JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Units & Measurement

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In an experiment of measuring the refractive index of a glass slab using travelling microscope in physics lab, a student measures real thickness of the glass slab as 5.25 mm and apparent thickness of the glass slab at 5.00 mm. Travelling microscope has 20 divisions in one cm on main scale and 50 divisions on Vernier scale is equal to 49 divisions on main scale. The estimated uncertainty in the measurement of refractive index of the slab is

\frac{x}{10} \times 1 0^{- 3}

, where x is ______

Answer: 41

Solution

The least count of the travelling microscope is

1 MSD - 1 VSD

. One main-scale division is

\frac{1}{20}

cm, and

50

vernier divisions equal

49

main divisions, so one vernier division is

\frac{49}{50}

MSD, giving

L C = (1 - \frac{49}{50}) \frac{1}{20} = \frac{1}{1000}

= 0.01

mm. The refractive index is

μ = \frac{h}{h ^{'}}

, whose fractional uncertainty is

\frac{d μ}{μ} = \frac{d h}{h} + \frac{d h ^{'}}{h ^{'}}

with

d h = d h^{'} = 0.01

mm. Thus

d μ = \frac{5.25}{5.00} (\frac{0.01}{5.25} + \frac{0.01}{5.00}) = \frac{41}{10} \times 1 0^{- 3}

, so

x = 41

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Integer answer

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JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Units & Measurement

Solution

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