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 (a) 4.98 cm (b) 5.00 cm (c) 5.18 cm (d) 5.08 cm

Answer: (a). LC = 1 MSD - 1 VSD = 0.1 - 0.09 = 0.01 cm Zero of VS at +0.1 cm with jaws closed: positive zero error e = +0.1 cm. Reading = 5 + 8× 0.01 = 5.08 cm Corrected = 5.08 - 0.1 = 4.98 cm, answer (a).

Consider the diameter of a spherical object being measured with the… | NEET 2025

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Least Count, Vernier Callipers & Screw Gauge

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 $x = 0.1\text{ cm}$ 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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Answer: (a)

LC $= 1\text{ MSD} - 1\text{ VSD} = 0.1 - 0.09 = 0.01$ cm
Zero of VS at $+0.1$ cm with jaws closed: positive zero error $e = +0.1$ cm.
Reading $= 5 + 8\times 0.01 = 5.08$ cm
Corrected $= 5.08 - 0.1 = 4.98$ cm, answer (a).

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