The total length of a sonometer wire fixed between two bridges is 110 cm. Now, two more bridges are placed to divide the length of the wire in the ratio 6 : 3 : 2. If the tension in the wire is 400 N and the mass per unit length of the wire is 0.01 kg m$$^{-1}$$, then the minimum common frequency with which all the three parts can vibrate, is:

The total length of the sonometer wire is 110 cm, which is 110 divided by 100, giving 1.10 meters. The wire is divided into three parts in the ratio 6:3:2. Let the common multiplier be $$ k $$. So, the lengths are $$ L_1 = 6k $$, $$ L_2 = 3k $$, and $$ L_3 = 2k $$. The sum of these lengths is $$ 6k + 3k + 2k = 11k $$, which equals 1.10 meters. Therefore, $$ 11k = 1.10 $$, and solving for $$ k $$, we get $$ k = \frac{1.10}{11} = 0.10 $$ meters.

Now, calculate each length:

First part: $$ L_1 = 6 \times 0.10 = 0.60 $$ meters,

Second part: $$ L_2 = 3 \times 0.10 = 0.30 $$ meters,

Third part: $$ L_3 = 2 \times 0.10 = 0.20 $$ meters.

The tension $$ T $$ is 400 N, and the mass per unit length $$ \mu $$ is 0.01 kg/m. The fundamental frequency for a stretched string is given by $$ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $$. Since $$ T $$ and $$ \mu $$ are the same for all parts, compute the constant $$ \sqrt{\frac{T}{\mu}} $$:

$$ \sqrt{\frac{T}{\mu}} = \sqrt{\frac{400}{0.01}} = \sqrt{40000} = 200 \text{ m/s}. $$

So, the frequency formula simplifies to $$ f = \frac{200}{2L} = \frac{100}{L} $$ Hz.

Now, find the fundamental frequency for each segment:

For the first segment ($$ L_1 = 0.60 $$ m): $$ f_1 = \frac{100}{0.60} = \frac{1000}{6} = \frac{500}{3} $$ Hz,

For the second segment ($$ L_2 = 0.30 $$ m): $$ f_2 = \frac{100}{0.30} = \frac{1000}{3} $$ Hz,

For the third segment ($$ L_3 = 0.20 $$ m): $$ f_3 = \frac{100}{0.20} = 500 $$ Hz.

Each segment can vibrate at its fundamental frequency or at harmonics, which are integer multiples of the fundamental frequency. To find the minimum common frequency for all three segments, we need the smallest frequency that is a common multiple of $$ f_1 $$, $$ f_2 $$, and $$ f_3 $$. This means there exist integers $$ n_1 $$, $$ n_2 $$, and $$ n_3 $$ such that:

$$ f = n_1 \cdot f_1 = n_1 \cdot \frac{500}{3}, $$

$$ f = n_2 \cdot f_2 = n_2 \cdot \frac{1000}{3}, $$

$$ f = n_3 \cdot f_3 = n_3 \cdot 500. $$

Set the expressions equal:

$$ n_1 \cdot \frac{500}{3} = n_2 \cdot \frac{1000}{3} = n_3 \cdot 500. $$

Multiply through by 3 to eliminate denominators:

$$ 500 n_1 = 1000 n_2 = 1500 n_3. $$

From $$ 500 n_1 = 1000 n_2 $$, divide both sides by 500: $$ n_1 = 2 n_2 $$.

From $$ 1000 n_2 = 1500 n_3 $$, divide both sides by 500: $$ 2 n_2 = 3 n_3 $$.

So, $$ n_1 = 2 n_2 $$ and $$ 2 n_2 = 3 n_3 $$. Solving for $$ n_2 $$ in terms of $$ n_3 $$: $$ n_2 = \frac{3}{2} n_3 $$. Since $$ n_2 $$ must be an integer, $$ n_3 $$ must be even. Let $$ n_3 = 2k $$ for some integer $$ k \geq 1 $$. Then:

$$ n_2 = \frac{3}{2} \times 2k = 3k, $$

$$ n_1 = 2 \times n_2 = 2 \times 3k = 6k. $$

The smallest positive integers occur when $$ k = 1 $$:

$$ n_3 = 2 \times 1 = 2, $$

$$ n_2 = 3 \times 1 = 3, $$

$$ n_1 = 6 \times 1 = 6. $$

Now, compute the common frequency $$ f $$:

Using $$ f = n_3 \cdot f_3 = 2 \times 500 = 1000 $$ Hz,

or $$ f = n_2 \cdot f_2 = 3 \times \frac{1000}{3} = 1000 $$ Hz,

or $$ f = n_1 \cdot f_1 = 6 \times \frac{500}{3} = 1000 $$ Hz.

Thus, the minimum common frequency is 1000 Hz.

Hence, the correct answer is Option A.