Question

In an unusual vernier, 9 vernier scale divisions coincide with 8 main scale division, then what is the least count of the vernier? (A) \((8 / 9) \mathrm{mm}\) (B) \((1 / 9) \mathrm{mm}\) (C) \((1 / 17) \mathrm{mm}\) (D) \((1 / 8) \mathrm{mm}\)

Short answer

The least count of the unusual vernier scale is \(\frac{1}{9}\) mm, which corresponds to option (B).

Step-by-step solution

Determine the values of the main scale and vernier scale divisions

We know that 9 vernier scale divisions coincide with 8 main scale divisions. Let's assume the length of one main scale division is M, and one vernier scale division is V. Since 8 main scale divisions are equivalent to 9 vernier scale divisions, we can write the equation: \(8M = 9V\)

Find the least count (LC) of the vernier scale

The least count of a vernier scale is defined as the difference between the values of one main scale division (M) and one vernier scale division (V). Least Count (LC) = M - V We will now use the equation from step 1 to express V in terms of M: \(V = \frac{8}{9}M\)

Substitute the value of V in the least count formula

Now that we have found the value of V in terms of M, we can substitute it back into the least count formula: LC = M - \(\frac{8}{9}M\)

Calculate the least count

To calculate the least count, we need to solve for LC: LC = \(\frac{9}{9}M - \frac{8}{9}M\) LC = \(\frac{1}{9}M\) Since the main scale is given in millimeters (mm), the least count will also be in millimeters. LC = \(\frac{1}{9}\) mm The least count of the vernier scale is \(\frac{1}{9}\) mm, which corresponds to option (B).