A spectrometer gives the following reading when used to measure the angle of a prism. 

Main scale reading : $58.5$ degree

Vernier scale reading : $09$ divisions

Given that $1$ division on main scale corresponds to $0 .5$ degree. Total divisions on the Vernier scale is $30$ and match with $29$ divisions of the main scale. The angle of the prism from the above data ....... $degree$

There are $100$ divisions on the circular scale of a screw gauge of pitch $1 \mathrm{~mm}$. With no measuring quantity in between the jaws, the zero of the circular scale lies $5$ divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found the $4$ linear scale divisions are clearly visible while $60$ divisions on circular scale coincide with the reference line. The diameter of the wire is :

A screw gauge has some zero error but its value is unknown. We have two identical rods. When the first rod is inserted in the screw, the state of the instrument is shown by diagram $(I).$ When both the rods are inserted together in series then the state is shown by the diagram $(II).$ What is the zero error of the instrument ? .......... $mm$

$1\,M.S.D. = 100\, C.S.D. = 1\, mm $ 

A travelling microscope is used to determine the refractive index of a glass slab. If $40$ divisions are there in $1 \; cm$ on main scale and $50$ Vernier scale divisions are equal to $49$ main scale divisions, then least count of the travelling microscope is $\dots \; \times 10^{-6} \; m$

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: