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 $mm$.

The figure below shows a particular position of the Vernier calipers on a centimetre scale. In this position,the value of $x$ shown in the figure is .......... $cm$ (figure is not to scale).

Diameter of a steel ball is measured using a Vernier callipers which has divisions of $0.1\,cm$ on its main scale $(MS)$ and $10$ divisions of its vernier scale $(VS)$ match $9$ divisions on the main scale. Three such measurements for a ball are given as:

$S$.No.	$MS\;(cm)$	$VS$ divisions
$(1)$	$0.5$	$8$
$(2)$	$0.5$	$4$
$(3)$	$0.5$	$6$

If the zero error is $-0.03\,cm,$ then the mean corrected diameter is ........... $cm$.

$A$ 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 the main scale corresponds to $100$ divisions of the circular scale. The diameter of the wire from the above data is: (in $cm$)

The internal and external diameters of a hollow cylinder measured with vernier calipers are $(5.73 \pm 0.01) \text{ cm}$ and $(6.01 \pm 0.01) \text{ cm}$ respectively. Then the thickness of the cylinder wall is

The vernier constant of Vernier callipers is $0.1 \,mm$ and it has zero error of $(-0.05) \,cm$. While measuring the diameter of a sphere,the main scale reading is $1.7 \,cm$ and the coinciding vernier division is $5$. The corrected diameter will be ........... $\times 10^{-2} \,cm$.