Asseretion $A:$ If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5\, {mm}$ and there are $50$ total divisions on circular scale, then least count is $0.001\, {cm}$.

Reason $R:$ Least Count $=\frac{\text { Pitch }}{\text { Total divisions on circular scale }}$

In the light of the above statements, choose the most appropriate answer from the options given below:

Which of the following is the most precise device for measuring length:

$(a)$ a vernier callipers with $20$ divisions on the sliding scale

$(b)$ a screw gauge of pitch $1\; mm$ and $100$ divisions on the circular scale

$(c)$ an optical instrument that can measure length to within a wavelength of light?

$N$ divisions on the main scale of a vernier calliper coincide with $(N + 1 )$ divisions of the vernier scale. If each division of main scale is $a$ units , then the least count of the instrument is

In a vernier callipers, each $cm$ on the main scale is divided into $20$ equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be $\dots \; \times 10^{-2} \;mm$

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 zero of circular scale division coincides with $25$ divisions above the reference level. If screw gauge has a zero error of $-0.004 \,cm,$ the correct diameter of the ball is