A bullet loses $$\left(\frac{1}{n}\right)^{th}$$ of its velocity passing through one plank. Considering uniform retardation, the number of such planks that are required to stop the bullet can be:

A bullet loses $$\left(\frac{1}{n}\right)^{\text{th}}$$ of its velocity when passing through one plank. This means if the initial velocity is $$u$$, after passing through one plank, the velocity becomes $$u - \frac{u}{n} = u \left(1 - \frac{1}{n}\right) = u \left(\frac{n-1}{n}\right)$$. The retardation is uniform, so we can use the equation of motion $$v^2 = u^2 + 2as$$, where $$v$$ is the final velocity, $$u$$ is the initial velocity, $$a$$ is the acceleration (which is negative for retardation), and $$s$$ is the distance traveled (thickness of one plank).

For one plank, initial velocity $$u$$, final velocity $$v = u \frac{n-1}{n}$$, acceleration $$a = -a$$ (where $$a > 0$$ is the magnitude of retardation), and distance $$s$$. Substituting into the equation:

$$\left(u \frac{n-1}{n}\right)^2 = u^2 + 2(-a)s$$

$$u^2 \frac{(n-1)^2}{n^2} = u^2 - 2as$$

Bring all terms to one side:

$$u^2 \frac{(n-1)^2}{n^2} - u^2 = -2as$$

$$u^2 \left( \frac{(n-1)^2}{n^2} - 1 \right) = -2as$$

Simplify the expression inside the parentheses:

$$\frac{(n-1)^2}{n^2} - 1 = \frac{(n-1)^2 - n^2}{n^2} = \frac{n^2 - 2n + 1 - n^2}{n^2} = \frac{-2n + 1}{n^2}$$

So:

$$u^2 \left( \frac{-2n + 1}{n^2} \right) = -2as$$

Multiply both sides by $$-1$$:

$$u^2 \left( \frac{2n - 1}{n^2} \right) = 2as$$

Solve for $$s$$:

$$s = \frac{u^2 (2n - 1)}{2a n^2}$$

Now, to stop the bullet, the final velocity must be zero. Let $$N$$ be the number of planks required. The total distance covered is $$N s$$. Using the equation of motion again, with initial velocity $$u$$, final velocity $$0$$, acceleration $$-a$$, and distance $$N s$$:

$$0^2 = u^2 + 2(-a)(N s)$$

$$0 = u^2 - 2a N s$$

$$2a N s = u^2$$

$$N = \frac{u^2}{2a s}$$

Substitute the expression for $$s$$:

$$N = \frac{u^2}{2a \left( \frac{u^2 (2n - 1)}{2a n^2} \right)} = \frac{u^2 \cdot 2a n^2}{2a u^2 (2n - 1)} = \frac{n^2}{2n - 1}$$

Thus, the number of planks required to stop the bullet is $$\frac{n^2}{2n - 1}$$. Comparing with the options:

A. Infinite
B. $$n$$
C. $$\frac{n^2}{2n-1}$$
D. $$\frac{2n^2}{n-1}$$

The expression $$\frac{n^2}{2n-1}$$ matches option C. Therefore, the correct answer is option C.

Hence, the correct answer is Option C.