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

$A$ Vernier calipers has $1 \,mm$ marks on the main scale. It has $20$ equal divisions on the Vernier scale which match with $16$ main scale divisions. For this Vernier calipers, the least count is (in $\,mm$)

The least count of a screw gauge is $0.01 \ mm$. If the pitch is increased by $75\%$ and the number of divisions on the circular scale is reduced by $50\%$,the new least count will be . . . . . . $\times 10^{-3} \ mm$.

In a screw gauge,the zero of the main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are $100$ divisions on the circular scale and the pitch of the screw gauge is $0.1 \text{ mm}$. When the diameter of a sphere is measured,the reading of the main scale is $5 \text{ mm}$ and the $50^{th}$ division of the circular scale coincides with the reference line of the main scale. The diameter of the sphere is . . . . . . $\text{mm}$.

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 $\text{ cm}$)

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 $cm$)