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Physics · Units, Dimensions & Error Analysis

NEET (UG) 2025 — Question 23

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\mathrm{x}=0.1 \mathrm{~cm} when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is M=5 cmM=5 \mathrm{~cm} and the number of coinciding vernier division is 8, the measured diameter after zero error correction, is

  1. Option A:

    5.18 cm

  2. Option B:

    5.08 cm

  3. Option C:

    4.98 cm

    Correct
  4. Option D:

    5.00 cm

Answer: C

Step-by-step solution

10 V.S.D. = 9 M.S.D.

1 M.S.D. =0.1 cm=0.1 \mathrm{~cm} (Given)

1 V.S.D. = 0.9 M.S.D.

1 V.S.D. =0.9×0.1 cm=0.9 \times 0.1 \mathrm{~cm}

∴\therefore V.S.D. =0.09 cm=0.09 \mathrm{~cm}

L.C. =1=1 M.S.D. -1 V.S.D.

=(0.1−0.09)cm=(0.1-0.09) \mathrm{cm}

=0.01 cm=0.01 \mathrm{~cm}

Reading =5.08 cm=5.08 \mathrm{~cm}

Positive zero error =0.1 cm=0.1 \mathrm{~cm} (Given)

∴\therefore Final Corrected reading =(5.08−0.1)cm=(5.08-0.1) \mathrm{cm}

=4.98 cm=4.98 \mathrm{~cm}

Answer key and solution verified before publishing.

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Exam
NEET (UG) 2025
Paper
2025 paper
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Vernier Calipers and Screw Gauge

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