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 $cm$)

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

The diameter of a sphere is measured using a vernier caliper whose $9$ divisions of the main scale are equal to $10$ divisions of the vernier scale. The shortest division on the main scale is equal to $1 \,mm$. The main scale reading is $2 \,cm$ and the second division of the vernier scale coincides with a division on the main scale. If the mass of the sphere is $8.635 \,g$, the density of the sphere is:

Student $A$ and Student $B$ used two screw gauges of equal pitch and $100$ equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is $0.322 \, \text{cm}$. The absolute value of the difference between the final circular scale readings observed by the students $A$ and $B$ is .... .
Given pitch $= 0.1 \, \text{cm}$.

Using a screw gauge with a pitch of $0.1 \ cm$ and $50$ divisions on its circular scale,the thickness of an object is measured. How should the measurement be correctly recorded (in $cm$)?

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.