In a vernier callipers, one main scale division is $1$ mm and $10$ vernier scale divisions coincide with $9$ main scale divisions. With the jaws closed, the zero of the vernier scale lies to the left of the zero of the main scale and the $6^{\text{th}}$ vernier division coincides with a main scale division. When a sphere is held between the jaws, the main scale reading is $2.1$ cm and the $4^{\text{th}}$ vernier division coincides. The correct diameter of the sphere is
Answer: (A) $2.18$ cm
Concept. Least count $=$ 1 MSD $-$ 1 VSD. A vernier zero to the left of the main zero is a negative zero error; the correct reading is observed reading $-$ zero error.
Step 1. Least count.
$$\text{LC} = 1\ \text{mm} - 0.9\ \text{mm} = 0.1\ \text{mm}$$
Step 2. Zero error. The vernier zero lies between $-1$ mm and $0$; with the $6^{\text{th}}$ division coinciding,
$$e = -1\ \text{mm} + 6(0.1\ \text{mm}) = -0.4\ \text{mm}$$
Step 3. Observed reading. $21\ \text{mm} + 4(0.1\ \text{mm}) = 21.4$ mm.
Step 4. Correct diameter.
$$21.4 - (-0.4) = 21.8\ \text{mm} = 2.18\ \text{cm}$$
Answer. $2.18$ cm (Option A). Using $+0.6$ mm (6 × LC) as the error gives the trap $2.08$ cm or $2.20$ cm.
Solution by Sreeraj P, M.Sc Physics