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Physics · Units, Dimensions & Error Analysis

JEE Advanced 2021 — Paper 1 — Question 1

The smallest division on the main scale of a Vernier calipers is 0.1 cm . Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is

Question figure
  1. Option A:

    3.07 cm

  2. Option B:

    3.11 cm

  3. Option C:

    3.15 cm

    Correct
  4. Option D:

    3.17 cm

Answer: C

Step-by-step solution

Least count =(1−910)(0.1)=0.01 cm=\left(1-\frac{9}{10}\right)(0.1)=0.01 \mathrm{~cm} Zero error =−0.1+0.06=−0.04 cm=-0.1+0.06=-0.04 \mathrm{~cm}

Final reading =3.1+0.01×1=3.11 cm=3.1+0.01 \times 1=3.11 \mathrm{~cm} So correct measurement =3.11+0.04=3.15 cm=3.11+0.04=3.15 \mathrm{~cm}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2021
Paper
Paper 1
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Vernier Calipers and Screw Gauge
The smallest division on the main scale of a Vernier calipers is 0.1… | JEE Advanced 2021 PYQ with Solution · DhiX AI