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Units and Measurements question

2022 · 26 Jun · Shift 2 · Q71
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Units and Measurements question

2022 · 26 Jun · Shift 2 · Q71

JEE MainPhysicsUnits and MeasurementsNumerical+4 / −1
A travelling microscope is used to determine the refractive index of a glass slab. If 40 divisions are there in 1 cm on main scale and 50 Vernier scale divisions are equal to 49 main scale divisions, then least count of the travelling microscope is ‾×\underline{\hspace{2cm}}\times​× 10 −-− 6 m.
Numerical answer
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Correct answer: 5

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

Given: 40 divisions are there in 1 cm1\,\text{cm}1cm on the main scale.

So, 1 MSD=1 cm40=10−2 m40=2.5×10−4 m1\,\text{MSD} = \frac{1\,\text{cm}}{40} = \frac{10^{-2}\,\text{m}}{40} = 2.5\times 10^{-4}\,\text{m}1MSD=401cm​=4010−2m​=2.5×10−4m

  1. Relate Vernier scale divisions to main scale divisions

Given: 50 VSD=49 MSD50\,\text{VSD} = 49\,\text{MSD}50VSD=49MSD

Therefore, 1 VSD=4950 MSD1\,\text{VSD} = \frac{49}{50}\,\text{MSD}1VSD=5049​MSD

  1. Compute 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

Substitute: LC=1 MSD−4950 MSD=150 MSD\text{LC} = 1\,\text{MSD} - \frac{49}{50}\,\text{MSD} = \frac{1}{50}\,\text{MSD}LC=1MSD−5049​MSD=501​MSD

Now, LC=150×2.5×10−4\text{LC} = \frac{1}{50}\times 2.5\times 10^{-4}LC=501​×2.5×10−4 =5×10−6 m= 5\times 10^{-6}\,\text{m}=5×10−6m

  1. Match with the required form

The least count is written as: ‾×10−6 m\underline{\hspace{2cm}}\times 10^{-6}\,\text{m}​×10−6m

Hence the required integer is: 555

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