The least count of the main scale of a screw gauge is $1\, mm$. The minimum number of divisions on its circular scale required to measure $5\,\mu m$ diameter of a wire is

Length of $9 \ MSD$ of a vernier caliper are equal to $10 \ VSD$ and $1 \ MSD = 1 \ mm$. For measuring the length of a rod,the reading on the main scale is $6.4 \ cm$ and the $8^{th}$ division on the vernier scale is in line with a marking on the main scale. If there is no zero error,find the length of the rod: (in $cm$)

$A$ student measured the diameter of a small steel ball using a screw gauge of least count $0.001 \, cm$. The main scale reading is $5 \, mm$ and the circular scale division coinciding with the reference level is $25$. If the screw gauge has a zero error of $-0.004 \, cm$,the correct diameter of the ball is: (in $, cm$)

In a screw gauge,the fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are $50$ divisions on the circular scale,and the main scale moves by $0.5 \, mm$ on a complete rotation. For a particular observation,the reading on the main scale is $5 \, mm$ and the $20^{th}$ division of the circular scale coincides with the reference line. Calculate the true reading in $mm$.

In a vernier callipers, $50$ vernier scale divisions are equal to $48$ main scale divisions. If one main scale division $= 0.05 \ mm$, then the least count of the vernier callipers is . . . . . . $mm$.

$A$ 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 when 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?