Moment of Inertia (M.I.) of four bodies having same mass $$M$$ and radius $$2R$$ are as follows
$$I_1$$ = M.I. of solid sphere about its diameter
$$I_2$$ = M.I. of solid cylinder about its axis
$$I_3$$ = M.I. of solid circular disc about its diameter
$$I_4$$ = M.I. of thin circular ring about its diameter
If $$2(I_2 + I_3) + I_4 = xI_1$$ then the value of $$x$$ will be ______.

All four bodies have mass $$M$$ and radius $$2R$$. The standard moment of inertia formulas (with radius $$r$$) are:

Solid sphere about diameter: $$\frac{2}{5}Mr^2$$. So $$I_1 = \frac{2}{5}M(2R)^2 = \frac{8}{5}MR^2$$.

Solid cylinder about its axis: $$\frac{1}{2}Mr^2$$. So $$I_2 = \frac{1}{2}M(2R)^2 = 2MR^2$$.

Solid circular disc about its diameter: $$\frac{1}{4}Mr^2$$. So $$I_3 = \frac{1}{4}M(2R)^2 = MR^2$$.

Thin circular ring about its diameter: $$\frac{1}{2}Mr^2$$. So $$I_4 = \frac{1}{2}M(2R)^2 = 2MR^2$$.

Now we compute $$2(I_2 + I_3) + I_4 = 2(2MR^2 + MR^2) + 2MR^2 = 2(3MR^2) + 2MR^2 = 6MR^2 + 2MR^2 = 8MR^2$$.

Setting this equal to $$xI_1$$: $$8MR^2 = x \cdot \frac{8}{5}MR^2$$.

Dividing both sides by $$\frac{8}{5}MR^2$$: $$x = \frac{8}{\frac{8}{5}} = 5$$.

The value of $$x$$ is $$\mathbf{5}$$.