There are $100$ divisions on the circular scale of a screw gauge of pitch $1 \mathrm{~mm}$. With no measuring quantity in between the jaws, the zero of the circular scale lies $5$ divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found the $4$ linear scale divisions are clearly visible while $60$ divisions on circular scale coincide with the reference line. The diameter of the wire is :
$4.65 \mathrm{~mm}$
$4.55 \mathrm{~mm}$
$4.60 \mathrm{~mm}$
$3.35 \mathrm{~mm}$
Two full turns of the circular scale of a screw gauge cover a distance of $1\ mm$ on its main scale. The total number of divisions on the circular scale is $50$. Further, it is found that the screw gauge has a zero error of $- 0.03\ mm$. While measuring the diameter of a thin wire, a student notes the main scale reading of $3\ mm$ and the number of circular scale divisions in line with the main scale as $35$. The diameter of the wire is ....... $mm$
The diameter of a wire is measured with a screw gauge having least count $0.01\;mm$. Which of the following correctly expresses the diameter?
Student $A$ and Student $B$ used two screw gauges of equal pitch and $100$ equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is $0.322\, {cm}$. The absolute value of the difference between the final circular scale readings observed by the students $A$ and $B$ is .... .
[Figure shows position of reference $'O'$ when jaws of screw gauge are closed]
Given pitch $=0.1 \,{cm}$.