Question

Least count of a vernier caliper is \( \frac{1}{20N} \, \text{cm} \).The value of one division on the main scale is \(1 \, \text{mm}\). Then the number of divisions of the main scale that coincide with \(N\) divisions of the vernier scale is:

• A. \( \frac{2N - 1}{20N} \)
• B. \( \frac{2N - 1}{2} \)
• C. \( 2N - 1 \)
• D. \( \frac{2N - 1}{2N} \)

Answer 1

The Correct Option is B

The least count (LC) of a vernier caliper is given by: \[ \text{LC} = 1 \text{ main scale division} - 1 \text{ vernier scale division} \]

Given that the least count is \(\frac{1}{20N}\) cm and one main scale division is 1 mm = 0.1 cm, we can write:

\[\frac{1}{20N} \text{ cm} = 0.1 \text{ cm} - \frac{N}{x} \text{ cm}\]

where 'x' represents the total number of vernier scale divisions.

We can assume that 'x' vernier scale divisions are equal to N main scale divisions. Thus, \(\frac{N}{x} \text{ cm}\) represents the length of N vernier divisions. Solving for \(\frac{N}{x}\), we get:

\[\frac{N}{x} = 0.1 - \frac{1}{20N} = \frac{2N - 1}{20N} \text{ cm}\]

Now, we know that N vernier scale divisions coincide with (n) main scale divisions. Then, the length of N vernier divisions equals the length of n main scale divisions:

\[\frac{N}{x} \text{ cm} = n \times 0.1 \text{ cm}\]

Therefore:

\[n = \frac{N}{x} \times \frac{1}{0.1} = \frac{2N - 1}{20N} \times 10 = \frac{2N - 1}{2}\]

Thus, the number of main scale divisions that coincide with N vernier scale divisions is \(\frac{2N - 1}{2}\).