In an experiment the angles are required to be measured using an instrument, $29$ divisions of the main scale exactly coincide with the $30$ divisions of the vernier scale. If the smallest division of the main scale is half- a degree $(= 0.5^o  )$, then the least count of the instrument 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 density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is $0.5 \ mm$ and there are $50$ divisions on the circular scale. The reading on the main scale is $2.5 \ mm$ and that on the circular scale is $20$ divisions. If the measured mass of the ball has a relative error of $2 \%$, the relative percentage error in the density is

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 wen 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 travelling microscope has $20$ divisions per $cm$ on the main scale while its Vernier scale has total $50$ divisions and $25$ Vernier scale divisions are equal to $24$ main scale divisions, what is the least count of the travelling microscope $..........\,cm$

Least count of a vernier caliper is $\frac{1}{20 \mathrm{~N}} \mathrm{~cm}$. The value of one division on the main scale is $1 \mathrm{~mm}$. Then the number of divisions of main scale that coincide with $\mathrm{N}$ divisions of vernier scale is :