In a vernier caliper, 9 divisions of main scale match with 9+n divisions of vernier scale. What should be the value of n so that count of the instrument remains at minimum possible value?

Options

A. 1
B. 9
C. 2
D. None of these

Explanation

To solve the problem, we need to determine the value of \( n \) such that the least count of the vernier caliper remains at its minimum possible value. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between main scale divisions (MSD) and vernier scale divisions (VSD)
In the given problem, it is stated that 9 divisions of the main scale (MS) match with \( 9 + n \) divisions of the vernier scale (VS).

### Step 2: Define the least count (LC) of the vernier caliper
The least count of a vernier caliper is given by the formula:
\[ \text{Least Count (LC)} = \text{1 MSD} - \text{1 VSD} \]

### Step 3: Express VSD in terms of MSD
From the problem, we can express the value of one VSD in terms of one MSD:
\[ \text{1 VSD} = \frac{9 \text{ MSD}}{9 + n} \]

### Step 4: Substitute VSD into the least count formula
Now, we can substitute this expression into the least count formula:
\[ \text{LC} = \text{1 MSD} - \frac{9 \text{ MSD}}{9 + n} \]

### Step 5: Simplify the least count expression
Factoring out \( \text{MSD} \) from the least count expression gives:
\[ \text{LC} = \text{MSD} \left( 1 - \frac{9}{9 + n} \right) \]
\[ = \text{MSD} \left( \frac{(9 + n) - 9}{9 + n} \right) \]
\[ = \text{MSD} \left( \frac{n}{9 + n} \right) \]

### Step 6: Minimize the least count
To minimize the least count, we need to find the smallest integer value of \( n \). The least count will be minimized when \( n \) is at its smallest possible value.

### Step 7: Test values of \( n \)
1. If \( n = 1 \):
\[ \text{LC} = \text{MSD} \left( \frac{1}{10} \right) \]
2. If \( n = 2 \):
\[ \text{LC} = \text{MSD} \left( \frac{2}{11} \right) \]
3. If \( n = 3 \):
\[ \text{LC} = \text{MSD} \left( \frac{3}{12} \right) \]

As we can see, increasing \( n \) leads to a larger least count. Therefore, the minimum value of \( n \) that keeps the least count at its minimum is \( n = 1 \).

### Conclusion
The least value of \( n \) that keeps the count of the instrument at its minimum possible value is:
\[ \boxed{1} \]