Consider the diameter of-a spherical object being measured with the help of a Vernier callipers. Suppose its \(10\) Vern…
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 \(x=0.1\mathrm{cm}\) when the jaws,of Vernier callipers are closed. If the main scale reading for the diameter is \(\mathrm{M}=5\mathrm{cm}\) and the number of coinciding vernier division is \(8\) , the measured diameter after zero error correction, is
\(4.98\mathrm{cm}\)
Least count = 1MSD - 1VSD
\(1MSD−\frac{9}{10}MSD\)\(=\frac{1}{10}MSD\)
\(=\frac{1}{10}\times 0.1\text{ cm}=0.01\text{cm}\)
Zero error = +0.1 cm
Main scale reading = 8 × 0.01 = 0.08 cm
Final measurement of diameter
= 5 + 0.08 - 0.1 = 4.98 cm
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