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NEET PHYSICSUNITS AND MEASUREMENTSMedium

Question

The main scale of a vernier calliper has nn divisions/cm. nn divisions of the vernier scale coincide with (n−1)(n-1) divisions of the main scale. The least count of the vernier calliper is:

A

1(n+1)(n−1) cm\frac{1}{(n+1)(n-1)}\text{ cm}

B

1n cm\frac{1}{n}\text{ cm}

C

1n2 cm\frac{1}{n^2}\text{ cm}

D

1n(n+1) cm\frac{1}{n(n+1)}\text{ cm}

Step-by-Step Solution

  1. Find the value of 1 Main Scale Division (MSD): Since there are nn divisions per cm on the main scale, the length of 1 MSD=1n cm1 \text{ MSD} = \frac{1}{n}\text{ cm}.
  2. Find the relation between MSD and VSD: It is given that nn divisions of the vernier scale (VSD) coincide with (n−1)(n-1) divisions of the main scale (MSD). Therefore, n VSD=(n−1) MSDn \text{ VSD} = (n-1) \text{ MSD}, which implies 1 VSD=n−1n MSD1 \text{ VSD} = \frac{n-1}{n} \text{ MSD}.
  3. Calculate the Least Count (LC): The least count of a vernier calliper is defined as the difference between one main scale division and one vernier scale division. LC=1 MSD−1 VSD\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} LC=1 MSD−n−1n MSD=(1−n−1n) MSD=1n MSD\text{LC} = 1 \text{ MSD} - \frac{n-1}{n} \text{ MSD} = \left(1 - \frac{n-1}{n}\right) \text{ MSD} = \frac{1}{n} \text{ MSD}.
  4. Substitute the value of 1 MSD: LC=1n×(1n cm)=1n2 cm\text{LC} = \frac{1}{n} \times \left(\frac{1}{n}\text{ cm}\right) = \frac{1}{n^2}\text{ cm}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from UNITS AND MEASUREMENTS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSUNITS AND MEASUREMENTSverniercalliperdivisionscmdivisionsvernier

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