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:

The vernier scale of a travelling microscope has $50$ divisions which coincide with $49$ main scale divisions. If each main scale division is $0.5 \ mm$,calculate the minimum inaccuracy in the measurement of distance. (in $mm$)

Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its $10$ Vernier Scale Divisions $(V.S.D.)$ are equal to its $9$ Main Scale Divisions $(M.S.D.)$. The least division on the $M.S.$ is $0.1 \ cm$ and the zero of $V.S.$ is at $x=0.1 \ cm$ when the jaws of the Vernier callipers are closed. If the main scale reading for the diameter is $M=5 \ cm$ and the number of the coinciding vernier division is $8$,the measured diameter after zero error correction is: (in $cm$)

In an experiment,the angles are required to be measured using an instrument in which $29$ divisions of the main scale exactly coincide with the $30$ divisions of the vernier scale. If the smallest division of the main scale is half a degree $(=0.5^{\circ})$,then the least count of the instrument is

The smallest division on the main scale of a Vernier calipers is $0.1 \text{ cm}$. Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is (in $\text{ cm}$)

In a vernier callipers,$50$ vernier scale divisions are equal to $48$ main scale divisions. If one main scale division $= 0.05 \ mm$,then the least count of the vernier callipers is . . . . . . $mm$.