The circular scale of a micrometer has $200$ divisions and a pitch of $2 \, mm$. Find the measured value of the thickness of a thin sheet in $mm$.

There are two Vernier calipers,both of which have $1 \ cm$ divided into $10$ equal divisions on the main scale. The Vernier scale of one of the calipers $(C_1)$ has $10$ equal divisions that correspond to $9$ main scale divisions. The Vernier scale of the other caliper $(C_2)$ has $10$ equal divisions that correspond to $11$ main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in $cm$) by calipers $C_1$ and $C_2$,respectively,are

The vernier scale used for measurement has a positive zero error of $0.2\, mm$. If while taking a measurement it was noted that the '$0$' on the vernier scale lies between $8.5\, cm$ and $8.6\, cm$ and the vernier coincidence is $6$,then the correct value of measurement is ............. $cm$. (Least count $= 0.01\, cm$)

The one division of main scale of vernier callipers reads $1\,mm$ and $10$ divisions of Vernier scale is equal to the $9$ divisions on main scale. When the two jaws of the instrument touch each other the $zero$ of the Vernier lies to the right of $zero$ of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws,the $zero$ of the Vernier scale lies in between $4.1\,cm$ and $4.2\,cm$ and $6^{\text{th}}$ Vernier division coincides with a main scale division. The diameter of the bob will be $.............10^{-2}\,cm$

When the gap is closed without placing any object in a screw gauge whose least count is $0.005 \ mm$,the $5^{th}$ division on its circular scale coincides with the reference line on the main scale. When a small sphere is placed,the reading on the main scale advances by $4$ divisions,whereas the circular scale reading advances by five times the corresponding reading when no object was placed. There are $200$ divisions on the circular scale. The radius of the sphere is .......... $mm$.

$A$ 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 $cm$)