Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its $10$ Vernier Scale Divisions (V.S.D.) are equal to its $9$ Main Scale Divisions (M.S.D.). The least division in the M.S. is $0.1$ cm and the zero of V.S. is at $x = 0.1$ cm when the jaws of Vernier callipers are closed.
If the main scale reading for the diameter is $M = 5$ cm and the number of coinciding vernier division is $8$, the measured diameter after zero error correction, is
Answer: (C) $4.98$ cm
Least count $= 1\ \text{MSD} - 1\ \text{VSD} = 0.1 - \dfrac{9}{10}(0.1) = 0.01$ cm.
With the jaws closed, the vernier zero is at $+0.1$ cm, so the zero error is $+0.1$ cm.
Observed reading $= 5 + 8 \times 0.01 = 5.08$ cm.
Corrected diameter $= 5.08 - 0.1 = 4.98$ cm.
Solution by Sreeraj P, M.Sc Physics