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

The diameter of a wire is measured with a screw gauge having least count $0.01\;mm$. Which of the following correctly expresses the diameter?

If in a Vernier callipers $10 \,VSD$ coincides with $8 \,MSD$, then the least count of Vernier calliper is ............ $m$  [given $1 \,MSD =1 \,mm ]$

A screw gauge gives the following reading when used to measure the diameter of a wire.

Main scale reading : $0\ mm$

Circular scale reading : $52\ divisions$

Given that $1\ mm$ on main scale corresponds to $100$ divisions of the circular scale. The diameter of wire from the above data 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.