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

When both jaws of vernier callipers touch each other,the zero mark of the vernier scale is to the right of the zero mark of the main scale,and the $4^{\text{th}}$ mark on the vernier scale coincides with a certain mark on the main scale. While measuring the length of a cylinder,the observer observes $15$ divisions on the main scale and the $5^{\text{th}}$ division of the vernier scale coincides with a main scale division. The measured length of the cylinder is . . . . . . $mm$. (Least count of Vernier calliper $= 0.1 \ mm$)

$A$ screw gauge gives the following readings when used to measure the diameter of a wire:
Main scale reading: $0 \, mm$
Circular scale reading: $52$ divisions
Given that $1 \, mm$ on the main scale corresponds to $100$ divisions on the circular scale. The diameter of the wire from the above data is ...... $cm$.

The vernier scale of a travelling microscope has $50$ divisions which coincide with $49$ main scale divisions. If each main scale division is $0.5 \ mm$,calculate the minimum inaccuracy in the measurement of distance. (in $mm$)

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 $\text{ cm}$)

The diagrams show readings of a screw gauge. Figure $(i)$ shows the zero error reading when the screw gauge is closed,and figure $(ii)$ shows the reading when the screw gauge is being used to measure the diameter of a ball-bearing. What is the diameter of the ball-bearing in $mm$? There are $50$ divisions on the circular scale.