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

If the screw on a screw-gauge is given six rotations, it moves by $3\; \mathrm{mm}$ on the main scale. If there are $50$ divisions on the circular scale the least count of the screw gauge 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 

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