If $50$ Vernier divisions are equal to $49$ main scale divisions of a travelling microscope and one smallest reading of main scale is $0.5 \,mm$, the Vernier constant of travelling microscope is:

$A$ student measured the diameter of a small steel ball using a screw gauge of least count $0.001 \, cm$. The main scale reading is $5 \, mm$ and the circular scale division coinciding with the reference level is $25$. If the screw gauge has a zero error of $-0.004 \, cm$,the correct diameter of the ball is: (in $, cm$)

The one division of main scale of vernier callipers reads $1\,mm$ and $10$ divisions of Vernier scale is equal to the $9$ divisions on main scale. When the two jaws of the instrument touch each other the $zero$ of the Vernier lies to the right of $zero$ of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws,the $zero$ of the Vernier scale lies in between $4.1\,cm$ and $4.2\,cm$ and $6^{\text{th}}$ Vernier division coincides with a main scale division. The diameter of the bob will be $.............10^{-2}\,cm$

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

$A$ screw gauge has a least count of $0.01\, mm$ and there are $50$ divisions on its circular scale. The pitch of the screw gauge is $........\, mm$.

In a vernier callipers,$10$ divisions of vernier scale coincide with $9$ divisions of main scale. One division of main scale is of $0.1 \ cm$. If in the measurement of inner diameter of a cylinder,the zero of the vernier scale lies between $1.3 \ cm$ and $1.4 \ cm$ of the main scale and the $2^{\text{nd}}$ division of the vernier scale coincides with a main scale division,then the diameter will be: (in $cm$)