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 Topic:
This problem belongs to "Units and Measurements," specifically focusing on error analysis and measuring instruments. The "Least Count" is the smallest value that can be accurately measured by a measuring instrument. For Vernier callipers, this is determined by the slight mismatch between the main scale and the sliding vernier scale.
Step 2: Key Formulas and Approach:
The Least Count (L.C.) is the difference between the magnitude of one main scale division (MSD) and one vernier scale division (VSD):
$L.C. = 1 \text{ MSD} - 1 \text{ VSD}$.
If $n$ divisions of the Vernier scale coincide with $m$ divisions of the main scale, then $1 \text{ VSD} = (m/n) \text{ MSD}$.
$L.C. = \left( 1 - \frac{m}{n} \right) \text{ MSD}$.

Step 3: Detailed Explanation:

Extract given data: $1 \text{ MSD} = 1 \text{ mm}$. The number of Vernier scale divisions $n = 20$. The number of main scale divisions $m = 16$.
Calculate the value of 1 VSD: Since 20 VSD equals 16 MSD: \[ 1 \text{ VSD} = \frac{16}{20} \text{ MSD} = 0.8 \text{ MSD} \]
Since $1 \text{ MSD} = 1 \text{ mm}$, then $1 \text{ VSD} = 0.8 \text{ mm}$.
Find the Least Count in mm: \[ L.C. = 1 \text{ MSD} - 1 \text{ VSD} = 1 \text{ mm} - 0.8 \text{ mm} = 0.2 \text{ mm} \]
Convert to the required unit (cm): Most options are in cm, so we convert mm to cm by dividing by 10. \[ L.C. = \frac{0.2}{10} \text{ cm} = 0.02 \text{ cm} \]
Step 4: Final Answer:
The least count of the vernier callipers is 0.02 cm.