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:

$A$ 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 $........\, mm$.

If the screw on a screw gauge is given six rotations,it moves by $3\; mm$ on the main scale. If there are $50$ divisions on the circular scale,the least count of the screw gauge is:

The main scale division of a vernier caliper is $1 \ mm$. The vernier scale divisions are in an arithmetic progression $(A.P.)$; the $1^{st}$ division is $0.95 \ mm$, the $2^{nd}$ division is $0.90 \ mm$, and so on. When an object is placed between the jaws of the vernier caliper, the zero of the vernier scale lies between $3.1 \ cm$ and $3.2 \ cm$, and the $4^{th}$ division of the vernier scale coincides with a main scale division. The reading of the vernier caliper is .......... $cm$.

$A$ travelling microscope has $20$ divisions per $cm$ on the main scale while its Vernier scale has total $50$ divisions and $25$ Vernier scale divisions are equal to $24$ main scale divisions. What is the least count of the travelling microscope in $cm$?

In a vernier calliper,when both jaws touch each other,the zero of the vernier scale shifts towards the left and its $4^{\text{th}}$ division coincides exactly with a certain division on the main scale. If $50$ vernier scale divisions $(VSD)$ are equal to $49$ main scale divisions $(MSD)$ and the zero error in the instrument is $0.04 \text{ mm}$,then how many main scale divisions are there in $1 \text{ cm}$?