The circular scale of a micrometer has $200$ divisions and a pitch of $2 \, mm$. Find the measured value of the thickness of a thin sheet in $mm$.

In a vernier callipers,$10$ divisions of vernier scale coincide with $9$ divisions of main scale. One division of main scale is of $0.1 \ cm$. If in the measurement of inner diameter of a cylinder,the zero of the vernier scale lies between $1.3 \ cm$ and $1.4 \ cm$ of the main scale and the $2^{\text{nd}}$ division of the vernier scale coincides with a main scale division,then the diameter will be: (in $cm$)

In a vernier callipers, $50$ vernier scale divisions are equal to $48$ main scale divisions. If one main scale division $= 0.05 \ mm$, then the least count of the vernier callipers is . . . . . . $mm$.

The main scale of a vernier caliper reads in millimeters and its vernier scale is divided into $8$ divisions,which coincide with $5$ divisions of the main scale. When the two jaws of the instrument touch each other,the zero of the vernier scale coincides with the zero of the main scale. $A$ rod is placed between the two jaws. It is observed that the zero of the vernier scale lies just to the left of the $36^{th}$ division of the main scale and the fourth division of the vernier scale coincides with a main scale division. The measured value is .......... $cm$.

The main scale division of a vernier caliper is $1 \ mm$. The vernier scale divisions are in an arithmetic progression $(A.P.)$; the $1^{st}$ division is $0.95 \ mm$, the $2^{nd}$ division is $0.90 \ mm$, and so on. When an object is placed between the jaws of the vernier caliper, the zero of the vernier scale lies between $3.1 \ cm$ and $3.2 \ cm$, and the $4^{th}$ division of the vernier scale coincides with a main scale division. The reading of the vernier caliper is .......... $cm$.

The least count of the main scale of a screw gauge is $1\, mm$. The minimum number of divisions on its circular scale required to measure $5\,\mu m$ diameter of a wire is