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 l

Question

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 on the M.S. is 0.1 cm and the zero of the V.S. is at $x = 0.1$ cm when the jaws of the Vernier callipers are closed. If the main scale reading for the diameter is $M = 5$ cm and the number of the coinciding Vernier division is 8, the measured diameter after zero-error correction is:

A.5.18 cm
B.5.08 cm
C.4.98 cm✓
D.5.00 cm

Solution

1 VSD $= 0.9$ MSD, so the least count is $0.1 \times 0.1 = 0.01$ cm. Reading $= 5 + 8 \times 0.01 = 5.08$ cm; the zero error is $+ 0.1$ cm, so the corrected diameter is $5.08 - 0.10 = 4.98$ cm.

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Vishal Gupta · Accepted Answer

C. 1 VSD = 0.9 MSD, so the least count is 0.1 	imes 0.1 = 0.01 cm. Reading = 5 + 8 	imes 0.01 = 5.08 cm; the zero error is +0.1 cm, so the corrected diameter is 5.08 - 0.10 = 4.98 cm.

Units and Measurements

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