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Units and Measurements question

2015 · Shift 1 · Q45
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Units and Measurements question

2015 · Shift 1 · Q45

JEE AdvancedPhysicsUnits and MeasurementsMultiple correct+4 / −2
Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
  1. A
    If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.
  2. B
    If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.
  3. C
    If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.
  4. D
    If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.
View written solutionFree

Correct answer: B, C

  1. Find the least count of the Vernier callipers

    In the main scale, 1 cm1\text{ cm}1 cm is divided into 888 equal divisions.

    So, one main scale division (MSD) is 1 MSD=1 cm8=0.125 cm=1.25 mm.1\text{ MSD} = \frac{1\text{ cm}}{8} = 0.125\text{ cm} = 1.25\text{ mm}.1 MSD=81 cm​=0.125 cm=1.25 mm.

    Given: 5 VSD=4 MSD5\text{ VSD} = 4\text{ MSD}5 VSD=4 MSD so, 1 VSD=45 MSD.1\text{ VSD} = \frac{4}{5}\text{ MSD}.1 VSD=54​ MSD.

    Therefore, Vernier least count is LCV=1 MSD−1 VSD\text{LC}_{V} = 1\text{ MSD} - 1\text{ VSD}LCV​=1 MSD−1 VSD =(1−45)MSD=15MSD.= \left(1 - \frac{4}{5}\right)\text{MSD} = \frac{1}{5}\text{MSD}.=(1−54​)MSD=51​MSD.

    Hence, LCV=15×1.25 mm=0.25 mm.\text{LC}_{V} = \frac{1}{5}\times 1.25\text{ mm} = 0.25\text{ mm}.LCV​=51​×1.25 mm=0.25 mm.

  2. Data for the screw gauge

    Circular scale has 100100100 divisions.

    One complete rotation moves the screw by two divisions on the linear scale.

    Let the least count of one linear scale division of the screw gauge be xxx. Then pitch of screw gauge is Pitch=2x.\text{Pitch} = 2x.Pitch=2x.

    Also, LCS=Pitch100.\text{LC}_{S} = \frac{\text{Pitch}}{100}.LCS​=100Pitch​.

  1. Check Options A and B

    Given condition: pitch of screw gauge is twice the least count of Vernier callipers.

    Since LCV=0.25 mm,\text{LC}_{V} = 0.25\text{ mm},LCV​=0.25 mm, therefore, Pitch=2×0.25=0.50 mm.\text{Pitch} = 2\times 0.25 = 0.50\text{ mm}.Pitch=2×0.25=0.50 mm.

    So screw gauge least count is LCS=0.50100=0.005 mm.\text{LC}_{S} = \frac{0.50}{100} = 0.005\text{ mm}.LCS​=1000.50​=0.005 mm.

    • Option A says 0.01 mm0.01\text{ mm}0.01 mm → False
    • Option B says 0.005 mm0.005\text{ mm}0.005 mm → True

  1. Check Options C and D

    Given condition: least count of the linear scale of screw gauge is twice the least count of Vernier callipers.

    So one linear scale division is x=2×0.25=0.50 mm.x = 2\times 0.25 = 0.50\text{ mm}.x=2×0.25=0.50 mm.

    Since one complete rotation moves it by two linear scale divisions, Pitch=2x=2×0.50=1.0 mm.\text{Pitch} = 2x = 2\times 0.50 = 1.0\text{ mm}.Pitch=2x=2×0.50=1.0 mm.

    Therefore, LCS=1.0100=0.01 mm.\text{LC}_{S} = \frac{1.0}{100} = 0.01\text{ mm}.LCS​=1001.0​=0.01 mm.

    • Option C says 0.01 mm0.01\text{ mm}0.01 mm → True
    • Option D says 0.005 mm0.005\text{ mm}0.005 mm → False

  1. Final answer

    Correct options are: B, C\boxed{\text{B, C}}B, C​

  2. Comparison with stored correct answer

    Stored correct answer: B, C

    My derived answer matches the stored answer.

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