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Units and Measurements question

2025 · 4 Apr · Shift 2 · Q61
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Units and Measurements question

2025 · 4 Apr · Shift 2 · Q61

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm . The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm ) :
  1. A
    0.002
  2. B
    0.0025
  3. C
    0.0005
  4. D
    0.001
View written solutionFree

Correct answer: A

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

The main scale has 300 divisions equal to 15 cm15\text{ cm}15 cm.

So, 1 MSD=15300=0.05 cm1\,\text{MSD} = \frac{15}{300} = 0.05\text{ cm}1MSD=30015​=0.05 cm

  1. Relate vernier scale divisions to main scale divisions

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

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

  1. Calculate least count

For a direct vernier, LC=1 MSD−1 VSD\text{LC} = 1\,\text{MSD} - 1\,\text{VSD}LC=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=0.002 cm

  1. Match with options

Thus, the least count is 0.002 cm\boxed{0.002\text{ cm}}0.002 cm​

So the correct option is A.

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