Figure $1$ shows the configuration of the main scale and Vernier scale before measurement. Figure $2$ shows the configuration corresponding to the measurement of the diameter $D$ of a tube. The measured value of $D$ is (in $cm$)

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

In an experiment to find out the diameter of a wire using a screw gauge,the following observations were noted:
$(a)$ The screw moves $0.5\,mm$ on the main scale in one complete rotation.
$(b)$ Total divisions on the circular scale $= 50$.
$(c)$ Main scale reading is $2.5\,mm$.
$(d)$ The $45^{\text{th}}$ division of the circular scale is on the pitch line.
$(e)$ The instrument has a $0.03\,mm$ negative zero error.
Then the diameter of the wire is $...........\,mm$.

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 $= 0.1 \ mm$)

Length of $9 \ MSD$ of a vernier caliper are equal to $10 \ VSD$ and $1 \ MSD = 1 \ mm$. For measuring the length of a rod,the reading on the main scale is $6.4 \ cm$ and the $8^{th}$ division on the vernier scale is in line with a marking on the main scale. If there is no zero error,find the length of the rod: (in $cm$)

$A$ 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?