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 . . . . . . $mm$.

The pitch of the screw gauge is $1\, mm$ and there are $100$ divisions on the circular scale. When nothing is put in between the jaws,the zero of the circular scale lies $8$ divisions below the reference line. When a wire is placed between the jaws,the first linear scale division is clearly visible while $72^{nd}$ division on the circular scale coincides with the reference line. The radius of the wire is.........$mm$

The vernier calliper used for measurement has a positive zero error of $0.3 \ mm$. While taking measurement for the internal diameter of a vessel,it was observed that the zero of the vernier scale lies between $9.5 \ cm$ and $9.6 \ cm$ of the main scale and the $6^{th}$ division of the vernier scale coincides with a main scale division. If the least count of the vernier calliper is $0.01 \ cm$,the correct value of the diameter will be: (in $cm$)

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 . . . . . . $\text{mm}$.

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 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 . . . . . . $\times 10^{-3} \ mm$.