There are 100 divisions on the circular scale of a screw gauge of pitch $$1 \mathrm{~mm}$$. With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that 4 linear scale divisions are clearly visible while 60 divisions on circular scale coincide with the reference line. The diameter of the wire is :

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: