$A$ student measured the diameter of a wire using a screw gauge with the least count $0.001\, cm$ and listed the measurements. The measured value should be recorded as (in $, cm$)

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 .......... $cm$.

Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its $10$ Vernier Scale Divisions $(V.S.D.)$ are equal to its $9$ Main Scale Divisions $(M.S.D.)$. The least division on the $M.S.$ is $0.1 \ cm$ and the zero of $V.S.$ is at $x=0.1 \ cm$ when the jaws of the Vernier callipers are closed. If the main scale reading for the diameter is $M=5 \ cm$ and the number of the coinciding vernier division is $8$,the measured diameter after zero error correction is: (in $cm$)

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

$A$ screw gauge has some zero error but its value is unknown. We have two identical rods. When the first rod is inserted in the screw gauge,the state of the instrument is shown by diagram $(I)$. When both the rods are inserted together in series,the state is shown by diagram $(II)$. What is the zero error of the instrument in $mm$? Given: $1 \, M.S.D. = 100 \, C.S.D. = 1 \, mm$.

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 $\text{ cm}$)