Question

A vernier callipers has 20 divisions on the vernier scale, which coincides with 19^th division on the main scale. The least count of the instrument is 0.1 mm. One main scale division is equal to ____mm.

• A. 1
• B. 0.5
• C. 2
• D. 5

Answer 1

The Correct Option is C

Given:
- 20 vernier scale divisions (VSD) coincide with 19 main scale divisions (MSD).
- Least count (L.C.) of the instrument is \( 0.1 \, \text{mm} \).

Step 1: Relation Between VSD and MSD
From the given data:

\[ 20 \, \text{VSD} = 19 \, \text{MSD}. \]

The value of one vernier scale division (1 VSD) is:

\[ 1 \, \text{VSD} = \frac{19}{20} \, \text{MSD}. \]

Step 2: Calculating the Least Count
The least count (L.C.) is given by:

\[ \text{L.C.} = 1 \, \text{MSD} - 1 \, \text{VSD}. \]

Substituting the value of 1 VSD:

\[ \text{L.C.} = 1 \, \text{MSD} - \frac{19}{20} \, \text{MSD} = \frac{1}{20} \, \text{MSD}. \]

Given that the least count is \( 0.1 \, \text{mm} \):

\[ 0.1 \, \text{mm} = \frac{1}{20} \, \text{MSD}. \]

Multiplying both sides by 20:

\[ 1 \, \text{MSD} = 2 \, \text{mm}. \]

Therefore, one main scale division is equal to \( 2 \, \text{mm} \).

Answer 2

Step 1: Given data.
Number of vernier scale divisions (n) = 20
These 20 vernier divisions coincide with 19 main scale divisions.
Least count (LC) = 0.1 mm

Step 2: Formula for least count.
The least count of a vernier callipers is given by:
\[ LC = 1 \, \text{main scale division (MSD)} - 1 \, \text{vernier scale division (VSD)} \]

Step 3: Express 1 vernier scale division in terms of main scale divisions.
Since 20 VSD = 19 MSD,
\[ 1 \, \text{VSD} = \frac{19}{20} \, \text{MSD} \]

Step 4: Substitute in the least count formula.
\[ LC = 1 \, \text{MSD} - \frac{19}{20} \, \text{MSD} = \frac{1}{20} \, \text{MSD} \]
Given LC = 0.1 mm,
\[ 0.1 = \frac{1}{20} \times \text{MSD} \]
\[ \text{MSD} = 0.1 \times 20 = 2 \, \text{mm} \]

Step 5: Final Answer.
One main scale division (MSD) = 2 mm

Final Answer: 2