Main scale division of vernier calliper is $1\ mm$ and vernier scale division are in $A.P. ; 1^{st}$ division is $0.95\ mm$ ; $2^{nd}$ division is $0.9\ mm$ and so on. When an object is placed between jaws of vernier calliper, zero of vernier lies between $3.1\ cm$ and $3.2\ cm$ and $4^{th}$ division of vernier coincide with main scale division. Reading of vernier is  .......... $cm$

A Vernier calipers has $1 \mathrm{~mm}$ 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 finding out refractive index of glass slab the following observations were made through travelling microscope $50$ vernier scale division $=$ $49 \mathrm{MSD} ; 20$ divisions on main scale in each $\mathrm{cm}$ For mark on paper

$\mathrm{MSR}=8.45 \mathrm{~cm}, \mathrm{VC}=26$

For mark on paper seen through slab

$\mathrm{MSR}=7.12 \mathrm{~cm}, \mathrm{VC}=41$

For powder particle on the top surface of the glass slab

$\mathrm{MSR}=4.05 \mathrm{~cm}, \mathrm{VC}=1$

$(\mathrm{MSR}=$ Main Scale Reading, $\mathrm{VC}=$ Vernier Coincidence)

Refractive index of the glass slab is:

The vernier scale of a travelling microscope has $50$ divisions which coincide with $49$ main scale divisions. If each main scale division is $0.5\, mm$, calculate the minimum inaccuracy in the measurement of distance.

In a screw gauge, $5$ complete rotations of circular scale give $1.5\, mm$ reading on linear scale. Circular scale has $50$ divisions. Least count of the screw gauge is

In a screw gauge, $5$ complete rotations of the screw cause it to move a linear distance of $0.25\, cm$. There are $100$ circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of $4$ main scale divisions and $30$ circular scale divisions . Assuming negligible zero error, the thickness of the wire is