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Units and Measurements question

2024 · 5 Apr · Shift 2 · Q73
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  5. /2024 · 5 Apr · Shift 2 · Q73

Units and Measurements question

2024 · 5 Apr · Shift 2 · Q73

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
A vernier callipers has 20 divisions on the vernier scale, which coincides with 19th 19^{\text {th }}19th  division on the main scale. The least count of the instrument is 0.1 mm0.1 \mathrm{~mm}0.1 mm. One main scale division is equal to ‾\underline{\hspace{2cm}}​ mm.
  1. A
    5
  2. B
    2
  3. C
    1
  4. D
    0.5
View written solutionFree

Correct answer: B

  1. Let one main scale division be MMM mm.

  2. Given: 20 vernier divisions coincide with 19 main scale divisions. Hence, 20 VSD=19 MSD20\,\text{VSD} = 19\,\text{MSD}20VSD=19MSD so 1 VSD=1920M1\,\text{VSD} = \frac{19}{20}M1VSD=2019​M

  3. For a direct vernier callipers, least count is L.C.=1 MSD−1 VSD\text{L.C.} = 1\,\text{MSD} - 1\,\text{VSD}L.C.=1MSD−1VSD

    Therefore, L.C.=M−1920M=120M\text{L.C.} = M - \frac{19}{20}M = \frac{1}{20}ML.C.=M−2019​M=201​M

  4. Given least count = 0.10.10.1 mm, so 120M=0.1\frac{1}{20}M = 0.1201​M=0.1 M=0.1×20=2 mmM = 0.1 \times 20 = 2\,\text{mm}M=0.1×20=2mm

  5. Thus one main scale division is 2 mm2\,\text{mm}2mm

  6. Checking options:

    • A: 555 mm ✗
    • B: 222 mm ✓
    • C: 111 mm ✗
    • D: 0.50.50.5 mm ✗

Hence, the correct option is B.

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