Mathematics in Physics

JEE Main level, numerical answer, medium

76 of 121

Question 76 of 121

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 5.25mm5.25mm and apparent thickness of the glass slab at 5.00mm5.00mm. Travelling microscope has 2020 divisions in one cmcm on main scale and 5050 divisions on Vernier scale is equal to 4949 divisions on main scale. The estimated uncertainty in the measurement of refractive index of the slab is x10×10−3\dfrac{x}{10}\times {10}^{-3}, where xx is ______

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Answer: 41

Given:
50VSD=49MSD⇒1VSD=4950MSD50VSD=49MSD \Rightarrow 1VSD=\dfrac{49}{50}MSD
So least count will be
=1MSD−1VSD=120cm−4950⋅120cm=1MSD-1VSD =\dfrac{1}{20}cm-\dfrac{49}{50}\cdot \dfrac{1}{20}cm
=150×120cm=0.01mm=\dfrac{1}{50}\times \dfrac{1}{20}cm =0.01mm
As we know, μ=hh′=5.255.00\mu =\dfrac{h}{h'}=\dfrac{5.25}{5.00}
Applying log on both sides, we get
lnμ=lnh–lnh’ln\mu =lnh–lnh’
Differentiating both sides, we get
dμμ=dhh−dh′h′\dfrac{d\mu }{\mu }=\dfrac{dh}{h}-\dfrac{dh'}{h'}
For calculation, all errors will be added, therefore
dμ=[0.015.25+0.015.00]μd\mu =\left[\dfrac{0.01}{5.25}+\dfrac{0.01}{5.00}\right]\mu
=4110×10−3=\dfrac{41}{10}\times {10}^{-3}
Hence, x=41x=41

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