The smallest division on the main scale of a Vernier calipers is $0.1 \text{ cm}$. Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is (in $\text{ cm}$)

$N$ divisions on the main scale of a vernier calliper coincide with $(N + 1)$ divisions of the vernier scale. If each division of the main scale is $a$ units,then the least count of the instrument is:

The least count of a screw gauge is $0.01 \ mm$. If the pitch is increased by $75\%$ and the number of divisions on the circular scale is reduced by $50\%$,the new least count will be . . . . . . $\times 10^{-3} \ mm$.

$A$ vernier callipers used by a student has $20$ divisions in $1\;cm$ on the main scale. $10$ vernier divisions coincide with $9$ main scale divisions. When the jaws are closed,the zero of the main scale is to the left of the zero of the vernier scale,and the $6^{th}$ division of the vernier scale coincides with a main scale division. The student places a wooden cylinder between the jaws to measure its length. The zero of the vernier scale is to the right of the $3.20\;cm$ mark,and the $8^{th}$ vernier division coincides with a main scale division. When measuring the thickness (diameter) of the cylinder,the zero of the vernier scale lies to the right of the $1.50\;cm$ mark,and the $6^{th}$ vernier division coincides with a main scale division. The correct values of the measured length and diameter are respectively:

In a screw gauge,$5$ complete rotations of the circular scale give a $1.5 \, mm$ reading on the linear scale. The circular scale has $50$ divisions. The least count of the screw gauge is: (in $, mm$)

$A$ 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?