Least count of a vernier caliper is $\frac{1}{20 \mathrm{~N}} \mathrm{~cm}$. The value of one division on the main scale is $1 \mathrm{~mm}$. Then the number of divisions of main scale that coincide with $\mathrm{N}$ divisions of vernier scale is :

A screw gauge with a pitch of $0.5 \ mm$ and a circular scale with $50$ divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the two jaws of the screw gauge are brought in contact, the $45^{th} $division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet (in $mm$) if the main scale reading is $0.5\ mm$ and the $25^{th}$ division coincides with the main scale line 

In a vernier callipers, $10$ divisions of vernier scale coincides with $9$ divisions of main scale, the least count of which is $0.1\,cm$. If in the measurement of inner diameter of cylinder zero of vernier scale lies between $1.3\,cm$ and $1.4\, cm$ of main scale and $2^{nd}$ division of vernier scale coincides with main scale division then diameter will be ………. $cm$

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 main scale corresponds to $100$ divisions of the circular scale. The diameter of wire from the above data is:

There are two Vernier calipers both of which have $1 \mathrm{~cm}$ divided into $10$ equal divisions on the main scale. The Vernier scale of one of the calipers $\left(C_1\right)$ has $10$ squal divisions that correspond to $9$ main scale divisions. The Vernier scale of the other caliper $\left(C_2\right)$ 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 $\mathrm{cm}$ ) by calipers $C_1$ and $C_2$, respectively, are