To solve the problem, we need to find the minimum common frequency with which all three parts of a sonometer wire can vibrate. The wire is divided into three segments in the ratio of 6:3:2, and we have the total length, tension, and mass per unit length provided.
### Step-by-Step Solution:
1. **Determine the Total Length of the Wire:**
The total length of the sonometer wire is given as:
\[
L = 110 \text{ cm} = 1.1 \text{ m}
\]
2. **Divide the Length According to the Given Ratio:**
The wire is divided in the ratio of 6:3:2. Let's denote the lengths of the three segments as \(L_1\), \(L_2\), and \(L_3\).
\[
L_1 + L_2 + L_3 = L
\]
Let \(L_1 = 6x\), \(L_2 = 3x\), and \(L_3 = 2x\). Then,
\[
6x + 3x + 2x = 1.1 \implies 11x = 1.1 \implies x = 0.1 \text{ m}
\]
Therefore, the lengths of the segments are:
\[
L_1 = 0.6 \text{ m}, \quad L_2 = 0.3 \text{ m}, \quad L_3 = 0.2 \text{ m}
\]
3. **Use the Formula for Frequency:**
The frequency of vibration for a segment of wire is given by:
\[
f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}
\]
where \(T\) is the tension and \(\mu\) is the mass per unit length.
4. **Calculate the Frequencies for Each Segment:**
Given:
- Tension \(T = 400 \text{ N}\)
- Mass per unit length \(\mu = 0.01 \text{ kg/m}\)
Now, calculate the frequencies:
- For \(L_1 = 0.6 \text{ m}\):
\[
f_1 = \frac{1}{2 \times 0.6} \sqrt{\frac{400}{0.01}} = \frac{1}{1.2} \sqrt{40000} = \frac{1}{1.2} \times 200 = \frac{200}{1.2} \approx 166.67 \text{ Hz}
\]
- For \(L_2 = 0.3 \text{ m}\):
\[
f_2 = \frac{1}{2 \times 0.3} \sqrt{\frac{400}{0.01}} = \frac{1}{0.6} \sqrt{40000} = \frac{1}{0.6} \times 200 = \frac{200}{0.6} \approx 333.33 \text{ Hz}
\]
- For \(L_3 = 0.2 \text{ m}\):
\[
f_3 = \frac{1}{2 \times 0.2} \sqrt{\frac{400}{0.01}} = \frac{1}{0.4} \sqrt{40000} = \frac{1}{0.4} \times 200 = \frac{200}{0.4} = 500 \text{ Hz}
\]
5. **Find the Minimum Common Frequency:**
The frequencies we calculated are:
\[
f_1 \approx 166.67 \text{ Hz}, \quad f_2 \approx 333.33 \text{ Hz}, \quad f_3 = 500 \text{ Hz}
\]
To find the minimum common frequency, we need to find the least common multiple (LCM) of the frequencies expressed in a common form.
The frequencies can be expressed as:
\[
f_1 = \frac{1000}{6} \text{ Hz}, \quad f_2 = \frac{1000}{3} \text{ Hz}, \quad f_3 = \frac{1000}{2} \text{ Hz}
\]
The LCM of the denominators (6, 3, 2) is 6. Thus, the minimum common frequency is:
\[
f = \frac{1000}{6} \times 6 = 1000 \text{ Hz}
\]
### Final Answer:
The minimum common frequency with which all three parts can vibrate is:
\[
\boxed{1000 \text{ Hz}}
\]
