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

Answer: (B)

$\frac{2N - 1}{2}$

Answer 2

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 $\frac{2N - 1}{2}$.