The least count of a vernier caliper is $\frac{1}{20N} \text{ cm}$. The value of one division on the main scale is $1 \text{ mm}$. Then the number of divisions of the main scale that coincide with $N$ divisions of the vernier scale is:

The smallest division on the main scale of a vernier callipers is $1 \ mm$,and $10$ vernier divisions coincide with $9$ main scale divisions. While measuring the diameter of a sphere,the zero mark of the vernier scale lies between $2.0 \ cm$ and $2.1 \ cm$ of the main scale,and the fifth division of the vernier scale coincides with a main scale division. The diameter of the sphere is:

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	$(b)$ $0.001\,cm$
	$(c)$ $0.0001\,cm$

$A$ travelling microscope has $20$ divisions per $cm$ on the main scale while its Vernier scale has total $50$ divisions and $25$ Vernier scale divisions are equal to $24$ main scale divisions. What is the least count of the travelling microscope in $cm$?

$A$ tiny metallic rectangular sheet has a length and breadth of $5 \ mm$ and $2.5 \ mm$,respectively. Using a specially designed screw gauge which has a pitch of $0.75 \ mm$ and $15$ divisions on the circular scale,you are asked to find the area of the sheet. In this measurement,the maximum fractional error will be $\frac{x}{100}$ where $x$ is . . . . . .

The pitch of a screw gauge is $0.5 \ mm$ and there are $100$ divisions on its circular scale. The instrument reads $+2$ divisions on the circular scale when nothing is placed between its jaws. In measuring the diameter of a wire,there are $8$ divisions on the main scale and the $83^{rd}$ division of the circular scale coincides with the reference line. Then the diameter of the wire is (in $mm$)