The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is $$n$$ times higher than the moment of inertia of the given ring. Here, n=_________Consider all the bodies have equal masses.

To compare the moments of inertia of a disc, ring, and sphere (all of mass $$M$$ and radius $$R$$) rotating about a diameter, we first recall the standard formulas for each.

For a ring about its diameter we have $$I_{\text{ring}} = \frac{1}{2}MR^2$$, for a disc about its diameter $$I_{\text{disc}} = \frac{1}{4}MR^2$$, and for a sphere about its diameter $$I_{\text{sphere}} = \frac{2}{5}MR^2$$.

Next, we check the condition given in the problem by computing the ratio

$$ \frac{I_{\text{disc}}}{I_{\text{ring}}} = \frac{MR^2/4}{MR^2/2} = \frac{1}{2} $$.

Since the problem states that $$I_{\text{disc}} = 2.5 \times I_{\text{ring}}$$, this standard result does not hold unless the disc and ring have different radii. We are told, however, that the sphere has the same radius as the disc.

Let us denote the ring’s radius by $$r$$ and the disc’s (and sphere’s) radius by $$R$$. Imposing $$I_{\text{disc}} = 2.5 \times I_{\text{ring}}$$ gives

$$ \frac{1}{4}MR^2 = 2.5 \times \frac{1}{2}Mr^2 = \frac{5}{4}Mr^2 $$.

From this we find

$$ R^2 = 5r^2 $$.

Since the sphere has the same radius as the disc, its moment of inertia is

$$ I_{\text{sphere}} = \frac{2}{5}MR^2 = \frac{2}{5}M(5r^2) = 2Mr^2 $$.

Consequently, the ratio of the sphere’s moment of inertia to the ring’s is

$$ n = \frac{I_{\text{sphere}}}{I_{\text{ring}}} = \frac{2Mr^2}{\frac{1}{2}Mr^2} = 4 $$.

Hence the answer is $$n = 4$$.