The pitch and the number of divisions on the circular scale for a given screw gauge are $0.5\,mm$ and $100$ respectively. When the screw gauge is fully tightened without any object,the zero of its circular scale lies $3$ divisions below the mean line. The readings of the main scale and the circular scale for a thin sheet are $5.5\,mm$ and $48$ respectively. The thickness of this sheet is: (in $,mm$)

The vernier scale used for measurement has a positive zero error of $0.2\, mm$. If while taking a measurement it was noted that the '$0$' on the vernier scale lies between $8.5\, cm$ and $8.6\, cm$ and the vernier coincidence is $6$,then the correct value of measurement is ............. $cm$. (Least count $= 0.01\, cm$)

In a vernier callipers, $50$ vernier scale divisions are equal to $48$ main scale divisions. If one main scale division $= 0.05 \ mm$, then the least count of the vernier callipers is . . . . . . $mm$.

Identify the physical quantity that cannot be measured using a spherometer:

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

In an experiment to find out the diameter of a wire using a screw gauge,the following observations were noted:
$(a)$ The screw moves $0.5\,mm$ on the main scale in one complete rotation.
$(b)$ Total divisions on the circular scale $= 50$.
$(c)$ Main scale reading is $2.5\,mm$.
$(d)$ The $45^{\text{th}}$ division of the circular scale is on the pitch line.
$(e)$ The instrument has a $0.03\,mm$ negative zero error.
Then the diameter of the wire is $...........\,mm$.