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 mm. The main scale reading is 2 cm and second division of vernier scale coincides with a division on main scale. If mass of the sphere is 8.635 g, the density of the sphere is:

Solution

Physics sir · 4 steps

First find the least count, then read the diameter, and only then use mass over volume.

1. Least count. 10 vernier divisions equal 9 main scale divisions, so 1 VSD = 0.9 mm. The least count is 1 MSD - 1 VSD = 1 - 0.9 = 0.1 mm = 0.01 cm.
2. Read the diameter. The main scale reading is 2 cm and the 2nd vernier division coincides. So the diameter is 2 + 2 × 0.01 = 2.02 cm, and the radius is r = 1.01 cm.

   ρ = m/(4/3π r³) = 8.635/(4/3 × 3.1416 × (1.01)³) g cm⁻³

3. Put in numbers. (1.01)³ = 1.0303, so the volume is 4.1888 × 1.0303 = 4.316 cm³. Then ρ = 8.635 / 4.316 = 2.0 g cm⁻³.
4. So. The density of the sphere is 2.0 g/cm³, option (b).

The idea to keep. For any vernier, least count = 1 MSD − 1 VSD. Reading = main scale reading + (coinciding division × least count). Halve the diameter before you cube it.

Answer from the NTA final answer key. Solution by PrepWiser.