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Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 8 April, Shift 2 — Question 43

There are 100 divisions on the circular scale of a screw gauge of pitch 1 mm . With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found the 4 linear scale divisions are clearly visible while 60 divisions on circular scale coincide with the reference line. The diameter of the wire is :

  1. Option A:

    4.65 mm

  2. Option B:

    4.55 mm

    Correct
  3. Option C:

    4.60 mm

  4. Option D:

    3.35 mm

Answer: B

Step-by-step solution

Least count =1100 mm=0.01 mm=\frac{1}{100} \mathrm{~mm}=0.01 \mathrm{~mm}

zero error =+0.05 mm=+0.05 \mathrm{~mm}

Reading =4×1 mm+60×0.01 mm−0.05 mm=4 \times 1 \mathrm{~mm}+60 \times 0.01 \mathrm{~mm}-0.05 \mathrm{~mm}

=4.55 mm=4.55 \mathrm{~mm}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
8 April, Shift 2
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Vernier Calipers and Screw Gauge

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