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

In an expression $$a \times 10^b$$ :

The diameter of a sphere is measured using a vernier caliper whose 9 divisions of main scale are equal to 10 divisions of vernier scale. The shortest division on the main scale is equal to $$1 \mathrm{~mm}$$. The main scale reading is $$2 \mathrm{~cm}$$ and second division of vernier scale coincides with a division on main scale. If mass of the sphere is 8.635 $$\mathrm{g}$$, the density of the sphere is:

In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its $$4^{\text {th }}$$ division coincides exactly with a certain division on main scale. If 50 vernier scale divisions equal to 49 main scale divisions and zero error in the instrument is $$0.04 \mathrm{~mm}$$ then how many main scale divisions are there in $$1 \mathrm{~cm}$$ ?