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JEE Main 2024 (6 Apr, Shift 2) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2024 (6 Apr, Shift 2)

Q25JEE Main 2024 · 6 APR · SHIFT 2Medium

Least Count, Vernier Callipers & Screw Gauge

In finding out refractive index of glass slab the following observations were made through travelling microscope 50 vernier scale division =49 MSD; 20= 49\text{ MSD};\ 20 divisions on main scale in each cm\text{cm} For mark on paper

MSR =8.45 cm, VC=26= 8.45\text{ cm},\ \text{VC} = 26

For mark on paper seen through slab

MSR=7.12 cm, VC=41\text{MSR} = 7.12\text{ cm},\ \text{VC} = 41

For powder particle on the top surface of the glass slab

MSR =4.05 cm, VC=1= 4.05\text{ cm},\ \text{VC} = 1

(MSR == Main Scale Reading, VC = Vernier Coincidence)

Refractive index of the glass slab is:

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Answer: (d)

  1. 1 MSD =120 cm=0.05 cm= \dfrac{1}{20}\text{ cm} = 0.05\text{ cm}; LC =0.0550=0.001 cm= \dfrac{0.05}{50} = 0.001\text{ cm}
  2. Readings: R1=8.45+0.026=8.476R_1 = 8.45 + 0.026 = 8.476, R2=7.12+0.041=7.161R_2 = 7.12 + 0.041 = 7.161, R3=4.05+0.001=4.051 cmR_3 = 4.05 + 0.001 = 4.051\text{ cm}
  3. Real depth =R1−R3=4.425 cm= R_1 - R_3 = 4.425\text{ cm}; apparent depth =R2−R3=3.110 cm= R_2 - R_3 = 3.110\text{ cm}
  4. μ=4.4253.110≈1.42\mu = \dfrac{4.425}{3.110} \approx 1.42. So the answer is (d).
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