$A$ student measured the diameter of a wire using a screw gauge with the least count $0.001\, cm$ and listed the measurements. The measured value should be recorded as (in $, cm$)

In a screw gauge,the fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are $50$ divisions on the circular scale,and the main scale moves by $0.5 \, mm$ on a complete rotation. For a particular observation,the reading on the main scale is $5 \, mm$ and the $20^{th}$ division of the circular scale coincides with the reference line. Calculate the true reading in $mm$.

In a vernier callipers, $20$ $VSD$ coincide with $16$ $MSD$ (each division of length $1 \text{ mm}$). The least count of the vernier callipers is: (in $\text{ cm}$)

$A$ Vernier calipers has $1 \,mm$ marks on the main scale. It has $20$ equal divisions on the Vernier scale which match with $16$ main scale divisions. For this Vernier calipers, the least count is (in $\,mm$)

$A$ vernier callipers has $20$ divisions on the vernier scale, which coincide with $19$ divisions on the main scale. The least count of the instrument is $0.1 \,mm$. One main scale division is equal to $...$ $mm$.

$A$ 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: $1 \, M.S.D. = 100 \, C.S.D. = 1 \, mm$.