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

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 diameter of a cylinder is measured using a vernier callipers with no zero error. It is found that the zero of the vernier scale lies between $5.10 \ cm$ and $5.15 \ cm$ of the main scale. The vernier scale has $50$ divisions equivalent to $2.45 \ cm$. The $24^{\text{th}}$ division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is: (in $cm$)

$A$ screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading : $0 \ mm$
Circular scale reading : $52 \ divisions$
Given that $1 \ mm$ on the main scale corresponds to $100$ divisions of the circular scale. The diameter of the wire from the above data is: (in $cm$)

The pitch of the screw gauge is $1\, mm$ and there are $100$ divisions on the circular scale. When nothing is put in between the jaws,the zero of the circular scale lies $8$ divisions below the reference line. When a wire is placed between the jaws,the first linear scale division is clearly visible while $72^{nd}$ division on the circular scale coincides with the reference line. The radius of the wire is.........$mm$