Question

N divisions on the main scale of a vernier calipers coin sides with \((\mathrm{N}+1)\) divisions on the vernier scale. If each division on the main scale is of a units, the least of count of instrument is................ (A) \(\\{\mathrm{a} /(\mathrm{N}+1)\\}\) (B) \(\\{\mathrm{a} /(\mathrm{N}-1)\\}\) (C) \(\\{(N+1) / a\\}\) (D) \(\\{(N-1) / a\\}\)

Short answer

The least count of the instrument is (A) \(\\{\mathrm{a} /(\mathrm{N}+1)\\}\).

Step-by-step solution

Finding the value of one division on the vernier scale

The least count can be calculated as the difference between one main scale reading and one vernier scale reading. From the given information, we know that N divisions on the main scale coincide with (N+1) divisions on the vernier scale. Let's denote the value of one division on the vernier scale as "b" units. Therefore: \(Nb = (N+1)b\)

Solve for the value of b

Now, in order to find the value of one vernier scale reading, we will solve for b in the above equation. Divide both sides of the equation by N(N+1): \(\frac{Nb}{N(N+1)} = \frac{(N+1)b}{N(N+1)}\) This simplifies to: \(\frac{b}{N+1} = a\) Now, solve for b: \(b = a(N+1)\)

Calculate the least count

The least count of the instrument is the difference between one main scale reading (a) and one vernier scale reading (b). Least count = \(|a - b|\) Substitute the value of b from step 2: Least count = \(|a - a(N+1)|\) Least count = \(|a(1 - (N+1))|\) Least count = \(|a(N+1 - N - 1)|\) Least count = \(a / (\mathrm{N} + 1)\) Comparing our result with the given options, we find that the correct option is: (A) \(\\{\mathrm{a} /(\mathrm{N}+1)\\}\)