The vernier constant of Vernier callipers is $0.1 \,mm$ and it has zero error of $(-0.05) \,cm$. While measuring the diameter of a sphere,the main scale reading is $1.7 \,cm$ and the coinciding vernier division is $5$. The corrected diameter will be ........... $\times 10^{-2} \,cm$.

In a vernier callipers,$10$ divisions of vernier scale coincide with $9$ divisions of main scale,the least count of which is $0.1\,cm$. If in the measurement of inner diameter of a cylinder,the zero of the vernier scale lies between $1.3\,cm$ and $1.4\,cm$ of the main scale and the $2^{nd}$ division of the vernier scale coincides with a main scale division,then the diameter will be .......... $cm$.

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

When both jaws of vernier callipers touch each other,the zero mark of the vernier scale is to the right of the zero mark of the main scale,and the $4^{\text{th}}$ mark on the vernier scale coincides with a certain mark on the main scale. While measuring the length of a cylinder,the observer observes $15$ divisions on the main scale and the $5^{\text{th}}$ division of the vernier scale coincides with a main scale division. The measured length of the cylinder is . . . . . . $mm$. (Least count of Vernier calliper $= 0.1 \ mm$)

$A$ specially designed Vernier calliper has the main scale least count of $1 \,mm$. On the Vernier scale,there are $10$ equal divisions and they match with $11$ main scale divisions. Then,the least count of the Vernier calliper is ........... $mm$.

$A$ screw gauge gives the following readings when used to measure the diameter of a wire:
Main scale reading: $0 \, mm$
Circular scale reading: $52$ divisions
Given that $1 \, mm$ on the main scale corresponds to $100$ divisions on the circular scale. The diameter of the wire from the above data is ...... $cm$.