$A$ student measured the length of a rod and wrote it as $3.50\;cm$. Which instrument did he use to measure it?

In a Vernier calipers, when both jaws touch each other, the zero of the Vernier scale is shifted to the right of the zero of the main scale and the $7^{th}$ Vernier division coincides with a main scale division. If the value of $1$ main scale division is $1 \text{ mm}$ and there are $10$ Vernier scale divisions, then the Vernier caliper has

$A$ 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 gauge,the state of the instrument is shown by diagram $(I)$. When both the rods are inserted together in series,the state is shown by diagram $(II)$. What is the zero error of the instrument in $mm$? Given: $1 \, M.S.D. = 100 \, C.S.D. = 1 \, mm$.

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: In a Vernier calliper,if a positive zero error exists,then while taking measurements,the reading taken will be more than the actual reading.
Reason $(R)$: The zero error in a Vernier calliper might have happened due to a manufacturing defect or due to rough handling.
In the light of the above statements,choose the correct answer from the options given below:

The least count of the main scale of a vernier callipers is $1\, mm$. Its vernier scale is divided into $10$ divisions and coincides with $9$ divisions of the main scale. When jaws are touching each other,the $7^{th}$ division of the vernier scale coincides with a division of the main scale and the zero of the vernier scale lies to the right side of the zero of the main scale. When this vernier is used to measure the length of a cylinder,the zero of the vernier scale is between $3.1\, cm$ and $3.2\, cm$ and the $4^{th}$ $VSD$ coincides with a main scale division. The length of the cylinder is $.....\, cm$. ($VSD$ is vernier scale division)

$A$ screw gauge gives the following reading 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 of the circular scale. The diameter of the wire from the above data is: (in $cm$)