$N$ divisions on the main scale of a vernier calliper coincide with $(N + 1)$ divisions of the vernier scale. If each division of the main scale is $a$ units,then the least count of the instrument is:

$A$ screw gauge has a pitch of $1.5\; mm$ and there is no zero error. The linear scale has markings at $MSD = 1\; mm$ and there are $100$ equal divisions on the circular scale. When the diameter of a sphere is measured with this instrument,the $2\; mm$ mark is visible on the linear scale,but the $3\; mm$ mark is not visible. The $76^{th}$ division of the circular scale is in line with the linear scale. What is the diameter of the sphere in $mm$?

In a vernier callipers,$10$ divisions of vernier scale coincide with $9$ divisions of main scale,the least count of which is $0.1\,cm$. 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 $2^{nd}$ division of the vernier scale coincides with a main scale division,then the diameter will be .......... $cm$.

The circular divisions of a screw gauge are $50$. It moves $0.5 \ mm$ on the main scale in one rotation. When the diameter of a wire is measured,the main scale reading is $3.5 \ mm$ and the circular scale reading is $32$. If the zero error (positive) in the screw gauge is $0.06 \ mm$,then the diameter of the wire is:

Thickness of a pencil measured by a screw gauge (least count $0.001 \ cm$) comes out to be $0.802 \ cm$. The percentage error in the measurement is (in $\%$)

Using a screw gauge with a pitch of $0.1 \ cm$ and $50$ divisions on its circular scale,the thickness of an object is measured. How should the measurement be correctly recorded (in $cm$)?