Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 1 February, Shift 1 — Question 39

10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is :

  1. Option A:

    12\frac{1}{2}

  2. Option B:

    1011\frac{10}{11}

  3. Option C:

    5011\frac{50}{11}

  4. Option D:

    511\frac{5}{11}

    Correct

Answer: D

Step-by-step solution

10MSD=11VSD10\text{MSD}=11\text{VSD}

\begin{array}{*{35}{r}}{} & 1\text{VSD}=\frac{10}{11}\text{MSD} \\{} & \text{LC}=1\text{MSD}-1\text{VSD} \\{} & ~=1\text{MSD}-\frac{10}{11}\text{MSD} \\{} & ~=\frac{1\text{MSD}}{11} \\{} & ~=\frac{5}{11}\text{ }\!\!~\!\!\text{ units }\!\!~\!\!\text{ } \\\end{array}

Answer key and solution verified before publishing.

Practise Units, Dimensions & Error Analysis

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2024
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Vernier Calipers and Screw Gauge
10 divisions on the main scale of a Vernier calliper coincide with 11… | JEE Main 2024 PYQ with Solution · DhiX AI