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Physics · Units, Dimensions & Error Analysis

JEE Main 2026 — 2 April, Evening Shift — Question 5

In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is 0.1mm0.1\mathrm{mm}. When diameter of a sphere is measured, the reading of main scale is 5mm5\mathrm{mm} and 50th50^{\mathrm{th}} division of circular scale coincides with the reference line of main scale. The diameter of sphere is mm.

  1. Option A:

    5.045

    Correct
  2. Option B:

    5.055

  3. Option C:

    5.45

  4. Option D:

    5.55

Answer: A

Step-by-step solution

Energy=hcλ=GM1M2r\mathrm{Energy} = \frac{\mathrm{hc}}{\lambda} = \frac{\mathrm{GM}_{1}\mathrm{M}_{2}}{\mathrm{r}} [ML−2T−2]=[h][T−1]=[G][M2L−2]\left[\mathrm{ML}^{-2}\mathrm{T}^{-2}\right] = \left[\mathrm{h}\right]\left[\mathrm{T}^{-1}\right] = \left[\mathrm{G}\right]\left[\mathrm{M}^{2}\mathrm{L}^{-2}\right] [G]=[M−2L−1T−1h−1]\left[\mathrm{G}\right] = \left[\mathrm{M}^{-2}\mathrm{L}^{-1}\mathrm{T}^{-1}\mathrm{h}^{-1}\right]

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
2 April, Evening Shift
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Vernier Calipers and Screw Gauge

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