Question

If nothing is kept between jaws, zero of Vernier scale lies right of 0 cm of main scale and \(4^{th}\) line of Vernier scale matches perfectly with any line of main scale. An object is kept between jaws and zero of Vernier scale crosses \(15^{th}\) division of main scale and \(5^{th}\) division of Vernier scale exactly matches with any line of main scale. (Least count = 0.1 mm and 1 MSD = 1mm). Find dimension of object :

• A. 15.1 mm
• B. 15.5 mm
• C. 15.4 mm
• D. 15.9 mm

Hint:

Remember: Positive zero error (Vernier zero to the right) is always subtracted from the observed reading. Negative zero error is added.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The actual reading of a Vernier Caliper is given by:
Actual Reading = (Measured Reading) \(-\) (Zero Error).
Step 2: Key Formula or Approach:
1. Zero Error = \(+(\text{Vernier scale matching division}) \times \text{Least Count}\) (if zero of Vernier is to the right).
2. Measured Reading = MSR + (VSR \(\times\) LC).
Step 3: Detailed Explanation:
1. Zero Error Calculation:
The zero of the Vernier scale is to the right (positive error). \(4^{th}\) division matches.
\[ \text{Zero Error} = +(4 \times 0.1 \text{ mm}) = +0.4 \text{ mm} \]
2. Measured Reading Calculation:
Main Scale Reading (MSR) = \(15^{th}\) division = 15 mm.
Vernier Scale matching division = \(5^{th}\).
\[ \text{Measured Reading} = 15 \text{ mm} + (5 \times 0.1 \text{ mm}) = 15.5 \text{ mm} \]
3. Actual Dimension:
\[ \text{Actual Reading} = 15.5 \text{ mm} - 0.4 \text{ mm} = 15.1 \text{ mm} \]
Step 4: Final Answer:
The dimension of the object is 15.1 mm.