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Physics · Units, Dimensions & Error Analysis

JEE Main 2026 — 24 January, Evening Shift — Question 27

In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division =0.05 mm=0.05 \mathrm{~mm}, then the least count of the vernier callipers is ____\_\_\_\_ mm .

  1. Option A:

    0.002

    Correct
  2. Option B:

    0.05

  3. Option C:

    0.02

  4. Option D:

    0.005

Answer: A

Step-by-step solution

LC=1MSD−1MSD=1MSD−4850MSD \mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{MSD}=1 \mathrm{MSD}-\frac{48}{50} \mathrm{MSD}

=250MSD=250×.05 mm=0.002 mm=\frac{2}{50} \mathrm{MSD}=\frac{2}{50} \times .05 \mathrm{~mm}=0.002 \mathrm{~mm}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Evening Shift
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Vernier Calipers and Screw Gauge
In a vernier callipers, 50 vernier scale divisions are equal to 48… | JEE Main 2026 PYQ with Solution · DhiX AI