$A$ screw gauge gives the following readings 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 on the circular scale. The diameter of the wire from the above data is ...... $cm$.

Which devices are used for the measurement of lengths of different orders?

The diagrams show readings of a screw gauge. Figure $(i)$ shows the zero error reading when the screw gauge is closed,and figure $(ii)$ shows the reading when the screw gauge is being used to measure the diameter of a ball-bearing. What is the diameter of the ball-bearing in $mm$? There are $50$ divisions on the circular scale.

Length of $9 \ MSD$ of a vernier caliper are equal to $10 \ VSD$ and $1 \ MSD = 1 \ mm$. For measuring the length of a rod,the reading on the main scale is $6.4 \ cm$ and the $8^{th}$ division on the vernier scale is in line with a marking on the main scale. If there is no zero error,find the length of the rod: (in $cm$)

There are $100$ divisions on the circular scale of a screw gauge of pitch $1 \,mm$. With no measuring quantity in between the jaws,the zero of the circular scale lies $5$ divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that $4$ linear scale divisions are clearly visible while $60$ divisions on the circular scale coincide with the reference line. The diameter of the wire is: (in $\,mm$)

While measuring the diameter of a wire using a screw gauge, the following readings were noted. The main scale reading is $1 \,mm$ and the circular scale reading is equal to $42$ divisions. The pitch of the screw gauge is $1 \,mm$ and it has $100$ divisions on the circular scale. The diameter of the wire is $\frac{x}{50} \,mm$. The value of $x$ is: