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 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 $1 \text{ cm}$?

Two full turns of the circular scale of a screw gauge cover a distance of $1 \ mm$ on its main scale. The total number of divisions on the circular scale is $50$. Further,it is found that the screw gauge has a zero error of $-0.03 \ mm$. While measuring the diameter of a thin wire,a student notes the main scale reading of $3 \ mm$ and the number of circular scale divisions in line with the main scale as $35$. The diameter of the wire is ....... $mm$

$A$ screw gauge has a pitch of $1.5\; mm$ and there is no zero error. The linear scale has markings at $MSD = 1\; mm$ and there are $100$ equal divisions on the circular scale. When the diameter of a sphere is measured with this instrument,the $2\; mm$ mark is visible on the linear scale,but the $3\; mm$ mark is not visible. The $76^{th}$ division of the circular scale is in line with the linear scale. What is the diameter of the sphere in $mm$?

Student $A$ and Student $B$ used two screw gauges of equal pitch and $100$ equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is $0.322 \, \text{cm}$. The absolute value of the difference between the final circular scale readings observed by the students $A$ and $B$ is .... .
Given pitch $= 0.1 \, \text{cm}$.

$A$ screw gauge has $50$ divisions on its circular scale. The circular scale is $4$ units ahead of the pitch scale marking,prior to use. Upon one complete rotation of the circular scale,a displacement of $0.5\, mm$ is noticed on the pitch scale. The nature of zero error involved,and the least count of the screw gauge,are respectively