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Units and Measurements question

2010 · Shift 2 · Q39
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Units and Measurements question

2010 · Shift 2 · Q39

JEE AdvancedPhysicsUnits and MeasurementsMCQ+4 / −1
A vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16 main scale divisions. For this Vernier calipers, the least count is
  1. A
    0.02 mm
  2. B
    0.05 mm
  3. C
    0.1 mm
  4. D
    0.2 mm
View written solutionFree

Correct answer: D

  1. Main scale division (MSD)

    The main scale has 1 mm1\text{ mm}1 mm marks, so 1 MSD=1 mm1\,\text{MSD} = 1\text{ mm}1MSD=1 mm

  2. Relation between Vernier scale and main scale

    Given: 20 VSD=16 MSD20\text{ VSD} = 16\text{ MSD}20 VSD=16 MSD

    Since 1 MSD=1 mm1\text{ MSD} = 1\text{ mm}1 MSD=1 mm, 20 VSD=16 mm20\text{ VSD} = 16\text{ mm}20 VSD=16 mm

    Therefore, 1 VSD=1620 mm=0.8 mm1\text{ VSD} = \frac{16}{20}\text{ mm} = 0.8\text{ mm}1 VSD=2016​ mm=0.8 mm

  3. Least count of Vernier calipers

    For a direct vernier, Least Count=1 MSD−1 VSD\text{Least Count} = 1\text{ MSD} - 1\text{ VSD}Least Count=1 MSD−1 VSD

    Substituting values: LC=1.0−0.8=0.2 mm\text{LC} = 1.0 - 0.8 = 0.2\text{ mm}LC=1.0−0.8=0.2 mm

  4. Match with options

    0.2 mm0.2\text{ mm}0.2 mm corresponds to Option D.

  5. Comparison with stored answer

    Stored correct answer: D

    Derived answer: D

    So, the derived answer agrees with the stored answer.

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