In vernier callipers, m divisions of main scale with `(m + 1)` divisions of vernier scale. If each division of main scale is d units, the least count of the instrument is

Options

A. `d//(m-1)`
B. `d//(m+1)`
C. `d//m`
D. `md//(m+1)`

Answer

To find the least count of the vernier calipers given the number of divisions on the main scale and the vernier scale, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Number of divisions on the main scale (MS) = \( m \)
- Number of divisions on the vernier scale (VS) = \( m + 1 \)
- Size of one main scale division (MSD) = \( d \) units

2. **Determine the Size of One Vernier Scale Division:**
- Let the size of one vernier scale division be \( x \).
- The total length represented by the main scale is \( m \cdot d \).
- The total length represented by the vernier scale is \( (m + 1) \cdot x \).
- Since both scales represent the same length, we can set up the equation:
\[ m \cdot d = (m + 1) \cdot x \]
- Rearranging gives:
\[ x = \frac{m \cdot d}{m + 1} \]

3. **Calculate the Least Count:**
- The least count (LC) of the instrument is defined as:
\[ \text{Least Count} = \text{Value of 1 MSD} - \text{Value of 1 VSD} \]
- Substituting the values we have:
\[ \text{LC} = d - x \]
- Substitute \( x \) from the previous step:
\[ \text{LC} = d - \frac{m \cdot d}{m + 1} \]
- To simplify, we can express \( d \) as:
\[ \text{LC} = \frac{d(m + 1)}{m + 1} - \frac{m \cdot d}{m + 1} \]
- This gives:
\[ \text{LC} = \frac{d(m + 1 - m)}{m + 1} = \frac{d}{m + 1} \]

4. **Final Result:**
- Therefore, the least count of the vernier calipers is:
\[ \text{Least Count} = \frac{d}{m + 1} \]

### Conclusion:
The least count of the instrument is \( \frac{d}{m + 1} \).