thin sheet are $5.5\,mm$ and $48$ respectively. The thickness of this sheet is: (in $,mm$)

The main scale of a vernier calliper has $n$ divisions per $cm$. $n$ divisions of the vernier scale coincide with $(n-1)$ divisions of the main scale. The least count of the vernier calliper is,

In a screw gauge, the zero of the main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are $100$ divisions on the circular scale and the pitch of the screw gauge is $0.1 \text{ mm}$. When the diameter of a sphere is measured, the reading of the main scale is $5 \text{ mm}$ and the $50^{th}$ division of the circular scale coincides with the reference line of the main scale. The diameter of the sphere is . . . . . . $\text{mm}$.

If in a Vernier callipers $10 \,VSD$ coincides with $8 \,MSD$,then the least count of the Vernier calliper is ............ $m$ [given $1 \,MSD = 1 \,mm$].