In a measuring instrument with a vernier scale the number of small divisions on the main scale that correspond to n2 divisions on the vernier scale is n1.here the length of one division on the main scale, like the least count is?

Question

1.n1-n2 2.n2-n1 3.n2/n2-n1 4.n1-n2/n2 5.n1/n2-n1

Answer 1

In a measuring instrument with a vernier scale, the least count can be calculated using the formula:

\[ \text{Least Count} = \text{Value of one main scale division (MSD)} - \text{Value of one vernier scale division (VSD)} \]

Given that the number of small divisions on the main scale corresponding to \(n_2\) divisions on the vernier scale is \(n_1\), we can derive the following:

1. Let \(L\) be the length of one main scale division (MSD).
2. The length of \(n_2\) divisions on the vernier scale (VSD) is \(n_2 \times \text{Length of one VSD}\).

Since \(n_1\) divisions of the main scale equal \(n_2\) divisions of the vernier scale, we can write:

\[ n_1 \times L = n_2 \times \text{VSD} \]

From this, the value of one vernier scale division can be determined as follows:

\[ \text{VSD} = \frac{n_1 \times L}{n_2} \]

Now substituting this into the least count formula:

\[ \text{Least Count} = L - \frac{n_1 \times L}{n_2} \]

Factoring out \(L\):

\[ \text{Least Count} = L \left( 1 - \frac{n_1}{n_2} \right) = L \left( \frac{n_2 - n_1}{n_2} \right) \]

To express the least count in terms of the number of divisions, we can set:

\[ \text{Length of one division on the main scale } (L) = \frac{n_1}{n_2 - n_1} \]

Thus the least count can be expressed as:

\[ \text{Least Count} = \frac{L \times (n_2 - n_1)}{n_2} \]

That simplifies to the format of values you have provided, and in terms of your options:

The least count is:

4. \(\frac{n_1 - n_2}{n_2}\)

(assuming we mean the certain configuration of divisions and format).