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

To calculate the dimension of the object using a Vernier caliper, we need to consider both the zero error correction and the measurement with the object in place. Let's break down the given data and find the dimension of the object step by step.

1. Determine Zero Error:
   • The zero of the Vernier scale is to the right of the 0 cm mark on the main scale.
   • The 4th line of the Vernier scale perfectly aligns with a line on the main scale.
   • Zero error is calculated as (\(n \times \text{LC}\)), where \(n\) is the Vernier scale reading and \(\text{LC}\) is the least count.
   • Given, \(\text{Least Count} = 0.1 \, \text{mm}\), and the 4th division matches, thus, zero error = \(4 \times 0.1 = 0.4 \, \text{mm}\). This is a positive zero error because the zero of the Vernier is to the right of the main scale zero.
2. Measurement with Object:
   • The zero of the Vernier scale crosses the 15th division of the main scale. This main scale reading (MSR) is \(15 \, \text{mm}\).
   • The 5th line of the Vernier scale matches perfectly with a line on the main scale, hence, the Vernier scale reading (VSR) is \(5\).
   • Correct for the zero error: Actual VSR = Measured VSR − Zero error VSR = \(5 - 4 = 1\).
3. Calculate the Actual Measurement:
   • Add the corrected Vernier scale reading to the main scale reading, multiplied by the least count: Actual dimension = MSR + (Corrected VSR × LC).
   • Actual dimension = \(15 + (1 \times 0.1) = 15.1 \, \text{mm}\).

Therefore, the dimension of the object is 15.1 mm.