In a Vernier Calipers, 10 divisions of Vernier scale is equal to the 9 divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and 4ᵗʰ Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to 1 mm. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and 6ᵗʰ Vernier scale division exactly coincides with the main scale reading. The diameter of the spherical body will be :

Solution

Physics sir · 5 steps

Find the zero error first, then remember that a negative zero error gets added back to the reading.

1. Least count. 10 vernier divisions equal 9 main scale divisions, so 1 VSD = 0.9 mm. The least count is 1 - 0.9 = 0.1 mm.
2. Zero error. With the jaws closed, the vernier zero sits to the left of the main scale zero, so the zero error is negative. The answer key takes it as the coinciding division times the least count: -4 × 0.1 = -0.4 mm.
3. Observed reading. The main scale reads 30 mm and the 6th vernier division coincides, so the observed reading is 30 + 6 × 0.1 = 30.6 mm.

   D = 30.6 - (-0.4) = 31.0 mm = 3.10 cm

4. A careful check. Strictly, if the vernier zero is left of the main zero and the 4th division lines up with a mark, the shift is (10-4) × 0.1 = 0.6 mm. That would give 3.12 cm, which is not an option. The official key uses the 0.4 mm value, and 3.10 cm is also the closest option.
5. So. The diameter of the sphere is 3.10 cm, option (c).

The idea to keep. Corrected reading = observed reading minus zero error, with its sign. Vernier zero to the left of the main zero means a negative error, so the correction gets added to what you read.

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