Four identical discs each of mass 'M' and diameter 'a' are arranged in a small plane as shown in figure. If the moment of inertia of the system about OO' is $$\frac{x}{4}Ma^2$$. Then, the value of $$x$$ will be _____.

We need to find the value of $$x$$ given that the total moment of inertia of the system of four identical discs about the axis $$OO'$$ is expressed as $$I = \frac{x}{4} M a^2$$.

Let each disc have a mass $$M$$ and a diameter $$a$$, which means the radius of each disc is $$r = \frac{a}{2}$$.

The moment of inertia of a single disc about an axis passing through its own center and perpendicular to its plane is given by: $$I_{cm} = \frac{1}{2} M r^2 = \frac{1}{2} M \left(\frac{a}{2}\right)^2 = \frac{1}{8} M a^2$$.

According to the theorem of parallel axes, the moment of inertia about a shifted parallel axis is given by: $$I = I_{cm} + M d^2$$, where $$d$$ is the distance between the two axes.

By symmetry, the center of each of the four discs lies at a distance equal to the radius $$r = \frac{a}{2}$$ from the central axis $$OO'$$. Therefore, the distance for all four discs is $$d = \frac{a}{2}$$.

The moment of inertia of one disc about the axis $$OO'$$ is: $$I_1 = \frac{1}{8} M a^2 + M \left(\frac{a}{2}\right)^2 = \frac{1}{8} M a^2 + \frac{1}{4} M a^2 = \frac{3}{8} M a^2$$.

Since the system consists of four identical discs identically arranged around the axis, the total moment of inertia of the system is: $$I_{total} = 4 \times I_1 = 4 \times \left(\frac{3}{8} M a^2\right) = \frac{3}{2} M a^2$$.

We can rewrite this expression to match the given form $$\frac{x}{4} M a^2$$ by multiplying the numerator and denominator by 2: $$I_{total} = \frac{6}{4} M a^2$$.

Comparing this with the given expression $$I = \frac{x}{4} M a^2$$, we find that $$x = 6$$.

Therefore, the correct value of $$x$$ is 6.