In a screw gauge, the zero of the circular scale lies $3$ divisions above the horizontal pitch line when their metallic studs are brought in contact. Using this instrument, the thickness of a sheet is measured. If the pitch scale reading is $1 \ mm$ and the circular scale reading is $51$, then the correct thickness of the sheet is . . . . . . $mm$. [Assume least count is $0.01 \ mm$]

In an experiment,the angles are required to be measured using an instrument in which $29$ divisions of the main scale exactly coincide with the $30$ divisions of the vernier scale. If the smallest division of the main scale is half a degree $(=0.5^{\circ})$,then the least count of the instrument is

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

In an experiment to find out the diameter of a wire using a screw gauge,the following observations were noted:
$(a)$ The screw moves $0.5\,mm$ on the main scale in one complete rotation.
$(b)$ Total divisions on the circular scale $= 50$.
$(c)$ Main scale reading is $2.5\,mm$.
$(d)$ The $45^{\text{th}}$ division of the circular scale is on the pitch line.
$(e)$ The instrument has a $0.03\,mm$ negative zero error.
Then the diameter of the wire is $...........\,mm$.

The pitch and the number of divisions on the circular scale for a given screw gauge are $0.5\,mm$ and $100$ respectively. When the screw gauge is fully tightened without any object,the zero of its circular scale lies $3$ divisions below the mean line. The readings of the main scale and the circular scale for a thin sheet are $5.5\,mm$ and $48$ respectively. The thickness of this sheet is: (in $,mm$)

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 ........... $cm$.