The diagrams show readings of a screw gauge. Figure $(i)$ shows the zero error reading when the screw gauge is closed,and figure $(ii)$ shows the reading when the screw gauge is being used to measure the diameter of a ball-bearing. What is the diameter of the ball-bearing in $mm$? There are $50$ divisions on the circular scale.

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 $= 0.1 \ mm$)

Using a screw gauge with a pitch of $0.1 \ cm$ and $50$ divisions on its circular scale,the thickness of an object is measured. How should the measurement be correctly recorded (in $cm$)?

$A$ 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 $\,mm$)

Match the instruments in Column-$I$ with their respective least counts in Column-$II$.

Column-$I$	Column-$II$
$(1)$ Traveling Microscope	$(a)$ $0.01\,cm$
$(2)$ Screw Gauge	$(b)$ $0.001\,cm$
	$(c)$ $0.0001\,cm$

In a vernier callipers,each $cm$ on the main scale is divided into $20$ equal parts. If the $10^{th}$ vernier scale division coincides with the $9^{th}$ main scale division,then the value of the vernier constant will be $\dots \; \times 10^{-2} \; mm$.