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.........$mm$

The least count of the main scale of a vernier callipers is $1\, mm$. Its vernier scale is divided into $10$ divisions and coincides with $9$ divisions of the main scale. When jaws are touching each other,the $7^{th}$ division of the vernier scale coincides with a division of the main scale and the zero of the vernier scale lies to the right side of the zero of the main scale. When this vernier is used to measure the length of a cylinder,the zero of the vernier scale is between $3.1\, cm$ and $3.2\, cm$ and the $4^{th}$ $VSD$ coincides with a main scale division. The length of the cylinder is $.....\, cm$. ($VSD$ is vernier scale division)

The internal and external radii of a hollow cylinder are measured with the help of a vernier callipers. Their values are $(4.23 \pm 0.01) \, cm$ and $(3.87 \pm 0.01) \, cm,$ respectively. The thickness of the wall of the cylinder is

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,

$A$ spectrometer gives the following reading when used to measure the angle of a prism.
Main scale reading : $58.5^{\circ}$
Vernier scale reading : $09$ divisions
Given that $1$ division on the main scale corresponds to $0.5^{\circ}$. The total number of divisions on the Vernier scale is $30$,which matches $29$ divisions of the main scale. The angle of the prism from the above data is ....... $degree$. (in $^{\circ}$)

Consider a Vernier callipers in which each $1 \ cm$ on the main scale is divided into $8$ equal divisions and a screw gauge with $100$ divisions on its circular scale. In the Vernier callipers,$5$ divisions of the Vernier scale coincide with $4$ divisions on the main scale and in the screw gauge,one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
$(A)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.01 \ mm$.
$(B)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.005 \ mm$.
$(C)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.01 \ mm$.
$(D)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.005 \ mm$.