The least count of the main scale of a vernier callipers is $1\, mm$. Its vernier scale is divided into $10$ divisions and coincide with $9$ divisions of the main scale. When jaws are touching each other, the $7$ th division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of a cylinder the zero of the vernier scale between $3.1\, cm$ and $3.2\, cm$ and $4^{th}$ $VSD$ coincides with a main scale division.The length of the cylinder is $.....cm$
($VSD$ is vernier scale division)

One main scale division of a vernier callipers is $a$ $cm$ and $n ^{\text {th }}$ division of the vernier scale coincide with $( n -1)^{\text {th }}$ division of the main scale. The least count of the callipers in $mm$ is

In a vernier callipers, $(N+1)$ divisions of vernier scale coincide with $N$ divisions of main scale. If $1 \mathrm{MSD}$ represents $0.1 \mathrm{~mm}$, the vernier constant (in $\mathrm{cm}$ ) is:

The vernier constant of Vernier callipers is $0.1 \,mm$ and it has zero error of $(-0.05) \,cm$. While measuring diameter of a sphere, the main scale reading is $1.7 \,cm$ and coinciding vernier division is $5$. The corrected diameter will be ........... $\times 10^{-2} \,cm$

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?

A screw gauge has $50$ divisions on its circular scale. The circular scale is $4$ units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of $0.5\, mm$ is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively