Question

In a vernier callipers, 20 VSD coincide with 16 MSD (each division of length 1 mm). The least count of the vernier callipers is: ____.

• A. 0.1 cm
• B. 0.02 cm
• C. 0.01 cm
• D. 0.2 cm

Hint:

A faster formula for Least Count is $L.C. = \left(1 - \frac{x}{y}\right) \times MSD$, where $x$ is the number of MSDs and $y$ is the number of VSDs. Here, $(1 - 16/20) \times 1 = 4/20 = 0.2$ mm.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The Least Count (L.C.) of a Vernier Calliper is the smallest distance that can be measured accurately. It is defined as the difference between one Main Scale Division (MSD) and one Vernier Scale Division (VSD).

Step 2: Key Formula or Approach:
1. $L.C. = 1 MSD - 1 VSD$ 2. Relationship: $n \cdot VSD = (n-m) \cdot MSD$

Step 3: Detailed Explanation:
Given: $1 MSD = 1$ mm, $20 VSD = 16 MSD$. 1. Calculate the value of 1 VSD: \[ 1 VSD = \frac{16}{20} MSD = 0.8 MSD \] 2. Since $1 MSD = 1$ mm: \[ 1 VSD = 0.8 \text{ mm} \] 3. Calculate Least Count: \[ L.C. = 1 MSD - 1 VSD = 1 \text{ mm} - 0.8 \text{ mm} = 0.2 \text{ mm} \] 4. Convert to cm: \[ L.C. = 0.02 \text{ cm} \]

Step 4: Final Answer:
The least count of the vernier callipers is 0.02 cm.