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Units and Measurements question

2022 · 29 Jul · Shift 1 · Q65
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Units and Measurements question

2022 · 29 Jul · Shift 1 · Q65

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
A travelling microscope has 20 divisions per cm\mathrm{cm}cm on the main scale while its vernier scale has total 50 divisions and 25 vernier scale divisions are equal to 24 main scale divisions, what is the least count of the travelling microscope?
  1. A
    0.001 cm
  2. B
    0.002 mm
  3. C
    0.002 cm
  4. D
    0.005 cm
View written solutionFree

Correct answer: C

  1. Find the value of one main scale division (MSD)

The main scale has 202020 divisions per cm.

So, 1 MSD=1 cm20=0.05 cm1\,\text{MSD} = \frac{1\,\text{cm}}{20} = 0.05\,\text{cm}1MSD=201cm​=0.05cm

  1. Use the relation between vernier scale divisions and main scale divisions

Given: 25 VSD=24 MSD25\,\text{VSD} = 24\,\text{MSD}25VSD=24MSD

Hence, 1 VSD=2425 MSD1\,\text{VSD} = \frac{24}{25}\,\text{MSD}1VSD=2524​MSD

Substitute 1 MSD=0.05 cm1\,\text{MSD} = 0.05\,\text{cm}1MSD=0.05cm: 1 VSD=2425×0.05=0.048 cm1\,\text{VSD} = \frac{24}{25} \times 0.05 = 0.048\,\text{cm}1VSD=2524​×0.05=0.048cm

  1. Find the least count

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

So, LC=0.05−0.048=0.002 cm\text{LC} = 0.05 - 0.048 = 0.002\,\text{cm}LC=0.05−0.048=0.002cm

  1. Match with the options

0.002 cm0.002\,\text{cm}0.002cm corresponds to Option C.

  1. Verification with stored answer

Stored correct answer is C, which matches our derived answer.

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