Question
Consider the diameter of a spherical object being measured with the help of a Vernier calipers. 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 $x = 0.1\text{ }cm$ when the jaws of Vernier callipers are closed.
If the main scale reading for the diameter is $M = 5\text{ }cm$ and the number of coinciding vernier division is 8 , the measured diameter after zero error correction, is
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Solution
Least count $= 1MSD - 1VSD$
$$\begin{matrix} & \ = 1MSD - \frac{9}{10}MSD \\ & \ = \frac{1}{10}MSD \\ & \ = \frac{1}{10} \times 0.1\text{ }cm = 0.01\text{ }cm \end{matrix}$$
Zero error = +0.1 cm
Main scale reading $= 5\text{ }cm$
Vernier scale reading $= 8 \times 0.01 = 0.08\text{ }cm$
Final measurement of diameter
$$= 5 + 0.08 - 0.1 = 4.98\text{ }cm$$
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