Question

100 Vernier scale divisions match with 99 main scale divisions of a slide caliper. If the value of each main scale division is 1 mm, then the Vernier constant is:

• A. 1 mm
• B. 100 µm
• C. 10 µm
• D. 1 µm

Hint:

The Vernier constant is calculated by dividing the length of one main scale division by the number of Vernier scale divisions.

Answer 1

The Correct Option is C

To determine the Vernier constant, we need to find the difference between one main scale division and one Vernier scale division.

Given:

• 1 main scale division (MSD) = 1 mm
• 100 Vernier scale divisions (VSD) = 99 main scale divisions

We need to find the length of 1 Vernier scale division (1 VSD):

\(1 \text{ VSD} = \frac{\text{value of 99 MSDs}}{100} = \frac{99 \text{ mm}}{100} = 0.99 \text{ mm}\)

Therefore, the difference between one main scale division and one Vernier scale division (the Vernier constant) is:

\[ \text{Vernier Constant} = 1 \text{ MSD} - 1 \text{ VSD} \]

\[ = 1 \text{ mm} - 0.99 \text{ mm} = 0.01 \text{ mm} \]

Since 0.01 mm is equivalent to 10 micrometers (µm), the Vernier constant is:

10 µm

Answer 2

The Vernier constant is the difference between one main scale division and one Vernier scale division: \[ \text{Vernier constant} = \frac{\text{Length of 1 main scale division}}{\text{Number of divisions on Vernier scale}} = \frac{1 \, \text{mm}}{100} = 0.01 \, \text{mm} = 10 \, \mu\text{m} \] Thus, the Vernier constant is 10 µm.

Answer 3

This question is about the Vernier constant (also called the least count), the smallest length a Vernier caliper can resolve. Checking each option against what 100 divisions matching 99 actually means settles it quickly.

1. 1 mm: This is the size of a single main scale division itself, not the extra precision the Vernier scale adds on top of it. If the least count equalled a whole main scale division, the Vernier scale would add no extra precision at all, which defeats the purpose of having one.
2. 100 µm: This equals 0.1 mm, a tenth of a main scale division. That would be the least count if 10 Vernier divisions matched 9 main scale divisions, a coarser caliper than the one described here, which packs 100 divisions into the same span.
3. 10 µm: Squeezing 100 Vernier divisions into the space of 99 main scale divisions means each Vernier division is very slightly shorter than a main scale division, by exactly a hundredth of a main scale division. Since 1 main scale division is 1 mm, that tiny leftover sliver is 1 mm divided by 100, which is 0.01 mm, or 10 micrometers.
4. 1 µm: This would need 1000 Vernier divisions matching 999 main scale divisions to get a leftover a thousand times smaller, far finer than the 100-to-99 caliper described in the question.

The Vernier constant comes from how much shorter each Vernier division is compared to a main scale division, and packing 100 of them into 99 main scale divisions makes that shortfall exactly a hundredth of a main scale division, or 10 micrometers.

So the correct answer is 10 µm.