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Physics · Units, Dimensions & Error Analysis

JEE Main 2025 — 4 April, Evening Shift — Question 48

For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm . The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm )

  1. Option A:

    0.0005

  2. Option B:

    0.001

  3. Option C:

    0.002

    Correct
  4. Option D:

    0.0025

Answer: C

Step-by-step solution

MSD=15300=0.05 cmM S D=\frac{15}{300}=0.05 \mathrm{~cm}

25VSD=24MSDMVSD=2425MSDLC=1MSD−1VSD=(1−2425)MSD=125×0.05 cm=0.002 cm\begin{aligned} & 25 \mathrm{VSD}=24 \mathrm{MSD} \\ & \quad \mathrm{MVSD}=\frac{24}{25} \mathrm{MSD} \\ & \mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD} \\ & =\left(1-\frac{24}{25}\right) \mathrm{MSD} \\ & =\frac{1}{25} \times 0.05 \mathrm{~cm} \\ & =0.002 \mathrm{~cm} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Evening Shift
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Vernier Calipers and Screw Gauge

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