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 $= 0.01\, cm$)

$A$ 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 when 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?

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

The circular divisions of a screw gauge are $50$. It moves $0.5 \ mm$ on the main scale in one rotation. When the diameter of a wire is measured,the main scale reading is $3.5 \ mm$ and the circular scale reading is $32$. If the zero error (positive) in the screw gauge is $0.06 \ mm$,then the diameter of the wire is:

For the determination of the refractive index of a glass slab,a travelling microscope is used whose main scale contains $300$ equal divisions equal to $15 \ cm$. The vernier scale attached to the microscope has $25$ divisions equal to $24$ divisions of the main scale. The least count $(LC)$ of the travelling microscope is (in $cm$):

In a screw gauge,the zero of the circular scale lies $3$ divisions above the horizontal pitch line when their metallic studs are brought in contact. Using this instrument,the thickness of a sheet is measured. If the pitch scale reading is $1 \ mm$ and the circular scale reading is $51$,then the correct thickness of the sheet is . . . . . . $mm$. [Assume least count is $0.01 \ mm$]