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Physics · Units, Dimensions & Error Analysis

JEE Main 2026 — 23 January, Morning Shift — Question 42

In a screw gauge, the zero of the circular scale lies 3 divisions above the horizontal pitch line when their metallic studs are brought in contact. Using this instrument thickness of a sheet is measured. If pitch scale reading is 1 mm and the circular scale reading is 51 then the correct thickness of the sheet is ____\_\_\_\_ mm. [Assume least count is 0.01 mm ]

  1. Option A:

    1.5

  2. Option B:

    1.48

  3. Option C:

    1.54

    Correct
  4. Option D:

    1.51

Answer: C

Step-by-step solution

Zero error e=−3×LC=−0.03 mm\mathrm{e}=-3 \times \mathrm{LC}=-0.03 \mathrm{~mm}

Reading taken =1 mm+51(0.01 mm)=1 \mathrm{~mm}+51(0.01 \mathrm{~mm}) =1.51 mm=1.51 \mathrm{~mm}

So, correct reading =1.51−(−0.03)=1.51-(-0.03) =1.54 mm=1.54 \mathrm{~mm}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
23 January, Morning Shift
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Vernier Calipers and Screw Gauge

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