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NEET 2026
Exam
NEET
Year
2026
Subject
Physics
Chapter
Chapter

NEET 2026 · Physics · Chapter

In a vernier callipers, 1 cm on the main scale is divided into M...

In a vernier callipers, 1 cm1\text{ cm} on the main scale is divided into MM equal parts. It is observed that PP divisions of the vernier scale coincide exactly with (P−2)(P-2) divisions of the main scale. The least count of the vernier callipers (in mm) is:

Options

  1. A2PM\dfrac{2}{PM}
  2. B20MP\dfrac{20M}{P}
  3. C10PM\dfrac{10}{PM}
  4. D20PM\dfrac{20}{PM}

Correct answer

D. 20PM\dfrac{20}{PM}

Step-by-step solution

Value of 11 Main Scale Division (MSD) = 1M cm=10M mm\dfrac{1}{M}\text{ cm} = \dfrac{10}{M}\text{ mm}

Given that PP Vernier Scale Divisions (VSD) coincide with (P−2)(P-2) Main Scale Divisions (MSD):

P×VSD=(P−2)×MSDP \times \text{VSD} = (P-2) \times \text{MSD}

1 VSD=P−2P MSD1\text{ VSD} = \dfrac{P-2}{P}\text{ MSD}

The Least Count (LC) of the vernier callipers is given by:

LC=1 MSD−1 VSD\text{LC} = 1\text{ MSD} - 1\text{ VSD}

LC=1 MSD−P−2P MSD\text{LC} = 1\text{ MSD} - \dfrac{P-2}{P}\text{ MSD}

LC=2P MSD\text{LC} = \dfrac{2}{P}\text{ MSD}

Substituting the value of 1 MSD1\text{ MSD} in mm:

LC=2P×10M mm=20PM mm\text{LC} = \dfrac{2}{P} \times \dfrac{10}{M}\text{ mm} = \dfrac{20}{PM}\text{ mm}

Answer: 20PM\dfrac{20}{PM}
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