In a screw gauge,when the circular scale is given five complete rotations,it moves linearly by $2.5 \text{ mm}$. If the circular scale has $100$ divisions,the least count of the screw gauge is . . . . . . $\text{mm}$.

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).

Answer the following:
$(a)$ You are given a thread and a metre scale. How will you estimate the diameter of the thread?
$(b)$ $A$ screw gauge has a pitch of $1.0\; mm$ and $200$ divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
$(c)$ The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of $100$ measurements of the diameter expected to yield a more reliable estimate than a set of $5$ measurements only?

$A$ student measured the diameter of a wire using a screw gauge with the least count $0.001\, cm$ and listed the measurements. The measured value should be recorded as (in $, cm$)

$A$ 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?

$A$ Vernier calipers has $1 \,mm$ marks on the main scale. It has $20$ equal divisions on the Vernier scale which match with $16$ main scale divisions. For this Vernier calipers, the least count is (in $\,mm$)